Unit 7 · Topic 7.7 · about 30 minutes · 7 parts · 13 scored checks, 4 of them multiple choice

AP Calculus AB 7.7: Finding Particular Solutions Using Initial Conditions and Separation of Variables

Q: When should I write the solution as an integral?

When dydx depends on x alone and you either cannot find an antiderivative, as with e-x2, or only know f from a graph or table. Then y = y0 + ax f(t)\,dt is the solution, and you evaluate it with areas or given values.

Q: How do I find the interval where the solution is defined?

Find every x where the formula, or dydx along it, breaks down: a zero denominator, a negative under a square root, a logarithm of something not positive. Those points cut the line into intervals, and the solution is the one interval that contains the initial x.

Q: What do I do with the plus or minus from a square root?

Look at the initial y. If y0 is negative, keep the negative root; if it is positive, keep the positive one. Both roots satisfy the squared equation, but only one passes through the point.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 7: 5 to 10% of the multiple-choice section

What you'll be able to do

Find the particular solution of a differential equation through a given point, write it as a starting value plus an integral when that is the clearest form, and state the interval on which it is a solution.

Key vocabulary

Term	What it means	The exam trap
General solution	The family of every function that satisfies a differential equation, usually written with a constant $C$ .	A family with $C$ in it is not yet an answer to a question that gives you a point.
Initial condition	A known point on the solution, written $y (a) = y_{0}$ : at $x = a$ , $y$ equals $y_{0}$ .	Keep the two numbers straight: one is an $x$ value and the other is a $y$ value.
Particular solution	The one solution of a differential equation that also satisfies a given initial condition.	Check an answer by putting the initial $x$ into it: it must give back the initial $y$ exactly.
Domain of a solution	The open interval, containing the initial $x$ , on which the particular solution is defined and differentiable.	Find it after you have the particular solution, since it depends on the initial condition as well as on the equation.

A slope field for dy/dx = y squared with three solution curves of the family y = 1/(C - x). The highlighted curve passes through (0, 1) and rises toward the vertical asymptote x = 1; it is the particular solution y = 1/(1 - x), defined only for x < 1. To the right of x = 1 the same formula gives a dashed curve below the x-axis, marked as not the solution. Two other curves, with C = -1 and C = 2.5, are other members of the general solution.

Part 1 Think first How tall at week 4?

Make a guess before the lesson starts. Committing to an answer, even a wrong one, helps the right one stick.

Predict first

Maya measures a bean plant every week. Its height $h$ , in centimeters, grows at the rate $\frac{d h}{d t} = 2 t + 1$ centimeters per week, where $t$ is the number of weeks since she started measuring. At $t = 2$ the plant is 7 centimeters tall. How tall is it at $t = 4$ ?

In short. A particular solution of a differential equation is the one member of the general solution that satisfies an initial condition $y (a) = y_{0}$ , and it is a solution on an open interval that contains $a$ .

Put the initial condition in right after you integrate, while $C$ still stands alone, and only then solve for $y$ .
When $\frac{d y}{d x} = f (x)$ , the particular solution with $y (a) = y_{0}$ is $y = y_{0} + \int_{a}^{x} f (t) d t$ : the starting value plus the change since $a$ .
A formula can be defined in places where it is not the solution. Keep the interval around the initial $x$ , and choose any $\pm$ sign from the initial condition.

Part 2 Learn it Three short ideas, each checked as you go

7a. One point picks one curve

The general solution of a differential equation is a family of curves, one for each value of $C$ . For $\frac{d y}{d x} = 6 x^{2} - 4$ it is $y = 2 x^{3} - 4 x + C$ . Every curve in that family has the right slopes, but only one of them passes through a given point. That point, written as an initial condition such as $y (1) = 3$ , picks out the particular solution. Put it in and solve for $C$ : $3 = 2 - 4 + C$ , so $C = 5$ and $y = 2 x^{3} - 4 x + 5$ .

In a separable equation, use the point right after you integrate, while $C$ still stands alone. Take $\frac{d y}{d x} = y \cos x$ with $y (0) = 5$ :

\int \frac{1}{y} d y = \int \cos x d x \Rightarrow \ln | y | = \sin x + C .

Now $x = 0$ , $y = 5$ gives $\ln 5 = 0 + C$ , so $C = \ln 5$ . Then $| y | = e^{\sin x + \ln 5} = 5 e^{\sin x}$ , and since $y (0) = 5$ is positive, $y = 5 e^{\sin x}$ .

If you solve for $y$ first, the constant changes shape: $e^{\sin x + C} = e^{C} e^{\sin x}$ , a factor in front, not something added on. Writing $y = e^{\sin x} + C$ there gives a curve that does not solve the equation at all.

Quick check · Multiple choice

No calculator. Which function is the solution of $\frac{d y}{d x} = 2 x y$ with the initial condition $y (0) = 3$ ?

7b. A particular solution written as an integral

When $\frac{d y}{d x} = f (x)$ depends on $x$ alone, the Fundamental Theorem of Calculus writes the particular solution for you. If $f$ is continuous, the solution with $y (a) = y_{0}$ is

F (x) = y_{0} + \int_{a}^{x} f (t) d t .

Two quick checks show why. The derivative of $\int_{a}^{x} f (t) d t$ is $f (x)$ , so $F^{'} (x) = f (x)$ . And at $x = a$ the integral runs over no width, so $F (a) = y_{0} + 0 = y_{0}$ .

Read it as the starting value plus the change since the start. That makes it useful in two cases. Some functions, such as $e^{- x^{2}}$ , have no antiderivative you can write with the usual functions, yet $\frac{d y}{d x} = e^{- x^{2}}$ with $y (0) = 2$ still has the solution $y = 2 + \int_{0}^{x} e^{- t^{2}} d t$ . And sometimes you only know $f$ through a graph or a table. If $y (1) = - 2$ and $\int_{1}^{4} f (t) d t = 7$ , then $y (4) = - 2 + 7 = 5$ , with no formula for $y$ at all.

For an $x$ to the left of $a$ , the integral runs backward: $\int_{a}^{x} f (t) d t = - \int_{x}^{a} f (t) d t$ .

Quick check · Fill the statement

No calculator. Tap a blank, then tap an answer.

The solution of $\frac{d y}{d x} = \sin (x^{2})$ with $y (1) = 4$ is $y =$ $+$ the integral from to $x$ of $d t$ .

7c. Where the solution lives

A solution has to be differentiable on an open interval that contains the initial $x$ , and it cannot jump across a place where it is not defined. So a particular solution can live on less than the whole line, even when its formula makes sense in more places.

The graph at the top shows $\frac{d y}{d x} = y^{2}$ with $y (0) = 1$ . Separating gives $- \frac{1}{y} = x + C$ , and the point gives $- 1 = 0 + C$ , so $C = - 1$ and $y = \frac{1}{1 - x}$ . The formula is defined for every $x$ except 1. But the solution starts at $x = 0$ and cannot pass $x = 1$ , where it blows up, so it is a solution only for $x < 1$ . The dashed piece to the right of 1 is the same formula and a different curve.

The initial condition also chooses a sign. For $\frac{d y}{d x} = \frac{x}{y}$ with $y (0) = - 3$ , separating gives $\frac{y^{2}}{2} = \frac{x^{2}}{2} + C$ , the point gives $C = \frac{9}{2}$ , and $y^{2} = x^{2} + 9$ . Taking a square root allows $\pm$ . Only the negative root passes through $(0, - 3)$ , so $y = - \sqrt{x^{2} + 9}$ .

Quick check · Multiple choice

No calculator. Let $y = f (x)$ be the solution of $\frac{d y}{d x} = y^{2}$ with $f (0) = - \frac{1}{2}$ . On which interval is $f$ a solution?

Common mistakes

The mistake	Why it is wrong	Do this instead
Solving for $y$ first, then writing $+ C$ on the end, as in $y = e^{x^{2}} + C$ .	Exponentiating turns the constant into a factor, $e^{x^{2} + C} = e^{C} e^{x^{2}}$ . An added $C$ gives a curve that does not solve the equation.	Use the initial condition on the line where $C$ first appears, right after integrating, and then solve for $y$ .
Giving the domain of the formula as the domain of the solution.	A solution cannot cross a point where it is not defined, so pieces of the formula on the far side belong to other curves.	Find where the formula breaks down, then keep the open interval that contains the initial $x$ .
Taking the positive square root by habit.	Both $\sqrt{}$ and $- \sqrt{}$ satisfy $y^{2} = \dots$ , and only one passes through the initial point.	Check the sign of $y_{0}$ and keep the root with that sign.

Part 3 Watch it done A particular solution and its interval

Watch each step, then finish the last one yourself.

Find the solution of $\frac{d y}{d x} = 3 x^{2} e^{- y}$ with the initial condition $y (0) = 0$ , and the interval on which it is a solution.

Step 1 · Separate
Multiply both sides by $e^{y} d x$ : $e^{y} d y = 3 x^{2} d x$ .
Step 2 · Integrate
$\int e^{y} d y = \int 3 x^{2} d x$ gives $e^{y} = x^{3} + C$ .
Step 3 · Find C now
Put in $x = 0$ , $y = 0$ : $e^{0} = 0 + C$ , so $C = 1$ and $e^{y} = x^{3} + 1$ .
Why this step?
Why put the point in before solving for $y$ ?
On this line $C$ stands alone, so the point gives it in one step. Solving first would put $C$ inside a logarithm, $y = \ln (x^{3} + C)$ . That still works here, but in other equations solving first turns $C$ into a factor or puts it under a root, where it is easy to mishandle.
Step 4 · Solve for y
Take the natural log of both sides: $y = \ln (x^{3} + 1)$ . The logarithm needs $x^{3} + 1 > 0$ .
Your step · Calculate

No calculator. Finish it. The solution is $y = \ln (x^{3} + 1)$ , and it is a solution on the open interval $x > k$ , which contains the initial $x = 0$ . What is $k$ ?

Your answer

Part 4 Try it A sign to choose, and a tank to fill

Try it · Multiple choice

No calculator. Which function is the solution of $\frac{d y}{d x} = - \frac{x}{y}$ with the initial condition $y (0) = - 2$ ?

Try it · Calculate

No calculator. A tank holds 40 gallons of water at $t = 0$ . Water flows in so that $\frac{d W}{d t} = 12 - 3 \sqrt{t}$ gallons per hour, where $W (t)$ is the amount in the tank after $t$ hours. How many gallons are in the tank at $t = 4$ ?

Your answer

Part 5 Explain it Write it as a free-response answer

Free response on the exam is written by hand, so do this one on paper. Nothing you type here leaves the page. When you are done, open the model answer and score yourself against each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. Consider the differential equation $\frac{d y}{d x} = 2 x y^{2}$ . Let $y = f (x)$ be the particular solution to the differential equation with the initial condition $f (1) = - 1$ .

(a) Write an equation for the line tangent to the graph of $f$ at $x = 1$ . Use the tangent line to approximate $f (1.1)$ .

(b) Find $y = f (x)$ , the particular solution to the differential equation with the initial condition $f (1) = - 1$ .

Your answer stays on this page. It is never sent or saved.

Model answer

(a) At $(1, - 1)$ , $\frac{d y}{d x} = 2 (1) (- 1)^{2} = 2$ . The tangent line is $y = - 1 + 2 (x - 1)$ , so $f (1.1) \approx - 1 + 2 (0.1) = - 0.8$ .

(b) $\frac{1}{y^{2}} d y = 2 x d x$ , so $\int y^{- 2} d y = \int 2 x d x$ and $- \frac{1}{y} = x^{2} + C$ . At $(1, - 1)$ : $- \frac{1}{- 1} = 1^{2} + C$ , so $1 = 1 + C$ and $C = 0$ . Then $- \frac{1}{y} = x^{2}$ , so $f (x) = - \frac{1}{x^{2}}$ .

(c) The interval is $x > 0$ . The solution has to be differentiable on an interval that contains the initial $x = 1$ , and $- \frac{1}{x^{2}}$ is not defined at $x = 0$ , so the interval cannot reach past 0.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

In (a), finds the slope 2 at $(1, - 1)$ , writes $y = - 1 + 2 (x - 1)$ , and approximates $f (1.1) \approx - 0.8$ .
In (b), separates the variables correctly: $y^{- 2} d y = 2 x d x$ .
In (b), finds both antiderivatives: $- \frac{1}{y}$ and $x^{2}$ .
In (b), includes a constant of integration, uses $f (1) = - 1$ to find $C = 0$ , and solves for $f (x) = - \frac{1}{x^{2}}$ .
In (c), states $x > 0$ and gives the reason: the interval must contain $x = 1$ , and the solution is not defined at $x = 0$ .

Part 6 Check yourself A solution read from a graph

All three case questions use this graph. Read every value you need from it.

The function $f$ is continuous on $0 \leq x \leq 8$ , and its graph is made of four line segments. Let $y = g (x)$ be the solution of the differential equation $\frac{d y}{d x} = f (x)$ with the initial condition $g (2) = 3$ .

The graph of f for x from 0 to 8, four line segments joining the points (0, -4), (2, 2), (4, 2), (6, 0) and (8, -2). It crosses the x-axis going up at x = 4/3, stays at height 2 from x = 2 to x = 4, falls to the x-axis at x = 6 and ends at -2 at x = 8. — The graph of f.

Case question 1 · Calculate

No calculator. What is $g (6)$ ?

Your answer

Case question 2 · Multiple choice

No calculator. Which expression gives $g (x)$ for $0 \leq x \leq 8$ ?

Case question 3 · Decide and justify

No calculator. A classmate finds that $\int_{0}^{2} f (t) d t = - 2$ and writes $g (0) = 3 + (- 2) = 1$ . Is the classmate right?

Which reason supports your answer?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the solution of $\frac{d y}{d x} = 3 x^{2} y$ with $y (0) = 4$ . Tap the first line with a mistake.

Check yourself · Sort it

No calculator. Each card gives a differential equation, an initial condition and the formula that satisfies both. Sort each card by where that formula is the particular solution.

For every real $x$

Only on part of the line

Part 7 Keep it Bring back earlier ideas

Every particular solution starts as a general solution from separation of variables, and the integral form rests on the Fundamental Theorem. Pulling both back up is what keeps them.

Bring it back · Match

No calculator. Tap a differential equation, then its general solution.

Bring it back · Calculate

No calculator. Evaluate $\int_{1}^{8} x^{- 2 / 3} d x$ .

Your answer

In three lines

A general solution is a family of curves; an initial condition picks the one particular solution through that point.
Use the point right after integrating, while C stands alone; for dy/dx = f(x), the solution is y0 plus the integral of f from a to x.
A solution lives on the open interval around the initial x where it stays defined, and the initial value picks any plus or minus sign.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

On the free-response question, show the separation with $d y$ and $d x$ on opposite sides, write both antiderivatives and one constant of integration, and use the initial condition on that same line before you solve for $y$ . Before you move on, put the initial $x$ into your answer and check that it gives back the initial $y$ .

Questions students ask

Do I have to use the initial condition before solving for y?

It is not a rule of algebra, but it is the safe habit. Right after integrating, $C$ stands alone and one substitution finds it. After you exponentiate or take a root, $C$ becomes a factor or sits under the root, where it is easy to mishandle.

Why is there only one particular solution through a point?

Putting the point into the general solution gives one equation for $C$ , and it has one answer. The slope field shows why: at each point the differential equation fixes one slope, so a curve through that point has only one way to go.

When should I write the solution as an integral?

When $\frac{d y}{d x}$ depends on $x$ alone and you either cannot find an antiderivative, as with $e^{- x^{2}}$ , or only know $f$ from a graph or table. Then $y = y_{0} + \int_{a}^{x} f (t) d t$ is the solution, and you evaluate it with areas or given values.

How do I find the interval where the solution is defined?

Find every $x$ where the formula, or $\frac{d y}{d x}$ along it, breaks down: a zero denominator, a negative under a square root, a logarithm of something not positive. Those points cut the line into intervals, and the solution is the one interval that contains the initial $x$ .

What do I do with the plus or minus from a square root?

Look at the initial $y$ . If $y_{0}$ is negative, keep the negative root; if it is positive, keep the positive one. Both roots satisfy the squared equation, but only one passes through the point.

Course alignment, for teachers

AP Calculus AB topic 7.7, Unit 7: Differential Equations.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps all 3 essential knowledge statements of the topic to the part that teaches each one: 3 are checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first How tall at week 4?

Part 2 Learn it Three short ideas, each checked as you go

7a. One point picks one curve

7b. A particular solution written as an integral

7c. Where the solution lives

Part 3 Watch it done A particular solution and its interval

Part 4 Try it A sign to choose, and a tank to fill

Part 5 Explain it Write it as a free-response answer

Part 6 Check yourself A solution read from a graph

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic