Applications of Integration and Differential Equations

Area Between Two Curves is an important application of integration. It allows us to find the exact amount of space enclosed by graphs on a coordinate plane.

When two curves form a closed region, the area of that region can often be found by subtracting one function from the other and integrating over the correct interval.

This lesson will show you how to identify the top and bottom curves, how to find intersection points, and how to set up and evaluate the integral for the enclosed area.

1. The Main Idea

Suppose two functions are given by \(y=f(x)\) and \(y=g(x)\). If \(f(x)\) is above \(g(x)\) on the interval from \(x=a\) to \(x=b\), then the area between the curves is

This works because for each value of \(x\), the vertical distance between the curves is

Adding all these thin vertical strips from \(x=a\) to \(x=b\) gives the total area.

2. Steps for Finding Area Between Two Curves

1. Sketch or imagine the graphs. This helps you see which curve is above the other.
2. Find the intersection points. Set the equations equal to each other. These points often give the limits of integration.
3. Identify the upper and lower functions. On the interval, determine which graph has the greater \(y\)-value.
4. Set up the integral. Use \(\int (\text{top} - \text{bottom})\,dx\).
5. Evaluate the integral. This gives the exact enclosed area.

3. Why We Use Top Minus Bottom

Area must be positive. If you subtract in the wrong order, the integral may become negative, even though the geometric area is positive.

So when integrating with respect to \(x\), always use

If the curves switch positions during the interval, then you must split the integral at the point where they cross.

4. Finding the Intersection Points

To find where two curves meet, set their equations equal:

Then solve for \(x\). These values are often the left and right boundaries of the enclosed region.

After finding the \(x\)-values, you can substitute them into either equation to get the corresponding points if needed.

Worked Example 1: A Basic Parabola and Line

Find the area enclosed by \(y=x+2\) and \(y=x^2\).

Step 1: Find the intersection points.

Set the equations equal:

So the curves intersect at \(x=-1\) and \(x=2\).

Step 2: Determine which curve is on top.

Choose a value between \(-1\) and \(2\), such as \(x=0\).

For \(y=x+2\), we get \(2\).

For \(y=x^2\), we get \(0\).

So \(y=x+2\) is above \(y=x^2\).

Step 3: Set up the integral.

Step 4: Evaluate.

Substitute \(x=2\):

Substitute \(x=-1\):

Now subtract:

Answer: The enclosed area is

Worked Example 2: Curves Symmetric About the y-Axis

Find the area enclosed by \(y=4-x^2\) and \(y=x^2\).

Step 1: Find the intersection points.

Step 2: Determine which curve is on top.

At \(x=0\),

\(y=4-x^2=4\) and \(y=x^2=0\), so \(y=4-x^2\) is above \(y=x^2\).

Step 3: Set up the integral.

Step 4: Evaluate.

Substitute \(x=\sqrt{2}\):

Since \((\sqrt{2})^3=2\sqrt{2}\), this becomes

Substitute \(x=-\sqrt{2}\):

Now subtract:

Answer:

Worked Example 3: When You Must Split the Area

Find the area enclosed by \(y=x^3\) and \(y=x\) from \(x=-1\) to \(x=1\).

Step 1: Find where the graphs intersect.

So the curves meet at \(x=-1\), \(x=0\), and \(x=1\).

Step 2: Check which graph is on top in each interval.

On \((-1,0)\), choose \(x=-\frac{1}{2}\):

\(x=-\frac{1}{2}\), but \(x^3=-\frac{1}{8}\). Since \(-\frac{1}{8} > -\frac{1}{2}\), we have \(x^3\) above \(x\).

On \((0,1)\), choose \(x=\frac{1}{2}\):

\(x=\frac{1}{2}\), but \(x^3=\frac{1}{8}\). Since \(\frac{1}{2} > \frac{1}{8}\), we have \(x\) above \(x^3\).

Step 3: Split the integral.

Step 4: Evaluate each part.

First part:

Second part:

Add the two areas:

Answer:

This example is important because it shows that one single subtraction does not always work over the whole interval. If the upper curve changes, the integral must be split.

5. When Boundaries Are Given Directly

Sometimes the region is enclosed not only by curves, but also by vertical lines such as \(x=1\) and \(x=3\).

In that case, use those values as the limits of integration, as long as you know which curve is on top throughout the interval.

Worked Example 4: Area with Given Vertical Boundaries

Find the area between \(y=2x+1\) and \(y=x^2\) from \(x=1\) to \(x=3\).

Step 1: Determine which graph is above.

At \(x=1\),

\(2x+1=3\) and \(x^2=1\), so the line is above there.

At \(x=3\),

\(2x+1=7\) and \(x^2=9\), so the parabola is above there.

This tells us the graphs switch places somewhere between \(1\) and \(3\), so we must find the intersection.

Step 2: Find the intersection point inside the interval.

In the interval \([1,3]\), the relevant point is \(x=1+\sqrt{2}\).

Step 3: Split the area.

From \(x=1\) to \(x=1+\sqrt{2}\), the line is above the parabola.

From \(x=1+\sqrt{2}\) to \(x=3\), the parabola is above the line.

This setup is the most important part. Once the integral is correct, the area can be evaluated carefully.

6. Common Mistakes to Avoid

• Forgetting to find intersection points. The correct limits are essential.
• Using bottom minus top. This may give a negative answer.
• Not splitting the integral when curves cross. If the upper curve changes, the area must be broken into parts.
• Confusing area with signed integral. A definite integral can be negative, but area must always be non-negative.

7. Quick Strategy Checklist

• Write both equations clearly.
• Find where they intersect.
• Check which curve is on top on each interval.
• Use \(A=\int (\text{top} - \text{bottom})\,dx\).
• Split into more than one integral if needed.
• Make sure your final area is positive.

8. Final Summary

To find the area between two curves, you integrate the vertical distance between them across the interval where they enclose a region.

When integrating with respect to \(x\), the formula is

The most important skills are finding the intersection points, deciding which graph is above the other, and splitting the integral if the curves switch order.

Once you master those steps, you can solve many geometric area problems using integration.

Volumes of Solids of Revolution is an important application of integration. It allows us to find the volume of a 3-dimensional solid formed when a 2-dimensional region is rotated around a line, called the axis of rotation.

For example, if a curve and the axes enclose a region, and that region is spun around the x-axis, the result is a solid. Integration helps us add up many very thin circular pieces to find the total volume.

In this lesson, you will learn how to use the Disk Method and the Washer Method to calculate these volumes.

1. The main idea behind solids of revolution

When a flat region is rotated about a horizontal or vertical line, each thin slice of the region creates a circular cross-section.

• If the slice reaches all the way to the axis of rotation, the cross-section is a disk.
• If there is a gap between the region and the axis of rotation, the cross-section is a washer, which looks like a ring.

The volume of the solid is found by adding the volumes of many thin disks or washers.

Since volume is cross-sectional area times thickness, the general idea is:

2. The Disk Method

The Disk Method is used when the region touches the axis of rotation, so each cross-section is a full circle.

The area of a circle is:

If the radius of each disk is given by a function, then the volume is:

if the slices are perpendicular to the x-axis.

Similarly, if the radius is written in terms of \(y\), then:

Disk Method about the x-axis

If a region under \(y=f(x)\) from \(x=a\) to \(x=b\) is rotated about the x-axis, the radius is \(f(x)\), so:

Disk Method about the y-axis

If a region between the y-axis and \(x=g(y)\) from \(y=c\) to \(y=d\) is rotated about the y-axis, the radius is \(g(y)\), so:

3. The Washer Method

The Washer Method is used when the solid has a hole in the middle. In that case, each cross-section is an outer circle with an inner circle removed.

The area of a washer is:

where:

• \(R\) is the outer radius
• \(r\) is the inner radius

So the volume formula becomes:

or, in terms of \(y\):

4. How to choose the correct method

Ask yourself these questions:

1. What line is the region rotating around?
2. Will each cross-section be a full circle or a ring?
3. What is the radius of the cross-section?
4. Should I integrate with respect to \(x\) or with respect to \(y\)?

If there is no hole, use the Disk Method.

If there is a hole, use the Washer Method.

5. Finding the radius correctly

The most important step is identifying the radius. The radius is always the distance from the axis of rotation to the curve.

For example:

• Rotating around the x-axis: radius is a vertical distance, often \(y\)
• Rotating around the y-axis: radius is a horizontal distance, often \(x\)
• Rotating around \(y=k\): radius is the distance between the curve and the line \(y=k\)
• Rotating around \(x=h\): radius is the distance between the curve and the line \(x=h\)

Distance should always be written as a positive expression.

6. Worked Example 1: Disk Method about the x-axis

Find the volume generated when the region under \(y=x^2\) from \(x=0\) to \(x=2\) is rotated about the x-axis.

Step 1: Identify the method

The region touches the x-axis, so there is no hole. Use the Disk Method.

Step 2: Write the radius

The radius is the distance from the x-axis to the curve:

Step 3: Set up the integral

Step 4: Integrate

Answer: The volume is $$\frac{32\pi}{5}$$ cubic units.

7. Worked Example 2: Washer Method about the x-axis

Find the volume generated when the region between \(y=x+1\) and \(y=1\) from \(x=0\) to \(x=3\) is rotated about the x-axis.

Step 1: Identify the method

The region does not go all the way to the x-axis because the lower curve is \(y=1\). This creates a hole, so use the Washer Method.

Step 2: Find outer and inner radii

• Outer radius: distance from x-axis to \(y=x+1\), so \(R(x)=x+1\)
• Inner radius: distance from x-axis to \(y=1\), so \(r(x)=1\)

Step 3: Set up the integral

Step 4: Simplify

Step 5: Integrate

Answer: The volume is $$18\pi$$ cubic units.

8. Worked Example 3: Disk Method about the y-axis

Find the volume generated when the region between the y-axis and \(x=\sqrt{y}\) from \(y=0\) to \(y=4\) is rotated about the y-axis.

Step 1: Identify the method

The region touches the y-axis, so use the Disk Method.

Step 2: Write the radius

The radius is the distance from the y-axis to the curve:

Step 3: Set up the integral

Step 4: Integrate

Answer: The volume is $$8\pi$$ cubic units.

9. Worked Example 4: Washer Method about a horizontal line

Find the volume generated when the region between \(y=x\) and the x-axis from \(x=0\) to \(x=2\) is rotated about the line \(y=3\).

Step 1: Identify the method

Since the axis of rotation is above the region, the cross-sections will have a hole. Use the Washer Method.

Step 2: Find the radii

The outer radius is the distance from \(y=3\) down to the x-axis:

The inner radius is the distance from \(y=3\) down to the line \(y=x\):

Step 3: Set up the integral

Step 4: Simplify

Step 5: Integrate

Answer: The volume is $$\frac{28\pi}{3}$$ cubic units.

10. Common mistakes to avoid

• Forgetting to square the radius: In both disk and washer formulas, the radius must be squared.
• Mixing up outer and inner radius: The outer radius is always the bigger distance.
• Using the wrong variable: If integrating with respect to \(x\), radii should usually be written in terms of \(x\).
• Ignoring the axis of rotation: If the axis is not the x-axis or y-axis, the radius must be measured carefully as a distance.
• Not checking for a hole: If the region does not touch the axis, you need washers, not disks.

11. Step-by-step strategy for solving problems

1. Sketch or imagine the region and the axis of rotation.
2. Decide whether the cross-sections are disks or washers.
3. Write the radius, or outer and inner radii, as distances.
4. Choose the correct integral formula.
5. Simplify the expression before integrating if possible.
6. Evaluate the definite integral and include units cubed.

12. Summary

Volumes of solids of revolution are found by rotating a region around a line and adding the areas of circular cross-sections.

Use the Disk Method when there is no hole:

Use the Washer Method when there is a hole:

The key to success is identifying the correct radius as the distance from the axis of rotation to the curve. Once the radii are correct, the integration is usually straightforward.

Lesson: Volumes by Known Cross-Sections

In many solids, the shape is too irregular to use a simple volume formula like the formula for a prism, cylinder, or cone. Instead, we find the volume by slicing the solid into many thin pieces and adding their volumes together. This is one of the most important applications of integration.

In this lesson, you will learn how to find the volume of a solid when you know the shape of its cross-sections. These cross-sections are slices of the solid taken perpendicular to an axis, and each slice forms a familiar 2D shape such as a square, rectangle, triangle, or semicircle.

The key idea is simple: if a slice at position \(x\) has area \(A(x)\), then the volume of the solid is found by integrating those areas across the interval:

This formula says that volume is the sum of the areas of infinitely many thin cross-sections.

1. The Basic Idea of Cross-Sectional Volume

Suppose a solid has a base in the \(xy\)-plane. At each point along the \(x\)-axis, a cross-section perpendicular to the \(x\)-axis is cut. If the cross-section has a known shape, then we can write a formula for its area.

To solve these problems, we usually follow these steps:

1. Describe the base region.
2. Find the length of the segment inside the base that determines the size of the cross-section.
3. Use the geometry of the cross-section to write its area \(A(x)\).
4. Integrate \(A(x)\) over the interval of the base.

2. Finding the Length of the Slice

Very often, the base is bounded by two curves: a top curve \(y=f(x)\) and a bottom curve \(y=g(x)\). Then the vertical length of the slice is:

This length becomes the side, base, diameter, or some other measurement of the cross-sectional shape.

If cross-sections are perpendicular to the \(y\)-axis instead, then we work with horizontal slices and write the width as a difference of \(x\)-values.

3. Common Cross-Section Shapes

Here are the most common area formulas you will use.

• Square with side \(s\): $$A = s^2$$
• Rectangle with base \(b\) and height \(h\): $$A = bh$$
• Equilateral triangle with side \(s\): $$A = \frac{\sqrt{3}}{4}s^2$$
• Isosceles right triangle with leg \(s\): $$A = \frac{1}{2}s^2$$
• Semicircle with diameter \(d\): $$A = \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{\pi}{8}d^2$$

The most important skill is identifying what measurement from the base corresponds to the shape. For example, if the problem says the cross-sections are squares and each square is built on the base region, then the side length of the square is usually the length of the slice in the base.

4. General Strategy

When solving a volume-by-cross-sections problem, ask yourself these questions:

• What variable am I integrating with respect to: \(x\) or \(y\)?
• What are the limits of integration?
• What is the length of the slice in the base?
• How does that length relate to the area of the cross-section?

Once you have the area formula, the rest is an integration problem.

Worked Example 1: Cross-sections are Squares

Problem: The base of a solid is the region bounded by \(y=x\) and \(y=x^2\). Cross-sections perpendicular to the \(x\)-axis are squares. Find the volume.

Step 1: Find the interval of the base.

The curves intersect where

so

Thus the interval is \(0\le x\le 1\).

Step 2: Find the side length of each square.

On \([0,1]\), the top curve is \(y=x\) and the bottom curve is \(y=x^2\). So the side length is

Step 3: Write the area function.

Since the cross-sections are squares,

Step 4: Integrate.

Expand first:

So,

Answer:

Worked Example 2: Cross-sections are Semicircles

Problem: The base of a solid is the region bounded by \(y=4-x^2\) and the \(x\)-axis. Cross-sections perpendicular to the \(x\)-axis are semicircles. Find the volume.

Step 1: Find the interval.

The parabola meets the \(x\)-axis where

So the limits are \(-2\) to \(2\).

Step 2: Find the diameter.

Each vertical slice runs from \(y=0\) to \(y=4-x^2\), so the diameter is

Step 3: Write the area function.

The area of a semicircle with diameter \(d\) is

Step 4: Integrate.

Expand:

Then

Now integrate:

Evaluating gives

Answer:

Worked Example 3: Cross-sections are Equilateral Triangles

Problem: The base of a solid is the region bounded by \(y=\sqrt{x}\), \(y=0\), and \(x=4\). Cross-sections perpendicular to the \(x\)-axis are equilateral triangles. Find the volume.

Step 1: Understand the base.

The region goes from \(x=0\) to \(x=4\), with top curve \(y=\sqrt{x}\) and bottom curve \(y=0\).

Step 2: Find the side length.

The vertical length of the slice is

Step 3: Write the area formula.

The area of an equilateral triangle with side \(s\) is

Step 4: Integrate.

Answer:

Worked Example 4: Cross-sections Perpendicular to the \(y\)-axis

Problem: The base of a solid is enclosed by \(x=y^2\) and \(x=4\). Cross-sections perpendicular to the \(y\)-axis are rectangles whose height is twice their base. Find the volume.

This example is important because the slices are taken perpendicular to the \(y\)-axis, so we integrate with respect to \(y\).

Step 1: Find the limits in \(y\).

The curves intersect where

So the interval is \(-2\le y\le 2\).

Step 2: Find the base of each rectangle.

At a given \(y\), the horizontal length goes from \(x=y^2\) to \(x=4\), so

Step 3: Use the rectangle condition.

The height is twice the base, so

Thus the cross-sectional area is

Step 4: Integrate.

Expand:

So

Evaluating gives

Answer:

5. Common Mistakes to Avoid

• Using the wrong top and bottom curves. Always check which function is larger on the interval.
• Forgetting that the slice length is not always the area. First find the slice length, then use the correct geometry formula.
• Mixing up radius and diameter. For circles and semicircles, problems often give a diameter, but the area formula usually uses radius.
• Integrating with the wrong variable. If slices are perpendicular to the \(y\)-axis, the integral is usually with respect to \(y\).
• Using incorrect limits of integration. Find where the bounding curves intersect.

6. Quick Problem-Solving Checklist

• Draw or imagine the base region.
• Determine whether slices are vertical or horizontal.
• Find the length of each slice from one boundary curve to the other.
• Turn that length into an area formula for the cross-section.
• Integrate over the correct interval.

7. Why This Works

A solid can be approximated by many very thin slabs. Each slab has volume about

Adding all the slabs gives an approximation:

When the slices become infinitely thin, the sum becomes the definite integral:

This is why integration is such a powerful tool for finding the volume of unusual solids.

Brief Summary

To find volume by known cross-sections, first determine the base region and the direction of slicing. Then find the slice length, use the cross-section shape to write an area function, and integrate that area across the interval. The whole process is built on the formula \(V=\int A(x)\,dx\) or \(V=\int A(y)\,dy\), depending on the direction of the slices.

Average Value of a Function is a way to find the “continuous average” of a function over an interval. You can think of it as the average height of the graph from one x-value to another.

In earlier grades, you found the average of a list of numbers by adding them and dividing by how many numbers there were. For a function, there are infinitely many values on an interval, so instead of adding separate numbers, we use an integral.

This idea is closely connected to the Mean Value Theorem for Integrals. That theorem says that if a function is continuous on an interval, then there is at least one point where the function’s value is exactly equal to its average value on that interval.

1. The Formula for Average Value

If a function \(f(x)\) is continuous on the interval \([a,b]\), then its average value on that interval is

This formula has two important parts:

• \(\int_a^b f(x)\,dx\) gives the net area under the curve from \(x=a\) to \(x=b\).
• Dividing by \(b-a\) spreads that total amount evenly across the interval length.

So the average value is like asking: If the function had one constant height across the whole interval, what height would give the same total area?

2. Why This Makes Sense

Suppose you want the average of the numbers 3, 5, and 7. You compute

For a continuous function, there are infinitely many outputs between \(a\) and \(b\). The integral acts like a continuous version of adding all those values, and dividing by the interval length \(b-a\) acts like dividing by how many values are being considered in a continuous sense.

3. Connection to the Mean Value Theorem for Integrals

If \(f\) is continuous on \([a,b]\), then there exists some number \(c\) in \([a,b]\) such that

This means there is at least one point on the graph where the function’s actual height equals its average height over the interval.

Geometrically, if you drew a horizontal line at the average value, the rectangle formed with width \(b-a\) would have the same area as the area under the curve over that interval.

4. How to Find the Average Value

1. Identify the function \(f(x)\) and the interval \([a,b]\).
2. Compute the definite integral \(\int_a^b f(x)\,dx\).
3. Divide the result by \(b-a\).
4. If asked, solve \(f(c)=f_{\text{avg}}\) to find where the function equals its average value.

5. Worked Example 1: A Basic Polynomial

Find the average value of \(f(x)=x^2\) on the interval \([0,2]\).

Step 1: Use the formula.

Step 2: Evaluate the integral.

Step 3: Divide by the interval length.

Answer: The average value of \(f(x)=x^2\) on \([0,2]\) is

This means the average height of the parabola from \(x=0\) to \(x=2\) is \(\frac{4}{3}\).

6. Worked Example 2: Finding the Value of \(c\)

For \(f(x)=x^2\) on \([0,2]\), find the value(s) of \(c\) such that \(f(c)\) equals the average value.

From Example 1, the average value is \(\frac{4}{3}\).

Set the function equal to this value:

Solve:

But the interval is \([0,2]\), so only the positive value belongs to the interval.

Answer:

This confirms the Mean Value Theorem for Integrals: there is a point where the graph reaches exactly its average height.

7. Worked Example 3: A Trigonometric Function

Find the average value of \(f(x)=\sin x\) on the interval \([0,\pi]\).

Use the formula:

Evaluate the integral:

Now divide by \(\pi\):

Answer:

So the average height of \(\sin x\) from \(0\) to \(\pi\) is \(\frac{2}{\pi}\).

8. Worked Example 4: A Function Below the x-axis Part of the Time

Find the average value of \(f(x)=x-3\) on the interval \([1,5]\).

Use the formula:

Evaluate the integral:

Substitute \(x=5\):

Substitute \(x=1\):

Now subtract:

So,

Answer:

This result makes sense because the line \(y=x-3\) is below the x-axis on part of the interval and above it on another part, and the positive and negative contributions balance out exactly.

9. Important Notes and Common Mistakes

• Do not forget to divide by \(b-a\). The integral alone is not the average value.
• Use the interval length, not the number of endpoints. The denominator is always \(b-a\).
• The average value can be negative, zero, or positive. It depends on the function’s net signed area.
• Average value is not always halfway between the maximum and minimum values. It depends on the shape of the graph.
• When finding \(c\), make sure your answer lies in the given interval.

10. Visual Meaning

If the average value of \(f(x)\) on \([a,b]\) is \(k\), then the rectangle with height \(k\) and width \(b-a\) has area

Since

multiplying both sides by \(b-a\) gives

So the rectangle has the same area as the signed area under the function. This is the main geometric idea behind average value.

11. Quick Practice Ideas

Try these on your own:

• Find the average value of \(f(x)=2x+1\) on \([0,3]\).
• Find the average value of \(f(x)=x^3\) on \([-1,1]\).
• For \(f(x)=\cos x\) on \([0,\tfrac{\pi}{2}]\), find the average value and then solve for \(c\).

12. Summary

The average value of a function on \([a,b]\) is found using

This gives the continuous arithmetic mean height of the function over the interval. The Mean Value Theorem for Integrals tells us that if the function is continuous, then there is some \(c\) in the interval such that \(f(c)=f_{\text{avg}}\).

When solving these problems, always remember: integrate first, then divide by the interval length, and if needed, set the function equal to the average value to find \(c\).

Introduction to Differential Equations

In many real-life situations, we do not just want to know a quantity itself—we want to know how fast it is changing. For example, a population may grow over time, a tank may fill with water, or a hot object may cool down. When the rate of change depends on the quantity that is changing, we use a differential equation.

A differential equation is an equation that contains an unknown function and one or more of its derivatives. In 12th Grade Maths, the most common starting point is a differential equation involving a function like \(y\) and its derivative \(\frac{dy}{dx}\).

This lesson introduces what differential equations are, how to write them using derivative notation, and how to recognize situations where a changing quantity depends on variable quantities.

1. What is a differential equation?

If \(y\) depends on \(x\), then \(\frac{dy}{dx}\) tells us the rate at which \(y\) changes with respect to \(x\). A differential equation connects \(y\), \(x\), and \(\frac{dy}{dx}\).

For example,

$$\frac{dy}{dx} = 3x$$

is a differential equation because it includes the derivative of \(y\).

Another example is

$$\frac{dy}{dx} = 2y$$

Here, the rate of change of \(y\) depends on the current value of \(y\) itself.

This idea is very important: in a differential equation, the change in a quantity is described in terms of the quantity, the independent variable, or both.

2. The meaning of derivative notation in differential equations

Suppose \(y\) is a function of \(x\). Then:

• \(\frac{dy}{dx}\) means the rate of change of \(y\) with respect to \(x\)
• \(\frac{dP}{dt}\) means the rate of change of population \(P\) with respect to time \(t\)
• \(\frac{dV}{dt}\) means the rate of change of volume \(V\) with respect to time \(t\)

The letter in the denominator tells us what variable is changing independently. The letter in the numerator tells us which quantity is responding to that change.

For example, if the amount of water in a tank is changing over time, then we would naturally use \(\frac{dV}{dt}\), not \(\frac{dy}{dx}\).

3. Translating words into a differential equation

To translate a situation into differential equation form, look for phrases that describe a rate of change. Common phrases include:

• changes at a rate of
• increases proportionally to
• decreases depending on
• the rate of growth is
• the rate of cooling depends on

These phrases usually tell you how to write the derivative.

For example:

• “The rate of change of \(y\) is proportional to \(y\)” means $$\frac{dy}{dx} = ky$$ for some constant \(k\).
• “The rate of change of population \(P\) is proportional to the population” means $$\frac{dP}{dt} = kP$$
• “The rate of change of temperature \(T\) depends on the difference between \(T\) and room temperature \(20\)” means $$\frac{dT}{dt} = k(T-20)$$

In each case, the derivative expresses how fast something changes, and the right side shows what that rate depends on.

4. Important vocabulary

• Dependent variable: the quantity that changes, such as \(y\), \(P\), or \(V\)
• Independent variable: the variable that drives the change, such as \(x\) or time \(t\)
• Rate of change: the derivative, such as \(\frac{dy}{dx}\) or \(\frac{dP}{dt}\)
• Constant of proportionality: a constant like \(k\) that tells how strongly quantities are related

5. General solution and particular solution

When we solve a differential equation, we usually get a family of functions. This is called the general solution.

For example, if

$$\frac{dy}{dx} = 3x$$

then integrating gives

$$y = \frac{3}{2}x^2 + C$$

This is the general solution, because many different curves have derivative \(3x\).

If we are also given a starting value, such as \(y=4\) when \(x=0\), we can find the exact value of \(C\). That gives a particular solution.

6. First-order differential equations

A first-order differential equation involves only the first derivative, such as \(\frac{dy}{dx}\). Most introductory examples are first-order equations.

Examples:

• $$\frac{dy}{dx} = x+2$$
• $$\frac{dy}{dx} = y$$
• $$\frac{dP}{dt} = 0.04P$$

These all describe how a quantity changes based on one simple rate rule.

7. Separable differential equations

One important type of differential equation is a separable differential equation. This means we can rearrange it so that all the \(y\)-terms are on one side and all the \(x\)-terms are on the other side.

A typical form is

$$\frac{dy}{dx} = g(x)h(y)$$

Then we separate the variables:

$$\frac{1}{h(y)}\,dy = g(x)\,dx$$

After that, we integrate both sides.

At this stage, your main goal is to understand that the equation describes a changing relationship, and that the derivative notation helps us write that relationship clearly.

8. Slope fields: a visual idea

A slope field is a picture that shows the slopes of solution curves at many points. If we know the differential equation

$$\frac{dy}{dx} = f(x,y)$$

then at each point \((x,y)\), we can compute the slope.

For example, if

$$\frac{dy}{dx} = y$$

then:

• when \(y=0\), slope \(=0\)
• when \(y=1\), slope \(=1\)
• when \(y=2\), slope \(=2\)
• when \(y=-1\), slope \(=-1\)

This tells us that the slope depends only on the height \(y\), not on \(x\). Slope fields help us see the behavior of solutions even before solving the equation exactly.

Worked Example 1: Writing a differential equation from a statement

Problem: The rate of change of a quantity \(y\) with respect to \(x\) is proportional to \(x\). Write a differential equation.

Step 1: “Rate of change of \(y\) with respect to \(x\)” means \(\frac{dy}{dx}\).

Step 2: “Is proportional to \(x\)” means it equals a constant times \(x\).

So the differential equation is

$$\frac{dy}{dx} = kx$$

where \(k\) is a constant.

Answer: $$\frac{dy}{dx} = kx$$

Worked Example 2: Translating a population model

Problem: A population \(P\) grows at a rate proportional to the current population. Write the differential equation.

Step 1: Since population changes with time, the derivative is \(\frac{dP}{dt}\).

Step 2: “Rate proportional to the current population” means the rate equals a constant times \(P\).

So,

$$\frac{dP}{dt} = kP$$

Answer: $$\frac{dP}{dt} = kP$$

This is one of the most common differential equations in science and economics.

Worked Example 3: Solving a simple differential equation

Problem: Solve

$$\frac{dy}{dx} = 4x$$

Step 1: Recognize that \(\frac{dy}{dx}\) is the derivative of \(y\).

Step 2: Integrate both sides with respect to \(x\):

$$y = \int 4x\,dx$$

$$y = 2x^2 + C$$

Answer: $$y = 2x^2 + C$$

This is the general solution.

Worked Example 4: Solving a separable differential equation

Problem: Solve

$$\frac{dy}{dx} = 3y$$

Step 1: Separate the variables by dividing both sides by \(y\):

$$\frac{1}{y}\frac{dy}{dx} = 3$$

This can be written as

$$\frac{1}{y}\,dy = 3\,dx$$

Step 2: Integrate both sides:

$$\int \frac{1}{y}\,dy = \int 3\,dx$$

$$\ln|y| = 3x + C$$

Step 3: Solve for \(y\) by rewriting in exponential form:

$$|y| = e^{3x+C}$$

Since \(e^C\) is just a positive constant, we write

$$y = Ce^{3x}$$

Answer: $$y = Ce^{3x}$$

This shows how a differential equation can describe exponential growth or decay.

9. How to identify the correct derivative notation

When translating a word problem, ask yourself:

1. What quantity is changing?
2. What variable is it changing with respect to?
3. What does the rate depend on?

For example:

• If distance changes with time, use \(\frac{ds}{dt}\)
• If temperature changes with time, use \(\frac{dT}{dt}\)
• If output changes with input \(x\), use \(\frac{dy}{dx}\)

This helps you turn verbal descriptions into mathematical equations accurately.

10. Common mistakes to avoid

• Mixing up variables: Use the correct independent variable. If change happens over time, use \(t\), not \(x\).
• Forgetting the constant of proportionality: “Proportional to” usually means multiply by a constant like \(k\).
• Leaving out the constant of integration: When solving, remember the \(+C\).
• Not identifying what the rate depends on: Read the statement carefully. Does the rate depend on \(x\), on the function itself, or on both?

11. Why differential equations matter

Differential equations are used to model changing systems. They appear in population growth, motion, electricity, chemistry, economics, and many other areas. Even at an introductory level, they help us describe how one quantity responds to another.

When you can translate a sentence like “the rate of growth depends on the current amount” into a differential equation such as \(\frac{dP}{dt}=kP\), you are turning a real-world pattern into mathematics.

Brief Summary

A differential equation is an equation that contains a function and its derivative. It is used when we want to describe how a quantity changes, especially when that rate of change depends on variables such as time, position, or the quantity itself.

To write a differential equation, identify the changing quantity, the variable it depends on, and what controls the rate of change. Introductory examples often use first-order equations such as \(\frac{dy}{dx}=kx\) or \(\frac{dP}{dt}=kP\), and some can be solved by integration or by separating variables.

Slope Fields (Direction Fields) are a visual way to understand a differential equation without solving it first.

When a differential equation tells us the derivative \(\frac{dy}{dx}\), it tells us the slope of a solution curve at each point \((x,y)\). A slope field shows this information by drawing many tiny line segments across the coordinate plane.

Each small segment has the slope given by the differential equation at that point. When we look at all the segments together, we can see the general shape of the whole family of solutions.

This is useful because it helps us:

• predict the behavior of solutions,
• estimate how solutions move through the plane,
• compare different starting values, and
• understand differential equations even when solving them exactly is difficult.

1. What a slope field represents

A differential equation of the form

assigns a slope to every point \((x,y)\).

For example, if

then at the point \((1,2)\), the slope is

So at \((1,2)\), we would draw a tiny line segment with slope \(3\).

If we repeat this at many points, we create the slope field.

Important idea: the slope field does not give one single graph. It shows the direction of all possible solution curves.

2. How to sketch a slope field

To sketch a slope field by hand:

1. Choose several points on a grid.
2. Compute the slope \(\frac{dy}{dx}\) at each point.
3. Draw a short line segment with that slope at each point.
4. Do not connect the small segments into one graph unless you are sketching a particular solution.

When drawing by hand, the segments should be short and centered at each point.

3. Reading a slope field

You can learn a lot from the pattern of the little segments:

• If the segments are horizontal, then the slope is \(0\).
• If the segments tilt upward from left to right, the slope is positive.
• If the segments tilt downward from left to right, the slope is negative.
• If the segments are very steep, the slope has large magnitude.

A solution curve should follow the direction of the segments. It should be tangent to them at every point.

4. Isoclines: places with the same slope

Sometimes many points share the same slope. A curve where the slope is constant is called an isocline.

For example, if

then slope \(0\) happens when

So along the line \(y=-x\), all the little segments are horizontal.

Similarly, slope \(1\) happens when

These isoclines help us sketch slope fields more quickly.

5. Worked Example 1: A slope field that depends only on \(x\)

Sketch the slope field for

Step 1: Notice that the slope depends only on \(x\), not on \(y\).

This means that all points with the same \(x\)-value have the same slope. So each vertical column of points will have matching line segments.

Step 2: Calculate some slopes:

• At \(x=-2\), slope \(=-2\)
• At \(x=-1\), slope \(=-1\)
• At \(x=0\), slope \(=0\)
• At \(x=1\), slope \(=1\)
• At \(x=2\), slope \(=2\)

Step 3: Draw short segments in each vertical column using these slopes.

So:

• left of the \(y\)-axis, segments slope downward,
• on the \(y\)-axis, segments are horizontal,
• right of the \(y\)-axis, segments slope upward.

What do the solutions look like? They should bend upward more and more as \(x\) increases. Since

the actual family of solutions is

which matches the shape suggested by the slope field.

6. Worked Example 2: A slope field that depends only on \(y\)

Sketch the slope field for

Step 1: Here the slope depends only on \(y\).

This means all points with the same \(y\)-value have the same slope. So each horizontal row will have matching line segments.

Step 2: Compute a few slopes:

• At \(y=-2\), slope \(=-2\)
• At \(y=-1\), slope \(=-1\)
• At \(y=0\), slope \(=0\)
• At \(y=1\), slope \(=1\)
• At \(y=2\), slope \(=2\)

Step 3: Draw the segments across each row.

So:

• below the \(x\)-axis, segments slope downward,
• on the \(x\)-axis, segments are horizontal,
• above the \(x\)-axis, segments slope upward.

Interpreting the field:

• If a solution starts above the \(x\)-axis, it rises and becomes steeper.
• If a solution starts below the \(x\)-axis, it falls and becomes steeper downward.
• The line \(y=0\) is a constant solution because its slope is always \(0\).

The actual solution family is

and the slope field gives the same behavior.

7. Worked Example 3: A slope field depending on both \(x\) and \(y\)

Sketch and interpret the slope field for

Now the slope changes from point to point in both directions.

Step 1: Calculate slopes at a few points:

• At \((0,0)\): \(\frac{dy}{dx}=0-0=0\)
• At \((1,0)\): \(\frac{dy}{dx}=1-0=1\)
• At \((0,1)\): \(\frac{dy}{dx}=0-1=-1\)
• At \((1,1)\): \(\frac{dy}{dx}=1-1=0\)
• At \((2,1)\): \(\frac{dy}{dx}=2-1=1\)
• At \((1,2)\): \(\frac{dy}{dx}=1-2=-1\)

Step 2: Look for where the slope is zero:

Along the line \(y=x\), all segments are horizontal.

Step 3: Decide where slopes are positive or negative:

• If \(x-y>0\), then slope is positive.
• If \(x-y<0\), then slope is negative.

That means:

• below the line \(y=x\), segments tilt upward,
• above the line \(y=x\), segments tilt downward.

Interpretation: Solution curves tend to level out when they approach the line \(y=x\), because that is where slope becomes zero.

8. Worked Example 4: Sketching a particular solution from an initial condition

Suppose the differential equation is

and we want the solution that passes through \((0,2)\).

Step 1: Understand the slope field.

Because the slope depends only on \(y\), every horizontal row has the same slope.

Find where the slope is zero:

So along the line \(y=1\), the segments are horizontal.

Above \(y=1\), slopes are positive. Below \(y=1\), slopes are negative.

Step 2: Start at the point \((0,2)\).

At \(y=2\), the slope is

So the curve begins by rising.

As it rises higher above \(y=1\), the value of \(y-1\) gets larger, so the slope gets steeper. This means the solution curve increases more and more rapidly.

Step 3: Sketch the curve so that it follows the tiny segments and passes through \((0,2)\).

If we solved it exactly, we would get an exponential-type curve. But even without solving, the slope field lets us predict the shape.

9. Equilibrium solutions

An equilibrium solution is a constant solution where the slope is always zero.

If

then equilibria happen when

These are important because nearby solution curves may move toward them or away from them.

For example, for

the equilibrium solution is

In the slope field, this appears as a horizontal line of flat segments.

10. Common patterns to notice

• If \(\frac{dy}{dx}=f(x)\): slopes repeat in vertical columns.
• If \(\frac{dy}{dx}=f(y)\): slopes repeat in horizontal rows.
• If \(\frac{dy}{dx}=0\): solution curves are horizontal lines.
• If slopes change sign across a line: solution behavior changes across that line.

11. How slope fields connect to solution families

A differential equation usually has many solutions. Each one corresponds to a different starting point, called an initial condition.

For example, if we know a point such as \((1,3)\), then we want the particular solution curve that passes through that point and follows the slope field everywhere.

So a slope field helps us see both:

• the general family of all solutions, and
• the specific solution for a given initial condition.

12. Common mistakes to avoid

• Using long line segments: slope field segments should be short, so they show direction at a point.
• Connecting all segments into one graph: the field represents many possible solutions, not just one.
• Forgetting that slope depends on both variables: in equations like \(\frac{dy}{dx}=x-y\), changing either \(x\) or \(y\) changes the slope.
• Drawing a solution curve that crosses the segments: a solution should follow the direction of the field and be tangent to the segments.
• Missing zero-slope lines: lines where \(\frac{dy}{dx}=0\) are often key to understanding the field.

13. Quick strategy for exam questions

When asked to sketch or interpret a slope field, try this:

1. Find where the slope is zero.
2. Check where the slope is positive and negative.
3. Compute slopes at a few easy points.
4. Look for repeating patterns by rows or columns.
5. Sketch possible solution curves that follow the field.

Brief Summary

A slope field is a picture of the slopes given by a differential equation at many points. Each small line segment shows the direction a solution curve would move at that point.

By studying where slopes are positive, negative, or zero, we can predict the shape of solution curves without solving the equation exactly. Slope fields are especially helpful for seeing families of solutions and sketching particular solutions from initial conditions.

Solving Separable Differential Equations is an important skill in differential equations. These equations describe how one quantity changes with respect to another, and in a separable equation, we can rearrange the variables so that all the terms involving one variable are on one side and all the terms involving the other variable are on the other side.

This method lets us turn a differential equation into two simpler integrals. After integrating, we solve for the unknown function and, if an initial condition is given, we can find the particular solution.

In this lesson, you will learn how to recognize a separable differential equation, how to separate the variables correctly, how to integrate both sides, and how to use an initial condition to determine the constant.

1. What is a separable differential equation?

A differential equation is separable if it can be written in the form

This is called separable because we can move all the terms involving \(y\) to one side and all the terms involving \(x\) to the other side:

Now the variables are separated, and each side can be integrated.

2. General steps for solving separable differential equations

1. Identify whether the equation is separable.
2. Rearrange so that all \(y\)-terms are with \(dy\), and all \(x\)-terms are with \(dx\).
3. Integrate both sides.
4. Solve for \(y\) if possible.
5. Use the initial condition, if one is given, to find the constant of integration.

In symbols, if

then we write

and integrate:

3. Why the separation step matters

The key idea is that integration works best when each side contains only one variable. If an equation mixes \(x\) and \(y\) on the same side, we usually cannot integrate directly. Separating the variables creates two ordinary integrals that you already know how to solve.

Be careful during rearrangement. You must move the factors correctly, just as you would when solving an algebraic equation. Think of \(dy/dx\) as a derivative that allows you to reorganize the expression into a form involving \(dy\) and \(dx\).

4. Worked Example 1: Basic separable equation

Solve

Step 1: Separate variables.

This equation already has all \(y\)-terms on one side, since there are actually no \(y\)-terms present. We can write

Step 2: Integrate both sides.

Solution:

This is the general solution because it contains an arbitrary constant \(C\).

5. Worked Example 2: Variable on both sides

Solve

Step 1: Separate variables.

We want all \(y\)-terms with \(dy\), so divide both sides by \(y\):

Now write this as

Step 2: Integrate both sides.

Step 3: Solve for \(y\).

To remove the natural logarithm, exponentiate both sides:

Since \(e^C\) is just another positive constant, we combine constants and write

where \(A\) can be any nonzero constant. Including the case \(A=0\), we usually write

Solution:

6. Worked Example 3: Finding a particular solution with an initial condition

Solve the differential equation

given that

Step 1: Separate variables.

Step 2: Integrate.

Step 3: Solve for \(y\).

Step 4: Use the initial condition.

Substitute \(x=0\) and \(y=5\):

So the particular solution is

This is called a particular solution because it satisfies both the differential equation and the given initial condition.

7. Worked Example 4: A slightly more challenging example

Solve

Step 1: Separate variables.

Multiply both sides by \(y\,dx\):

Step 2: Integrate both sides.

Step 3: Solve for \(y\).

Multiply by 2:

Taking square roots gives

This is the general solution in explicit form.

8. Using initial conditions carefully

An initial condition usually looks like \(y(x_0)=y_0\). After finding the general solution, substitute \(x_0\) for \(x\) and \(y_0\) for \(y\). Then solve for the constant.

For example, if the general solution is

and the condition is \(y(1)=2\), then

So the particular solution is

9. Common mistakes to avoid

• Not separating completely: Make sure one side has only \(y\) and \(dy\), and the other side has only \(x\) and \(dx\).
• Forgetting the constant of integration: After integrating, always include \(+C\).
• Dropping absolute value signs: Remember that \(\int \frac{1}{y}\,dy = \ln|y| + C\), not just \(\ln y + C\).
• Using the initial condition too early in a confusing way: It is usually easiest to first find the general solution, then plug in the initial condition.
• Algebra errors after integrating: Be careful when solving for \(y\), especially when exponentials or square roots appear.

10. Quick recognition tips

You may be able to solve an equation by separation if:

• \(\dfrac{dy}{dx}\) is equal to a product of a function of \(x\) and a function of \(y\)
• you can divide or multiply to place all \(y\)-terms on one side
• the equation can be rewritten in the form \(M(y)\,dy = N(x)\,dx\)

Examples of separable forms include:

In each case, the variables can be separated before integrating.

11. Practice thinking process

When you see a differential equation, ask yourself these questions:

1. Can I rewrite the equation so all \(y\)-terms are with \(dy\)?
2. Can I rewrite the equation so all \(x\)-terms are with \(dx\)?
3. Do I know how to integrate both sides?
4. Is there an initial condition to help find the constant?

If the answer to the first two questions is yes, then separation of variables is probably the correct method.

12. Brief summary

A separable differential equation can be rewritten so that the variables are on opposite sides of the equation. After separating, integrate both sides and solve for \(y\). If an initial condition is given, substitute it into the general solution to find the constant and get the particular solution.

The most important skills are separating correctly, integrating carefully, and using initial conditions accurately. With practice, this method becomes a reliable step-by-step process.

Exponential Growth and Decay Models are used when a quantity changes at a rate proportional to how much of the quantity is already present.

This idea appears in many real situations. A population may grow faster when it is already large, because more individuals can reproduce. A radioactive substance may decay faster when more atoms are present, because more atoms are available to break down.

In both situations, the same differential equation models the change:

Here:

• \(y\) is the amount at time \(t\),
• \(\frac{dy}{dt}\) is the rate of change of the amount,
• \(k\) is a constant of proportionality.

If \(k>0\), the model describes exponential growth. If \(k<0\), it describes exponential decay.

This lesson will show how to solve this differential equation, why the number \(e\) appears naturally, and how to apply the result to population and radioactivity problems.

1. Why the model makes sense

The equation $$\frac{dy}{dt}=ky$$ says that the rate of change is proportional to the current amount.

For example, if a bacteria culture doubles quickly when there are many bacteria, then the growth rate should depend on how many bacteria are already in the culture.

Similarly, if a radioactive sample has many unstable atoms, then more atoms can decay each second. As the sample shrinks, the number decaying each second also shrinks.

This is exactly what proportional change means.

2. Solving the differential equation \(\frac{dy}{dt}=ky\)

This is a separable differential equation. That means we can separate the variables \(y\) and \(t\) onto opposite sides.

Start with

Assume \(y\neq 0\) and divide both sides by \(y\):

Now write this in differential form:

Integrate both sides:

This gives

To solve for \(y\), rewrite in exponential form:

Use the rule \(e^{kt+C}=e^C e^{kt}\). Since \(e^C\) is just a positive constant, we combine signs and constants into one new constant \(A\):

So the general solution is

This is why the base \(e\) appears naturally. It comes from integrating \(\frac{1}{y}\).

3. Using an initial condition

Usually, a problem gives the amount at the starting time. This is called an initial condition.

If \(y(0)=y_0\), substitute \(t=0\) into the general solution:

So \(A=y_0\), and the solution becomes

This is the most common form of an exponential growth or decay model.

4. Interpreting the constant \(k\)

• If \(k>0\), then \(e^{kt}\) increases as \(t\) increases, so \(y\) grows.
• If \(k<0\), then \(e^{kt}\) decreases as \(t\) increases, so \(y\) decays.
• The larger the absolute value of \(k\), the faster the change.

For instance, \(k=0.08\) means growth, while \(k=-0.08\) means decay.

5. Important forms of the model

The standard exponential model is

where:

• \(y_0\) is the initial amount,
• \(k\) is the growth or decay constant,
• \(t\) is time.

Sometimes you need to find \(k\) from information given in the problem. If you know \(y\) at a certain time, substitute into the model and solve for \(k\).

6. Population growth model

When population growth is proportional to population size, we use

The solution is

where \(P_0\) is the initial population.

This model works best when the environment does not strongly limit growth during the time being studied.

Worked Example 1: Solving from a differential equation and initial condition

A population satisfies

Find the population model.

Step 1: Use the general exponential solution:

Step 2: Apply the initial condition \(P(0)=500\):

So \(A=500\).

Answer:

This means the population starts at 500 and grows continuously at a relative rate of 3% per unit time.

Worked Example 2: Finding a future population

A town has population 12,000 and grows according to the model

where \(t\) is measured in years. Find the population after 5 years.

Step 1: Substitute \(t=5\):

Step 2: Approximate:

Answer: After 5 years, the population is about

7. Radioactive decay model

Radioactive decay is modeled the same way, except the constant is negative:

The solution is

where:

• \(N\) is the amount remaining at time \(t\),
• \(N_0\) is the initial amount,
• \(k\) is negative because the amount decreases.

As time passes, the quantity gets smaller, but the model never becomes negative.

8. Half-life and decay constant

The half-life is the time it takes for a substance to decay to half its original amount.

If the half-life is \(h\), then

Using the model \(N=N_0e^{kt}\), substitute \(t=h\):

Divide by \(N_0\):

Take natural logarithms:

So

Since \(\ln(1/2)<0\), the value of \(k\) is negative, as expected for decay.

Worked Example 3: Radioactive decay with a known half-life

A radioactive sample starts with 80 grams and has a half-life of 6 years. Find a model for the amount remaining after \(t\) years.

Step 1: Use the form

Step 2: Use the half-life. After 6 years, 40 grams remain:

Divide by 80:

Take natural logs:

Step 3: Write the model:

An approximate value of \(k\) is

So an approximate model is

9. Finding the growth or decay constant from data

Sometimes a problem gives the starting amount and the amount at a later time. Then you solve for \(k\).

Start with

Substitute the known values, then isolate the exponential term, take natural logarithms, and solve for \(k\).

Worked Example 4: Finding \(k\) from two data values

A bacteria culture starts with 300 cells. After 4 hours, it has 540 cells. Assume exponential growth.

Find:

1. the growth model,
2. the number of cells after 7 hours.

Step 1: Start with the model

Step 2: Use the information \(N(4)=540\):

Divide by 300:

Take natural logs:

Approximate:

So the model is

Step 3: Find the amount after 7 hours:

Since \(e^{1.02865}\approx 2.7974\),

Answer:

• Growth model: $$\boxed{N=300e^{0.14695t}}$$
• After 7 hours: $$\boxed{839\text{ cells (approximately)}}$$

10. How to recognize an exponential growth or decay problem

You should think of the model \(\frac{dy}{dt}=ky\) when:

• the problem says the rate is proportional to the amount present,
• a population grows continuously,
• a radioactive substance decays,
• a quantity increases or decreases by a constant percent rate.

11. Common mistakes to avoid

• Mixing up growth and decay: growth has \(k>0\), decay has \(k<0\).
• Forgetting the natural log: when solving for \(k\), use \(\ln\), not ordinary subtraction or division.
• Using the wrong initial value: in \(y=y_0e^{kt}\), the number \(y_0\) is the amount when \(t=0\).
• Ignoring time units: if \(t\) is in years, then \(k\) must be per year. If \(t\) is in hours, then \(k\) must be per hour.
• Rounding too early: keep extra decimal places until the final answer.

12. Step-by-step strategy for solving problems

1. Write the model $$y=y_0e^{kt}$$
2. Identify whether the situation is growth or decay.
3. Substitute the given values.
4. If needed, solve for \(k\) using natural logarithms.
5. Use the completed model to find the unknown quantity.
6. Check that your answer makes sense in the context.

Brief Summary

Exponential growth and decay come from the differential equation $$\frac{dy}{dt}=ky$$ where the rate of change is proportional to the amount present.

By separating variables and integrating, we get the solution $$y=Ae^{kt}$$ and, with initial condition \(y(0)=y_0\), the model becomes $$y=y_0e^{kt}$$.

This model is used for population growth when \(k>0\) and radioactive decay when \(k<0\). To solve applications, substitute the known values, use natural logarithms when needed, and interpret the result in context.

Newton’s Law of Cooling is a differential equation model that describes how the temperature of an object changes over time when it is placed in surroundings with a constant temperature.

This law is useful in many real-life situations. For example, it can model a hot cup of tea cooling in a room, or a cold drink warming up on a table. In both cases, the key idea is that the speed of temperature change depends on how different the object’s temperature is from the surrounding temperature.

In 12th Grade Maths, this topic connects directly to separable differential equations. We use the model to set up an equation, solve it analytically, and then apply initial conditions to find a specific formula.

1. The main idea

Newton’s Law of Cooling says:

The rate of change of the temperature of an object is directly proportional to the difference between the object’s temperature and the ambient temperature.

Let:

• \(T(t)\) = temperature of the object at time \(t\)

• \(T_a\) = constant ambient (surrounding) temperature

• \(k\) = positive proportionality constant

Then the differential equation is:

The minus sign is important. It makes the model work correctly in both situations:

• If the object is hotter than the surroundings, then \(T - T_a > 0\), so \(\frac{dT}{dt} < 0\). The temperature decreases.

• If the object is colder than the surroundings, then \(T - T_a < 0\), so \(\frac{dT}{dt} > 0\). The temperature increases.

So the object’s temperature always moves toward the ambient temperature.

2. Understanding the equation

The expression \(T - T_a\) measures how far the object’s temperature is from the surrounding temperature. This is often called the temperature difference.

The law says that when this difference is large, the temperature changes quickly. When this difference is small, the temperature changes slowly.

That is why very hot coffee cools quickly at first, but much more slowly later as it gets closer to room temperature.

3. Solving the differential equation

We start with:

This is a separable differential equation. We separate the variables:

Now integrate both sides:

Exponentiating both sides gives:

Since \(e^C\) is just a positive constant, we write the solution more simply as:

So the general solution is:

This is the standard temperature model for Newton’s Law of Cooling.

4. Using the initial condition

If the initial temperature at time \(t=0\) is \(T(0)=T_0\), then substitute into the formula:

So the particular solution becomes:

This is the most useful form of the model.

5. What the graph looks like

The graph of \(T(t)\) is an exponential curve.

• If \(T_0 > T_a\), the graph decreases and levels off toward \(T_a\).

• If \(T_0 < T_a\), the graph increases and levels off toward \(T_a\).

The ambient temperature \(T_a\) is a horizontal asymptote. This means the object gets closer and closer to \(T_a\), but in the model it does not exactly reach it in finite time.

6. Slope field interpretation

Since this topic is also connected to slope fields, consider the differential equation:

The slope at any point depends only on \(T\), not directly on \(t\). So for each fixed temperature value, the slope is the same across the whole row of the slope field.

• When \(T = T_a\), we get \(\frac{dT}{dt}=0\), so the slope is horizontal.

• When \(T > T_a\), the slope is negative.

• When \(T < T_a\), the slope is positive.

• The farther \(T\) is from \(T_a\), the steeper the slope.

This matches the physical meaning: temperatures above the room temperature fall, and temperatures below the room temperature rise.

7. Worked Example 1: Find the formula

A cup of soup is placed in a room at \(20^\circ\text{C}\). Its initial temperature is \(80^\circ\text{C}\). The cooling constant is \(k=0.3\). Find the temperature function.

Step 1: Use the model

Here:

• \(T_a = 20\)

• \(T_0 = 80\)

• \(k = 0.3\)

Step 2: Substitute

Answer:

This means the soup cools from \(80^\circ\text{C}\) toward the room temperature of \(20^\circ\text{C}\).

8. Worked Example 2: Find the temperature after a given time

Using the model from Example 1, find the temperature after 5 minutes.

We use:

Substitute \(t=5\):

Using \(e^{-1.5} \approx 0.223\):

Answer: After 5 minutes, the temperature is about \(33.4^\circ\text{C}\).

9. Worked Example 3: Find the cooling constant from data

A metal object is removed from an oven and placed in a room at \(25^\circ\text{C}\). Its initial temperature is \(125^\circ\text{C}\). After 10 minutes, its temperature is \(65^\circ\text{C}\). Find the value of \(k\), then write the temperature model.

Step 1: Start with the formula

Substitute the known values:

Step 2: Use the data point \(T(10)=65\)

Take natural logarithms:

Using \(\ln(0.4) \approx -0.9163\):

Step 3: Write the model

Answer:

• \(k \approx 0.0916\)

• Temperature model: \(T(t) = 25 + 100e^{-0.0916t}\)

10. Worked Example 4: Find the time required

A drink starts at \(90^\circ\text{C}\) in a room at \(22^\circ\text{C}\). After 8 minutes, it has cooled to \(58^\circ\text{C}\). How long will it take to cool to \(30^\circ\text{C}\)?

Step 1: Write the general model

Step 2: Use \(T(8)=58\) to find \(k\)

Take natural logarithms:

Step 3: Now find when \(T(t)=30\)

Take natural logarithms:

Substitute the value of \(k\):

Then:

Using \(\ln(2/17) \approx -2.1401\):

Answer: It will take about 27 minutes for the drink to cool to \(30^\circ\text{C}\).

11. Common mistakes to avoid

• Forgetting the ambient temperature: The model is not just \(Ce^{-kt}\). It must include \(T_a\):

  $$T(t)=T_a+Ce^{-kt}$$

• Using the wrong initial value: Remember that \(T_0\) is the object’s temperature at \(t=0\), not the ambient temperature.

• Dropping the negative sign: The equation uses \(-k(T-T_a)\), not \(+k(T-T_a)\), when \(k\) is taken as positive.

• Solving for time incorrectly: When the variable is in the exponent, you must isolate the exponential term first and then use natural logarithms.

• Mixing up warming and cooling: The same model works for both. If the object starts below ambient temperature, the temperature rises toward \(T_a\).

12. Quick checklist for solving Newton’s Law of Cooling problems

1. Identify \(T_a\), \(T_0\), and any other given temperature data.

2. Use the model:

   $$T(t) = T_a + (T_0 - T_a)e^{-kt}$$

3. If \(k\) is unknown, substitute another known point to solve for it.

4. If asked for temperature at a certain time, substitute the time value.

5. If asked for the time to reach a temperature, set \(T(t)\) equal to that value and solve using logarithms.

13. Brief summary

Newton’s Law of Cooling models how an object’s temperature moves toward the ambient temperature. The rate of change is proportional to the temperature difference between the object and its surroundings.

The differential equation is:

Its solution is:

This formula allows us to find temperature, time, or the cooling constant \(k\). It also helps us understand the slope field: temperatures above ambient decrease, temperatures below ambient increase, and all solutions approach the ambient temperature over time.

Euler’s Method is a way to approximate solutions to differential equations when finding an exact formula is difficult or impossible.

It is especially useful for an initial value problem, where you know both a differential equation and a starting point.

A typical initial value problem looks like this:

$$\frac{dy}{dx}=f(x,y), \quad y(x_0)=y_0$$

This means the slope of the solution curve at any point is given by the function \(f(x,y)\), and the curve passes through the starting point \((x_0,y_0)\).

Euler’s Method uses the idea that if you know the slope at one point, you can move a small step forward along the tangent line to estimate the next point on the solution curve.

1. The Main Idea Behind Euler’s Method

Suppose you are at the point \((x_n,y_n)\). The differential equation tells you the slope there:

$$\frac{dy}{dx}=f(x_n,y_n)$$

If you move a small amount \(h\) in the \(x\)-direction, then the change in \(y\) is approximately:

$$\Delta y \approx f(x_n,y_n)\,h$$

So the next estimated value is:

$$y_{n+1}=y_n+h\,f(x_n,y_n)$$

At the same time, the next \(x\)-value is:

$$x_{n+1}=x_n+h$$

These two equations are the heart of Euler’s Method.

2. Euler’s Method Formula

Starting from \((x_0,y_0)\), repeat these steps:

1. Find the slope using the differential equation: \(f(x_n,y_n)\).
2. Multiply the slope by the step size \(h\).
3. Add that change to the current \(y\)-value.
4. Increase \(x\) by \(h\).

So the update rules are:

$$x_{n+1}=x_n+h$$

$$y_{n+1}=y_n+h\,f(x_n,y_n)$$

Here:

• \(h\) is the step size
• smaller \(h\) usually gives a more accurate answer
• larger \(h\) is faster but less accurate

3. Why It Works

The exact solution curve may be curved, but over a very short interval, a curve looks almost like a straight line.

Euler’s Method replaces the curve with a sequence of tiny tangent-line steps. Each step uses the slope at the current point to move forward.

This is why the method is also described as using local tangent line advancements.

4. How to Organize the Work

It is often easiest to use a table with columns for:

• \(n\)
• \(x_n\)
• \(y_n\)
• \(f(x_n,y_n)\)
• \(y_{n+1}=y_n+h f(x_n,y_n)\)

This helps avoid mistakes, especially when several steps are needed.

5. Worked Example 1: A Basic Euler’s Method Problem

Use Euler’s Method to approximate \(y(0.3)\) for the differential equation

$$\frac{dy}{dx}=x+y, \quad y(0)=1$$

using step size \(h=0.1\).

Step 1: Identify the formula

Here, \(f(x,y)=x+y\), and the starting point is \((0,1)\).

We need to move from \(x=0\) to \(x=0.3\), so we will take 3 steps of size \(0.1\).

Step 2: First step

Start with:

$$x_0=0, \quad y_0=1$$

Slope at the starting point:

$$f(x_0,y_0)=f(0,1)=0+1=1$$

Update \(y\):

$$y_1=y_0+h f(x_0,y_0)=1+0.1(1)=1.1$$

Update \(x\):

$$x_1=0+0.1=0.1$$

So the next point is:

$$ (x_1,y_1)=(0.1,1.1) $$

Step 3: Second step

Slope at \((0.1,1.1)\):

$$f(0.1,1.1)=0.1+1.1=1.2$$

Update \(y\):

$$y_2=1.1+0.1(1.2)=1.22$$

Update \(x\):

$$x_2=0.2$$

So:

$$ (x_2,y_2)=(0.2,1.22) $$

Step 4: Third step

Slope at \((0.2,1.22)\):

$$f(0.2,1.22)=0.2+1.22=1.42$$

Update \(y\):

$$y_3=1.22+0.1(1.42)=1.362$$

Update \(x\):

$$x_3=0.3$$

Therefore,

$$y(0.3)\approx 1.362$$

6. Worked Example 2: A Differential Equation with a Negative Slope

Use Euler’s Method to approximate \(y(1.5)\) for

$$\frac{dy}{dx}=2x-y, \quad y(1)=2$$

with step size \(h=0.25\).

Step 1: Set up

Here, \(f(x,y)=2x-y\).

We start at:

$$x_0=1, \quad y_0=2$$

To get to \(x=1.5\), we need 2 steps because:

$$1 + 2(0.25)=1.5$$

Step 2: First step

Slope at \((1,2)\):

$$f(1,2)=2(1)-2=0$$

Update \(y\):

$$y_1=2+0.25(0)=2$$

Update \(x\):

$$x_1=1.25$$

Step 3: Second step

Slope at \((1.25,2)\):

$$f(1.25,2)=2(1.25)-2=2.5-2=0.5$$

Update \(y\):

$$y_2=2+0.25(0.5)=2.125$$

Update \(x\):

$$x_2=1.5$$

Therefore,

$$y(1.5)\approx 2.125$$

7. Worked Example 3: Using a Table and More Steps

Approximate \(y(0.4)\) for

$$\frac{dy}{dx}=y-x^2, \quad y(0)=1$$

using \(h=0.1\).

Here, \(f(x,y)=y-x^2\).

After 4 steps, we have reached \(x=0.4\).

So,

$$y(0.4)\approx 1.44949$$

8. Worked Example 4: Comparing Two Step Sizes

Approximate \(y(0.2)\) for

$$\frac{dy}{dx}=x-y, \quad y(0)=1$$

first with \(h=0.2\), then with \(h=0.1\).

This example shows how step size affects accuracy.

Method A: Using \(h=0.2\)

Only one step is needed.

Start:

$$x_0=0, \quad y_0=1$$

Slope:

$$f(0,1)=0-1=-1$$

Update:

$$y_1=1+0.2(-1)=0.8$$

So:

$$y(0.2)\approx 0.8$$

Method B: Using \(h=0.1\)

Now two steps are needed.

First step:

$$f(0,1)=-1$$

$$y_1=1+0.1(-1)=0.9$$

$$x_1=0.1$$

Second step:

$$f(0.1,0.9)=0.1-0.9=-0.8$$

$$y_2=0.9+0.1(-0.8)=0.82$$

So:

$$y(0.2)\approx 0.82$$

What do we notice?

Using the smaller step size gave a different estimate. In general, the smaller step size is more reliable because it follows the curve more closely.

9. Connection to Slope Fields

A slope field shows small line segments representing the slope \(\frac{dy}{dx}=f(x,y)\) at many points.

Euler’s Method can be seen as moving from one point to the next by following these slope directions step by step.

So if you imagine a slope field, Euler’s Method is like tracing a path through it using short tangent-line moves.

10. Common Mistakes to Avoid

• Using the wrong formula: Remember that $$y_{n+1}=y_n+h f(x_n,y_n)$$ not just \(y_n+f(x_n,y_n)\).
• Forgetting to multiply by \(h\): The change in \(y\) is slope times step size.
• Using the new point too early: Always compute the slope from the current point \((x_n,y_n)\).
• Making arithmetic errors: Small mistakes early can affect all later steps.
• Using the wrong number of steps: Check how far you must move in \(x\) and divide by \(h\).

11. A Step-by-Step Strategy for Any Euler’s Method Question

1. Write the differential equation in the form \(\frac{dy}{dx}=f(x,y)\).
2. Identify the initial point \((x_0,y_0)\).
3. Identify the step size \(h\).
4. Determine how many steps are needed to reach the target \(x\)-value.
5. Repeat the Euler update formula: $$y_{n+1}=y_n+h f(x_n,y_n)$$
6. Keep track of each new \((x,y)\) pair carefully.
7. State the final approximation clearly.

12. When Euler’s Method Is Useful

Euler’s Method is useful when:

• an exact solution is hard to find
• you only need an approximate value
• you want to understand how solutions behave step by step
• you are connecting differential equations to slope fields and numerical methods

It is one of the simplest numerical methods in differential equations, and it builds the foundation for more accurate methods studied later.

13. Brief Summary

Euler’s Method approximates the solution to an initial value problem by moving forward in small steps using the slope given by the differential equation.

The key formulas are:

$$x_{n+1}=x_n+h$$

$$y_{n+1}=y_n+h f(x_n,y_n)$$

The method works by using the tangent line at the current point to estimate the next point. Smaller step sizes usually lead to better approximations.