7.3 Separation of variables

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is currently in development and is a work-in-progress.

Separation of variables

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What you’ll learn:

How to solve first-order differential equations using separation of variables
How to apply initial conditions to find a particular solution

So far, you’ve seen differential equations like

$\frac{d y}{d x} = 2 x$

which you can solve by integrating both sides with respect to $x$ . When the right-hand side depends on both $x$ and $y$ , a common method is separation of variables.

The idea is to rewrite the equation so that:

all $y$ -terms (and $d y$ ) are on one side
all $x$ -terms (and $d x$ ) are on the other side

This works when the differential equation can be written in the form

$\frac{d y}{d x} = f (x) \cdot g (y)$

If you can’t rewrite it as a product of a function of $x$ and a function of $y$ , then separation of variables won’t apply.

Find the particular solution to the differential equation

$\frac{d y}{d x} = e^{x - y}$

With initial condition: $y (0) = ln (5)$

Step 1: Rewrite the equation to separate the variables

Start by rewriting the exponential so the $x$ and $y$ parts are separated:

$\frac{d y}{d x} = e^{x - y} = \frac{e ^{x}}{e ^{y}}$

A reliable way to separate variables is:

multiply both sides by $d x$
move the $y$ -expression to the side with $d y$

Multiply both sides by $e^{y}$ :

$e^{y} d y = e^{x} d x$

Step 2: Integrate both sides

$\int e^{y} d y = \int e^{x} d x$

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

Step 3: Apply the initial condition if given

If no initial condition is given, you would solve for $y$ to get a general solution. Here, the initial condition is $(x, y) = (0, ln (5))$ , so substitute it to find $C$ :

$e^{l n (5)} = e^{0} + C$

$5 = 1 + C$

$C = 4$

So the equation becomes

$e^{y} = e^{x} + 4$

Now solve for $y$ by taking the natural log of both sides:

$ln (e^{y}) = ln (e^{x} + 4)$

$y = ln (e^{x} + 4)$

Examples

Find the general solution to

$\frac{d y}{d x} = y + y x^{2}$

Solution

(spoiler)

First, factor the right-hand side so it’s written as a product:

$\frac{d y}{d x} = y (1 + x^{2})$

Now separate the variables. Multiply by $d x$ , then divide by $y$ :

$\frac{1}{y} d y = (1 + x^{2}) d x$

Integrate both sides:

$\int \frac{1}{y} d y = \int (1 + x^{2}) d x$

$ln ∣ y ∣ = x + \frac{x ^{3}}{3} + C$

Solve for $y$ by exponentiating both sides:

$e^{l n ∣ y ∣} = e^{x + x^{3} /3 + C}$

$∣ y ∣ = e^{x + x^{3} /3 + C}$

Absorb the sign into the constant (since $\pm e^{C}$ is still an arbitrary nonzero constant). This gives the standard general-solution form:

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

Find the particular solution to

$\frac{d y}{d x} = x y^{2}$

with $y (0) = 1$ .

Solution

(spoiler)

Separate the variables:

$\frac{1}{y ^{2}} d y = x d x$

Integrate both sides:

$\int \frac{1}{y ^{2}} d y = \int x d x$

$- \frac{1}{y} = \frac{x ^{2}}{2} + C$

Use the initial condition $(x, y) = (0, 1)$ to find $C$ :

$- \frac{1}{1} = \frac{0 ^{2}}{2} + C$

$C = - 1$

Substitute $C = - 1$ and solve for $y$ :

$- \frac{1}{y} = \frac{x ^{2}}{2} - 1$

Multiply both sides by $y$ :

$- 1 = (\frac{x ^{2}}{2} - 1) y$

Divide:

$y = \frac{- 1}{\frac{x ^{2}}{2} - 1} = \frac{- 1}{\frac{x ^{2} - 2}{2}}$

$y = - \frac{2}{x ^{2} - 2}$

Find the particular solution to

$\frac{d y}{d x} = y + 1$

with $y (2) = 0$

Solution

(spoiler)

Separate the variables:

$\frac{1}{y + 1} d y = d x$

Integrate both sides. (On the left, you can use $u$ -substitution with $u = y + 1$ .)

$\int \frac{1}{y + 1} d y = \int 1 d x$

$ln (y + 1) = x + C$

Use the initial condition $(x, y) = (2, 0)$ to find $C$ :

$ln (0 + 1) = 2 + C$

$0 = 2 + C$

$C = - 2$

Substitute $C = - 2$ and solve for $y$ :

$ln (y + 1) = x - 2$

$e^{l n (y + 1)} = e^{x - 2}$

$y + 1 = e^{x - 2}$

$y = e^{x - 2} - 1$

If

$\frac{d y}{d t} = \frac{2 t + 1}{y}$

and $y = 2$ when $t = 0$ , find $y$ when $t = 1$ .

Solution

(spoiler)

Even though the variables are $y$ and $t$ (instead of $y$ and $x$ ), the separation steps are the same.

Separate the variables:

$y d y = (2 t + 1) d t$

Integrate both sides:

$\int y d y = \int (2 t + 1) d t$

$\frac{y ^{2}}{2} = t^{2} + t + C$

Use the initial condition $y = 2$ when $t = 0$ to find $C$ :

$\frac{2 ^{2}}{2} = 0^{2} + 0 + C$

$2 = C$

$\frac{y ^{2}}{2} = t^{2} + t + 2$

$y^{2} = 2 t^{2} + 2 t + 4$

$y = \pm 2 t^{2} + 2 t + 4$

Note: The initial condition $y (0) = 2$ tells you $y$ is positive at $t = 0$ , so you take the positive branch:

$y = 2 t^{2} + 2 t + 4$

Now evaluate at $t = 1$ :

$y = 2 (1)^{2} + 2 (1) + 4$

$= 2 + 2 + 4$

$= 22$

Lastly, here is a problem similar to a multi-part free-response question from the 2022 AB exam:

Consider the differential equation

$\frac{d y}{d x} = π cos (π x) y$

Let $y = f (x)$ be the particular solution to the differential equation with initial condition $f (1) = 4$ .

a) Write an equation for the line tangent to the solution curve at the point $(1, 4)$ and use it to approximate $f (0.9)$ .

b) Given $f^{''} (x) < 0$ for $- 1 \leq x \leq 1$ , is the approximation from part a) an overestimate or underestimate for $f (0.9)$ ?

c) Use separation of variables to find the particular solution to the differential equation with initial condition $f (1) = 4$ .

Solutions

a) Equation of tangent line at $(1, 4)$

(spoiler)

The equation of the tangent line to $y = f (x)$ at $(a, f (a))$ is

$y - f (a) = f^{'} (a) (x - a)$

Here, $(a, f (a)) = (1, 4)$ . To find $f^{'} (1)$ , evaluate the differential equation at $x = 1$ and $y = 4$ :

$\frac{d y}{d x}_{x = 1, y = 4} = π cos (π \cdot 1) 4$

$= π (- 1) (2)$

$= - 2 π$

So the tangent line is

$y - 4 = - 2 π (x - 1)$

To approximate $f (0.9)$ , plug $x = 0.9$ into the tangent line:

$y - 4 = - 2 π (0.9 - 1)$

$y = - 2 π (- 0.1) + 4$

$f (0.9) \approx 0.2 π + 4$

b) Underapproximate or overapproximate?

(spoiler)

Since $f^{''} (x) < 0$ on $- 1 \leq x \leq 1$ , the graph is concave down on that interval.

For a concave down function, the tangent line lies above the curve. That means the tangent-line (linear) approximation from part (a) is an overestimate of $f (0.9)$ .

c) Particular solution with $f (1) = 4$

(spoiler)

Separate the variables:

$\frac{1}{y} d y = π cos (π x) d x$

Integrate both sides:

$\int \frac{1}{y} d y = \int π cos (π x) d x$

Rewrite the left integrand as a power:

$\int y^{- 1/2} d y = sin (π x) + C$

$2 y^{1/2} = sin (π x) + C$

Use the initial condition $(1, 4)$ to find $C$ :

$24 = sin (π \cdot 1) + C$

$4 = 0 + C$

$C = 4$

Now solve for $y$ :

$2 y^{1/2} = sin (π x) + 4$

$y^{1/2} = \frac{1}{2} sin (π x) + 2$

$y = (\frac{1}{2} sin (π x) + 2)^{2}$

Separation of variables

Used for first-order differential equations of the form $\frac{d y}{d x} = f (x) g (y)$
Rearrange to isolate $y$ -terms with $d y$ and $x$ -terms with $d x$
Integrate both sides to solve

Applying initial conditions

Substitute given $(x, y)$ values after integrating to solve for the constant $C$
Plug $C$ back into the general solution for the particular solution

Example 1: $\frac{d y}{d x} = e^{x - y}$ , $y (0) = ln (5)$

Separate: $e^{y} d y = e^{x} d x$
Integrate: $e^{y} = e^{x} + C$
Apply initial condition: $C = 4$
Solution: $y = ln (e^{x} + 4)$

Example 2: $\frac{d y}{d x} = y + y x^{2}$

Factor: $y (1 + x^{2})$
Separate: $\frac{1}{y} d y = (1 + x^{2}) d x$
Integrate: $ln ∣ y ∣ = x + \frac{x ^{3}}{3} + C$
General solution: $y = C e^{x + x^{3} /3}$

Example 3: $\frac{d y}{d x} = x y^{2}$ , $y (0) = 1$

Separate: $\frac{1}{y ^{2}} d y = x d x$
Integrate: $- \frac{1}{y} = \frac{x ^{2}}{2} + C$
Apply initial condition: $C = - 1$
Solution: $y = - \frac{2}{x ^{2} - 2}$

Example 4: $\frac{d y}{d x} = y + 1$ , $y (2) = 0$

Separate: $\frac{1}{y + 1} d y = d x$
Integrate: $ln (y + 1) = x + C$
Apply initial condition: $C = - 2$
Solution: $y = e^{x - 2} - 1$

Example 5: $\frac{d y}{d t} = \frac{2 t + 1}{y}$ , $y (0) = 2$

Separate: $y d y = (2 t + 1) d t$
Integrate: $\frac{y ^{2}}{2} = t^{2} + t + C$
Apply initial condition: $C = 2$
Solution: $y = 2 t^{2} + 2 t + 4$
At $t = 1$ : $y = 22$

Tangent line approximation and concavity

Tangent line at $(x_{0}, y_{0})$ : $y - y_{0} = f^{'} (x_{0}) (x - x_{0})$
If $f^{''} (x) < 0$ , function is concave down
- Tangent line overestimates function near $x_{0}$

Example 6: $\frac{d y}{d x} = π cos (π x) y$ , $f (1) = 4$

Tangent line at $(1, 4)$ : $y - 4 = - 2 π (x - 1)$
Approximate $f (0.9)$ : $f (0.9) \approx 0.2 π + 4$
Concavity: $f^{''} (x) < 0$ implies tangent overestimates $f (0.9)$
Separate: $\frac{1}{y} d y = π cos (π x) d x$
Integrate: $2 y^{1/2} = sin (π x) + C$
Apply initial condition: $C = 4$
Solution: $y = (\frac{1}{2} sin (π x) + 2)^{2}$

Separation of variables

What you’ll learn:

How to solve first-order differential equations using separation of variables
How to apply initial conditions to find a particular solution

So far, you’ve seen differential equations like

$\frac{d y}{d x} = 2 x$

which you can solve by integrating both sides with respect to $x$ . When the right-hand side depends on both $x$ and $y$ , a common method is separation of variables.

The idea is to rewrite the equation so that:

all $y$ -terms (and $d y$ ) are on one side
all $x$ -terms (and $d x$ ) are on the other side

This works when the differential equation can be written in the form

$\frac{d y}{d x} = f (x) \cdot g (y)$

If you can’t rewrite it as a product of a function of $x$ and a function of $y$ , then separation of variables won’t apply.

Find the particular solution to the differential equation

$\frac{d y}{d x} = e^{x - y}$

With initial condition: $y (0) = ln (5)$

Step 1: Rewrite the equation to separate the variables

Start by rewriting the exponential so the $x$ and $y$ parts are separated:

$\frac{d y}{d x} = e^{x - y} = \frac{e ^{x}}{e ^{y}}$

A reliable way to separate variables is:

multiply both sides by $d x$
move the $y$ -expression to the side with $d y$

Multiply both sides by $e^{y}$ :

$e^{y} d y = e^{x} d x$

Step 2: Integrate both sides

$\int e^{y} d y = \int e^{x} d x$

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

Step 3: Apply the initial condition if given

If no initial condition is given, you would solve for $y$ to get a general solution. Here, the initial condition is $(x, y) = (0, ln (5))$ , so substitute it to find $C$ :

$e^{l n (5)} = e^{0} + C$

$5 = 1 + C$

$C = 4$

So the equation becomes

$e^{y} = e^{x} + 4$

Now solve for $y$ by taking the natural log of both sides:

$ln (e^{y}) = ln (e^{x} + 4)$

$y = ln (e^{x} + 4)$

Examples

Find the general solution to

$\frac{d y}{d x} = y + y x^{2}$

Solution

(spoiler)

First, factor the right-hand side so it’s written as a product:

$\frac{d y}{d x} = y (1 + x^{2})$

Now separate the variables. Multiply by $d x$ , then divide by $y$ :

$\frac{1}{y} d y = (1 + x^{2}) d x$

Integrate both sides:

$\int \frac{1}{y} d y = \int (1 + x^{2}) d x$

$ln ∣ y ∣ = x + \frac{x ^{3}}{3} + C$

Solve for $y$ by exponentiating both sides:

$e^{l n ∣ y ∣} = e^{x + x^{3} /3 + C}$

$∣ y ∣ = e^{x + x^{3} /3 + C}$

Absorb the sign into the constant (since $\pm e^{C}$ is still an arbitrary nonzero constant). This gives the standard general-solution form:

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

Find the particular solution to

$\frac{d y}{d x} = x y^{2}$

with $y (0) = 1$ .

Solution

(spoiler)

Separate the variables:

$\frac{1}{y ^{2}} d y = x d x$

Integrate both sides:

$\int \frac{1}{y ^{2}} d y = \int x d x$

$- \frac{1}{y} = \frac{x ^{2}}{2} + C$

Use the initial condition $(x, y) = (0, 1)$ to find $C$ :

$- \frac{1}{1} = \frac{0 ^{2}}{2} + C$

$C = - 1$

Substitute $C = - 1$ and solve for $y$ :

$- \frac{1}{y} = \frac{x ^{2}}{2} - 1$

Multiply both sides by $y$ :

$- 1 = (\frac{x ^{2}}{2} - 1) y$

Divide:

$y = \frac{- 1}{\frac{x ^{2}}{2} - 1} = \frac{- 1}{\frac{x ^{2} - 2}{2}}$

$y = - \frac{2}{x ^{2} - 2}$

Find the particular solution to

$\frac{d y}{d x} = y + 1$

with $y (2) = 0$

Solution

(spoiler)

Separate the variables:

$\frac{1}{y + 1} d y = d x$

Integrate both sides. (On the left, you can use $u$ -substitution with $u = y + 1$ .)

$\int \frac{1}{y + 1} d y = \int 1 d x$

$ln (y + 1) = x + C$

Use the initial condition $(x, y) = (2, 0)$ to find $C$ :

$ln (0 + 1) = 2 + C$

$0 = 2 + C$

$C = - 2$

Substitute $C = - 2$ and solve for $y$ :

$ln (y + 1) = x - 2$

$e^{l n (y + 1)} = e^{x - 2}$

$y + 1 = e^{x - 2}$

$y = e^{x - 2} - 1$

If

$\frac{d y}{d t} = \frac{2 t + 1}{y}$

and $y = 2$ when $t = 0$ , find $y$ when $t = 1$ .

Solution

(spoiler)

Even though the variables are $y$ and $t$ (instead of $y$ and $x$ ), the separation steps are the same.

Separate the variables:

$y d y = (2 t + 1) d t$

Integrate both sides:

$\int y d y = \int (2 t + 1) d t$

$\frac{y ^{2}}{2} = t^{2} + t + C$

Use the initial condition $y = 2$ when $t = 0$ to find $C$ :

$\frac{2 ^{2}}{2} = 0^{2} + 0 + C$

$2 = C$

$\frac{y ^{2}}{2} = t^{2} + t + 2$

$y^{2} = 2 t^{2} + 2 t + 4$

$y = \pm 2 t^{2} + 2 t + 4$

Note: The initial condition $y (0) = 2$ tells you $y$ is positive at $t = 0$ , so you take the positive branch:

$y = 2 t^{2} + 2 t + 4$

Now evaluate at $t = 1$ :

$y = 2 (1)^{2} + 2 (1) + 4$

$= 2 + 2 + 4$

$= 22$

Lastly, here is a problem similar to a multi-part free-response question from the 2022 AB exam:

Consider the differential equation

$\frac{d y}{d x} = π cos (π x) y$

Let $y = f (x)$ be the particular solution to the differential equation with initial condition $f (1) = 4$ .

a) Write an equation for the line tangent to the solution curve at the point $(1, 4)$ and use it to approximate $f (0.9)$ .

b) Given $f^{''} (x) < 0$ for $- 1 \leq x \leq 1$ , is the approximation from part a) an overestimate or underestimate for $f (0.9)$ ?

c) Use separation of variables to find the particular solution to the differential equation with initial condition $f (1) = 4$ .

Solutions

a) Equation of tangent line at $(1, 4)$

(spoiler)

The equation of the tangent line to $y = f (x)$ at $(a, f (a))$ is

$y - f (a) = f^{'} (a) (x - a)$

Here, $(a, f (a)) = (1, 4)$ . To find $f^{'} (1)$ , evaluate the differential equation at $x = 1$ and $y = 4$ :

$\frac{d y}{d x}_{x = 1, y = 4} = π cos (π \cdot 1) 4$

$= π (- 1) (2)$

$= - 2 π$

So the tangent line is

$y - 4 = - 2 π (x - 1)$

To approximate $f (0.9)$ , plug $x = 0.9$ into the tangent line:

$y - 4 = - 2 π (0.9 - 1)$

$y = - 2 π (- 0.1) + 4$

$f (0.9) \approx 0.2 π + 4$

b) Underapproximate or overapproximate?

(spoiler)

Since $f^{''} (x) < 0$ on $- 1 \leq x \leq 1$ , the graph is concave down on that interval.

For a concave down function, the tangent line lies above the curve. That means the tangent-line (linear) approximation from part (a) is an overestimate of $f (0.9)$ .

c) Particular solution with $f (1) = 4$

(spoiler)

Separate the variables:

$\frac{1}{y} d y = π cos (π x) d x$

Integrate both sides:

$\int \frac{1}{y} d y = \int π cos (π x) d x$

Rewrite the left integrand as a power:

$\int y^{- 1/2} d y = sin (π x) + C$

$2 y^{1/2} = sin (π x) + C$

Use the initial condition $(1, 4)$ to find $C$ :

$24 = sin (π \cdot 1) + C$

$4 = 0 + C$

$C = 4$

Now solve for $y$ :

$2 y^{1/2} = sin (π x) + 4$

$y^{1/2} = \frac{1}{2} sin (π x) + 2$

$y = (\frac{1}{2} sin (π x) + 2)^{2}$

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is currently in development and is a work-in-progress.

Separation of variables

7 min read

Font

Discuss

Feedback

What you’ll learn:

How to solve first-order differential equations using separation of variables
How to apply initial conditions to find a particular solution

So far, you’ve seen differential equations like

$\frac{d y}{d x} = 2 x$

which you can solve by integrating both sides with respect to $x$ . When the right-hand side depends on both $x$ and $y$ , a common method is separation of variables.

The idea is to rewrite the equation so that:

all $y$ -terms (and $d y$ ) are on one side
all $x$ -terms (and $d x$ ) are on the other side

This works when the differential equation can be written in the form

$\frac{d y}{d x} = f (x) \cdot g (y)$

If you can’t rewrite it as a product of a function of $x$ and a function of $y$ , then separation of variables won’t apply.

Find the particular solution to the differential equation

$\frac{d y}{d x} = e^{x - y}$

With initial condition: $y (0) = ln (5)$

Step 1: Rewrite the equation to separate the variables

Start by rewriting the exponential so the $x$ and $y$ parts are separated:

$\frac{d y}{d x} = e^{x - y} = \frac{e ^{x}}{e ^{y}}$

A reliable way to separate variables is:

multiply both sides by $d x$
move the $y$ -expression to the side with $d y$

Multiply both sides by $e^{y}$ :

$e^{y} d y = e^{x} d x$

Step 2: Integrate both sides

$\int e^{y} d y = \int e^{x} d x$

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

Step 3: Apply the initial condition if given

If no initial condition is given, you would solve for $y$ to get a general solution. Here, the initial condition is $(x, y) = (0, ln (5))$ , so substitute it to find $C$ :

$e^{l n (5)} = e^{0} + C$

$5 = 1 + C$

$C = 4$

So the equation becomes

$e^{y} = e^{x} + 4$

Now solve for $y$ by taking the natural log of both sides:

$ln (e^{y}) = ln (e^{x} + 4)$

$y = ln (e^{x} + 4)$

Examples

Find the general solution to

$\frac{d y}{d x} = y + y x^{2}$

Solution

(spoiler)

First, factor the right-hand side so it’s written as a product:

$\frac{d y}{d x} = y (1 + x^{2})$

Now separate the variables. Multiply by $d x$ , then divide by $y$ :

$\frac{1}{y} d y = (1 + x^{2}) d x$

Integrate both sides:

$\int \frac{1}{y} d y = \int (1 + x^{2}) d x$

$ln ∣ y ∣ = x + \frac{x ^{3}}{3} + C$

Solve for $y$ by exponentiating both sides:

$e^{l n ∣ y ∣} = e^{x + x^{3} /3 + C}$

$∣ y ∣ = e^{x + x^{3} /3 + C}$

Absorb the sign into the constant (since $\pm e^{C}$ is still an arbitrary nonzero constant). This gives the standard general-solution form:

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

Find the particular solution to

$\frac{d y}{d x} = x y^{2}$

with $y (0) = 1$ .

Solution

(spoiler)

Separate the variables:

$\frac{1}{y ^{2}} d y = x d x$

Integrate both sides:

$\int \frac{1}{y ^{2}} d y = \int x d x$

$- \frac{1}{y} = \frac{x ^{2}}{2} + C$

Use the initial condition $(x, y) = (0, 1)$ to find $C$ :

$- \frac{1}{1} = \frac{0 ^{2}}{2} + C$

$C = - 1$

Substitute $C = - 1$ and solve for $y$ :

$- \frac{1}{y} = \frac{x ^{2}}{2} - 1$

Multiply both sides by $y$ :

$- 1 = (\frac{x ^{2}}{2} - 1) y$

Divide:

$y = \frac{- 1}{\frac{x ^{2}}{2} - 1} = \frac{- 1}{\frac{x ^{2} - 2}{2}}$

$y = - \frac{2}{x ^{2} - 2}$

Find the particular solution to

$\frac{d y}{d x} = y + 1$

with $y (2) = 0$

Solution

(spoiler)

Separate the variables:

$\frac{1}{y + 1} d y = d x$

Integrate both sides. (On the left, you can use $u$ -substitution with $u = y + 1$ .)

$\int \frac{1}{y + 1} d y = \int 1 d x$

$ln (y + 1) = x + C$

Use the initial condition $(x, y) = (2, 0)$ to find $C$ :

$ln (0 + 1) = 2 + C$

$0 = 2 + C$

$C = - 2$

Substitute $C = - 2$ and solve for $y$ :

$ln (y + 1) = x - 2$

$e^{l n (y + 1)} = e^{x - 2}$

$y + 1 = e^{x - 2}$

$y = e^{x - 2} - 1$

If

$\frac{d y}{d t} = \frac{2 t + 1}{y}$

and $y = 2$ when $t = 0$ , find $y$ when $t = 1$ .

Solution

(spoiler)

Even though the variables are $y$ and $t$ (instead of $y$ and $x$ ), the separation steps are the same.

Separate the variables:

$y d y = (2 t + 1) d t$

Integrate both sides:

$\int y d y = \int (2 t + 1) d t$

$\frac{y ^{2}}{2} = t^{2} + t + C$

Use the initial condition $y = 2$ when $t = 0$ to find $C$ :

$\frac{2 ^{2}}{2} = 0^{2} + 0 + C$

$2 = C$

$\frac{y ^{2}}{2} = t^{2} + t + 2$

$y^{2} = 2 t^{2} + 2 t + 4$

$y = \pm 2 t^{2} + 2 t + 4$

Note: The initial condition $y (0) = 2$ tells you $y$ is positive at $t = 0$ , so you take the positive branch:

$y = 2 t^{2} + 2 t + 4$

Now evaluate at $t = 1$ :

$y = 2 (1)^{2} + 2 (1) + 4$

$= 2 + 2 + 4$

$= 22$

Lastly, here is a problem similar to a multi-part free-response question from the 2022 AB exam:

Consider the differential equation

$\frac{d y}{d x} = π cos (π x) y$

Let $y = f (x)$ be the particular solution to the differential equation with initial condition $f (1) = 4$ .

a) Write an equation for the line tangent to the solution curve at the point $(1, 4)$ and use it to approximate $f (0.9)$ .

b) Given $f^{''} (x) < 0$ for $- 1 \leq x \leq 1$ , is the approximation from part a) an overestimate or underestimate for $f (0.9)$ ?

c) Use separation of variables to find the particular solution to the differential equation with initial condition $f (1) = 4$ .

Solutions

a) Equation of tangent line at $(1, 4)$

(spoiler)

The equation of the tangent line to $y = f (x)$ at $(a, f (a))$ is

$y - f (a) = f^{'} (a) (x - a)$

Here, $(a, f (a)) = (1, 4)$ . To find $f^{'} (1)$ , evaluate the differential equation at $x = 1$ and $y = 4$ :

$\frac{d y}{d x}_{x = 1, y = 4} = π cos (π \cdot 1) 4$

$= π (- 1) (2)$

$= - 2 π$

So the tangent line is

$y - 4 = - 2 π (x - 1)$

To approximate $f (0.9)$ , plug $x = 0.9$ into the tangent line:

$y - 4 = - 2 π (0.9 - 1)$

$y = - 2 π (- 0.1) + 4$

$f (0.9) \approx 0.2 π + 4$

b) Underapproximate or overapproximate?

(spoiler)

Since $f^{''} (x) < 0$ on $- 1 \leq x \leq 1$ , the graph is concave down on that interval.

For a concave down function, the tangent line lies above the curve. That means the tangent-line (linear) approximation from part (a) is an overestimate of $f (0.9)$ .

c) Particular solution with $f (1) = 4$

(spoiler)

Separate the variables:

$\frac{1}{y} d y = π cos (π x) d x$

Integrate both sides:

$\int \frac{1}{y} d y = \int π cos (π x) d x$

Rewrite the left integrand as a power:

$\int y^{- 1/2} d y = sin (π x) + C$

$2 y^{1/2} = sin (π x) + C$

Use the initial condition $(1, 4)$ to find $C$ :

$24 = sin (π \cdot 1) + C$

$4 = 0 + C$

$C = 4$

Now solve for $y$ :

$2 y^{1/2} = sin (π x) + 4$

$y^{1/2} = \frac{1}{2} sin (π x) + 2$

$y = (\frac{1}{2} sin (π x) + 2)^{2}$

Separation of variables

Used for first-order differential equations of the form $\frac{d y}{d x} = f (x) g (y)$
Rearrange to isolate $y$ -terms with $d y$ and $x$ -terms with $d x$
Integrate both sides to solve

Applying initial conditions

Substitute given $(x, y)$ values after integrating to solve for the constant $C$
Plug $C$ back into the general solution for the particular solution

Example 1: $\frac{d y}{d x} = e^{x - y}$ , $y (0) = ln (5)$

Separate: $e^{y} d y = e^{x} d x$
Integrate: $e^{y} = e^{x} + C$
Apply initial condition: $C = 4$
Solution: $y = ln (e^{x} + 4)$

Example 2: $\frac{d y}{d x} = y + y x^{2}$

Factor: $y (1 + x^{2})$
Separate: $\frac{1}{y} d y = (1 + x^{2}) d x$
Integrate: $ln ∣ y ∣ = x + \frac{x ^{3}}{3} + C$
General solution: $y = C e^{x + x^{3} /3}$

Example 3: $\frac{d y}{d x} = x y^{2}$ , $y (0) = 1$

Separate: $\frac{1}{y ^{2}} d y = x d x$
Integrate: $- \frac{1}{y} = \frac{x ^{2}}{2} + C$
Apply initial condition: $C = - 1$
Solution: $y = - \frac{2}{x ^{2} - 2}$

Example 4: $\frac{d y}{d x} = y + 1$ , $y (2) = 0$

Separate: $\frac{1}{y + 1} d y = d x$
Integrate: $ln (y + 1) = x + C$
Apply initial condition: $C = - 2$
Solution: $y = e^{x - 2} - 1$

Example 5: $\frac{d y}{d t} = \frac{2 t + 1}{y}$ , $y (0) = 2$

Separate: $y d y = (2 t + 1) d t$
Integrate: $\frac{y ^{2}}{2} = t^{2} + t + C$
Apply initial condition: $C = 2$
Solution: $y = 2 t^{2} + 2 t + 4$
At $t = 1$ : $y = 22$

Tangent line approximation and concavity

Tangent line at $(x_{0}, y_{0})$ : $y - y_{0} = f^{'} (x_{0}) (x - x_{0})$
If $f^{''} (x) < 0$ , function is concave down
- Tangent line overestimates function near $x_{0}$

Example 6: $\frac{d y}{d x} = π cos (π x) y$ , $f (1) = 4$

Tangent line at $(1, 4)$ : $y - 4 = - 2 π (x - 1)$
Approximate $f (0.9)$ : $f (0.9) \approx 0.2 π + 4$
Concavity: $f^{''} (x) < 0$ implies tangent overestimates $f (0.9)$
Separate: $\frac{1}{y} d y = π cos (π x) d x$
Integrate: $2 y^{1/2} = sin (π x) + C$
Apply initial condition: $C = 4$
Solution: $y = (\frac{1}{2} sin (π x) + 2)^{2}$

Separation of variables

What you’ll learn:

How to solve first-order differential equations using separation of variables
How to apply initial conditions to find a particular solution

So far, you’ve seen differential equations like

$\frac{d y}{d x} = 2 x$

which you can solve by integrating both sides with respect to $x$ . When the right-hand side depends on both $x$ and $y$ , a common method is separation of variables.

The idea is to rewrite the equation so that:

all $y$ -terms (and $d y$ ) are on one side
all $x$ -terms (and $d x$ ) are on the other side

This works when the differential equation can be written in the form

$\frac{d y}{d x} = f (x) \cdot g (y)$

If you can’t rewrite it as a product of a function of $x$ and a function of $y$ , then separation of variables won’t apply.

Find the particular solution to the differential equation

$\frac{d y}{d x} = e^{x - y}$

With initial condition: $y (0) = ln (5)$

Step 1: Rewrite the equation to separate the variables

Start by rewriting the exponential so the $x$ and $y$ parts are separated:

$\frac{d y}{d x} = e^{x - y} = \frac{e ^{x}}{e ^{y}}$

A reliable way to separate variables is:

multiply both sides by $d x$
move the $y$ -expression to the side with $d y$

Multiply both sides by $e^{y}$ :

$e^{y} d y = e^{x} d x$

Step 2: Integrate both sides

$\int e^{y} d y = \int e^{x} d x$

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

Step 3: Apply the initial condition if given

If no initial condition is given, you would solve for $y$ to get a general solution. Here, the initial condition is $(x, y) = (0, ln (5))$ , so substitute it to find $C$ :

$e^{l n (5)} = e^{0} + C$

$5 = 1 + C$

$C = 4$

So the equation becomes

$e^{y} = e^{x} + 4$

Now solve for $y$ by taking the natural log of both sides:

$ln (e^{y}) = ln (e^{x} + 4)$

$y = ln (e^{x} + 4)$

Examples

Find the general solution to

$\frac{d y}{d x} = y + y x^{2}$

Solution

(spoiler)

First, factor the right-hand side so it’s written as a product:

$\frac{d y}{d x} = y (1 + x^{2})$

Now separate the variables. Multiply by $d x$ , then divide by $y$ :

$\frac{1}{y} d y = (1 + x^{2}) d x$

Integrate both sides:

$\int \frac{1}{y} d y = \int (1 + x^{2}) d x$

$ln ∣ y ∣ = x + \frac{x ^{3}}{3} + C$

Solve for $y$ by exponentiating both sides:

$e^{l n ∣ y ∣} = e^{x + x^{3} /3 + C}$

$∣ y ∣ = e^{x + x^{3} /3 + C}$

Absorb the sign into the constant (since $\pm e^{C}$ is still an arbitrary nonzero constant). This gives the standard general-solution form:

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

Find the particular solution to

$\frac{d y}{d x} = x y^{2}$

with $y (0) = 1$ .

Solution

(spoiler)

Separate the variables:

$\frac{1}{y ^{2}} d y = x d x$

Integrate both sides:

$\int \frac{1}{y ^{2}} d y = \int x d x$

$- \frac{1}{y} = \frac{x ^{2}}{2} + C$

Use the initial condition $(x, y) = (0, 1)$ to find $C$ :

$- \frac{1}{1} = \frac{0 ^{2}}{2} + C$

$C = - 1$

Substitute $C = - 1$ and solve for $y$ :

$- \frac{1}{y} = \frac{x ^{2}}{2} - 1$

Multiply both sides by $y$ :

$- 1 = (\frac{x ^{2}}{2} - 1) y$

Divide:

$y = \frac{- 1}{\frac{x ^{2}}{2} - 1} = \frac{- 1}{\frac{x ^{2} - 2}{2}}$

$y = - \frac{2}{x ^{2} - 2}$

Find the particular solution to

$\frac{d y}{d x} = y + 1$

with $y (2) = 0$

Solution

(spoiler)

Separate the variables:

$\frac{1}{y + 1} d y = d x$

Integrate both sides. (On the left, you can use $u$ -substitution with $u = y + 1$ .)

$\int \frac{1}{y + 1} d y = \int 1 d x$

$ln (y + 1) = x + C$

Use the initial condition $(x, y) = (2, 0)$ to find $C$ :

$ln (0 + 1) = 2 + C$

$0 = 2 + C$

$C = - 2$

Substitute $C = - 2$ and solve for $y$ :

$ln (y + 1) = x - 2$

$e^{l n (y + 1)} = e^{x - 2}$

$y + 1 = e^{x - 2}$

$y = e^{x - 2} - 1$

If

$\frac{d y}{d t} = \frac{2 t + 1}{y}$

and $y = 2$ when $t = 0$ , find $y$ when $t = 1$ .

Solution

(spoiler)

Even though the variables are $y$ and $t$ (instead of $y$ and $x$ ), the separation steps are the same.

Separate the variables:

$y d y = (2 t + 1) d t$

Integrate both sides:

$\int y d y = \int (2 t + 1) d t$

$\frac{y ^{2}}{2} = t^{2} + t + C$

Use the initial condition $y = 2$ when $t = 0$ to find $C$ :

$\frac{2 ^{2}}{2} = 0^{2} + 0 + C$

$2 = C$

$\frac{y ^{2}}{2} = t^{2} + t + 2$

$y^{2} = 2 t^{2} + 2 t + 4$

$y = \pm 2 t^{2} + 2 t + 4$

Note: The initial condition $y (0) = 2$ tells you $y$ is positive at $t = 0$ , so you take the positive branch:

$y = 2 t^{2} + 2 t + 4$

Now evaluate at $t = 1$ :

$y = 2 (1)^{2} + 2 (1) + 4$

$= 2 + 2 + 4$

$= 22$

Lastly, here is a problem similar to a multi-part free-response question from the 2022 AB exam:

Consider the differential equation

$\frac{d y}{d x} = π cos (π x) y$

Let $y = f (x)$ be the particular solution to the differential equation with initial condition $f (1) = 4$ .

a) Write an equation for the line tangent to the solution curve at the point $(1, 4)$ and use it to approximate $f (0.9)$ .

b) Given $f^{''} (x) < 0$ for $- 1 \leq x \leq 1$ , is the approximation from part a) an overestimate or underestimate for $f (0.9)$ ?

c) Use separation of variables to find the particular solution to the differential equation with initial condition $f (1) = 4$ .

Solutions

a) Equation of tangent line at $(1, 4)$

(spoiler)

The equation of the tangent line to $y = f (x)$ at $(a, f (a))$ is

$y - f (a) = f^{'} (a) (x - a)$

Here, $(a, f (a)) = (1, 4)$ . To find $f^{'} (1)$ , evaluate the differential equation at $x = 1$ and $y = 4$ :

$\frac{d y}{d x}_{x = 1, y = 4} = π cos (π \cdot 1) 4$

$= π (- 1) (2)$

$= - 2 π$

So the tangent line is

$y - 4 = - 2 π (x - 1)$

To approximate $f (0.9)$ , plug $x = 0.9$ into the tangent line:

$y - 4 = - 2 π (0.9 - 1)$

$y = - 2 π (- 0.1) + 4$

$f (0.9) \approx 0.2 π + 4$

b) Underapproximate or overapproximate?

(spoiler)

Since $f^{''} (x) < 0$ on $- 1 \leq x \leq 1$ , the graph is concave down on that interval.

For a concave down function, the tangent line lies above the curve. That means the tangent-line (linear) approximation from part (a) is an overestimate of $f (0.9)$ .

c) Particular solution with $f (1) = 4$

(spoiler)

Separate the variables:

$\frac{1}{y} d y = π cos (π x) d x$

Integrate both sides:

$\int \frac{1}{y} d y = \int π cos (π x) d x$

Rewrite the left integrand as a power:

$\int y^{- 1/2} d y = sin (π x) + C$

$2 y^{1/2} = sin (π x) + C$

Use the initial condition $(1, 4)$ to find $C$ :

$24 = sin (π \cdot 1) + C$

$4 = 0 + C$

$C = 4$

Now solve for $y$ :

$2 y^{1/2} = sin (π x) + 4$

$y^{1/2} = \frac{1}{2} sin (π x) + 2$

$y = (\frac{1}{2} sin (π x) + 2)^{2}$