7.3 Separation of variables

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Separation of variables

7 min read

Font

Discuss

Feedback

What you’ll learn

Separation of variables: Group variables and integrate to find a general solution.
Particular solutions: Use an initial condition to solve for the constant $C$ .

So far, you’ve seen differential equations like $\frac{d y}{d x} = 2 x$ , which can be solved by integrating both sides with respect to $x$ .

When the derivative 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

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

and rearranged into

$\frac{1}{g ( y )} d y = f (x) d x$

If the equation can’t be rewritten as a product of a function of $x$ and a function of $y$ , then separation of variables won’t apply.

General vs. particular solutions

To analyze a differential equation, check if you are given an initial starting point.

General solution: No initial condition is given. Your goal is to isolate $y$ , and the final answer will always contain $+ C$ .
Particular Solution: An initial condition $(x, y)$ is given. Use this point to solve for the exact value of $C$ .

AP tip:

Always substitute your initial condition to find $C$ immediately after integrating. Finding $C$ early keeps the algebra cleaner.

Case 1: General solutions (keeping $+ C$ )

Find the general solution to each differential equation:

1a) $\frac{d y}{d x} = 2 x y$

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

Solutions

1a) $\frac{d y}{d x} = 2 x y$

(spoiler)

1. Separate the variables:

Multiply by $d x$ and divide by $y$ :

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

2. Integrate both sides:

$\int \frac{1}{y} d y ln ∣ y ∣ = \int 2 x d x = x^{2} + C$

Because the integral is indefinite, add the constant of integration $C$ .

Conventionally, it’s added to the side with the independent variable (in this case, $x$ ), or the right hand-side.

3. Isolate $y$ :

Exponentiate both sides to undo the natural log:

$∣ y ∣ = e^{x^{2} + C}$

Using the product rule of exponents ( $e^{A + B} = e^{A} \cdot e^{B}$ ), rewrite the right side:

$∣ y ∣ = e^{C} \cdot e^{x^{2}}$

Since $e^{C}$ is just a constant, we can absorb the absolute value bars and replace $e^{C}$ with a new arbitrary constant, $C$ :

$y = C e^{x^{2}}$

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

(spoiler)

1. Separate the variables:

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

2. Integrate both sides:

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

3. Isolate $y$ :

Multiply by $- 1$ and take the reciprocal of both sides. The constant stays locked in the denominator:

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

Case 2: Particular solutions (solving for $C$ )

Find the particular solution to each differential equation given the initial condition.

2a) $\frac{d y}{d x} = \frac{x}{y}$ with initial condition $y (0) = - 3$

2b) $\frac{d y}{d x} = 1 - y$ with initial condition $(2, 0)$ .

Solutions

2a) $\frac{d y}{d x} = \frac{x}{y}$ with initial condition $y (0) = - 3$

(spoiler)

1. Separate the variables:

$y d y = x d x$

2. Integrate both sides:

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

3. Plug in the initial condition $y (0) = - 3$ to find $C$ :

$\frac{1}{2} (- 3)^{2} \frac{9}{2} = \frac{1}{2} (0)^{2} + C = C$

4. Substitute $C$ back into the equation and solve for $y$ :

$\frac{1}{2} y^{2} y^{2} y = \frac{1}{2} x^{2} + \frac{9}{2} = x^{2} + 9 = \pm x^{2} + 9$

Note: The initial condition states that $y = - 3$ (a negative value) when $x = 0$ . Therefore, select the negative root branch:

$y = - x^{2} + 9$

2b) $\frac{d y}{d x} = 1 - y$ with initial condition $(2, 0)$ .

(spoiler)

1. Separate the variables:

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

2. Integrate both sides:

On the left, $u$ -substitution with $u = 1 - y$ is used.

$\int \frac{1}{1 - y} d y - ln ∣1 - y ∣ = \int 1 d x = x + C$

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

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

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

$- ln ∣1 - y ∣ ln ∣1 - y ∣ e^{l n ∣1 - y ∣} 1 - y = x - 2 = 2 - x = e^{2 - x} = e^{2 - x}$

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

Method: Pre-separation algebra

You may have to factor or apply exponent rules before the variables can be separated.

Find the general solution to each differential equation:

3a) $\frac{d y}{d x} = x + x y^{2}$

3b) $\frac{d y}{d x} = e^{x - y}$

Solutions

3a) $\frac{d y}{d x} = x + x y^{2}$

(spoiler)

1. Factor the right-hand side:

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

2. Separate the variables:

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

3. Integrate both sides:

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

4. Isolate $y$ :

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

3b) $\frac{d y}{d x} = e^{x - y}$

(spoiler)

1. Rewrite the expression

By applying exponent rules,

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

2. Separate the variables:

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

3. Integrate both sides:

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

4. Isolate $y$ :

Take the natural log of both sides.

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

Finding a value from a particular solution

If

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

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

Solution

(spoiler)

In this case, $t$ is the independent variable (instead of $x$ ).

1. Separate the variables:

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

2. Integrate both sides:

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

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

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

*4. Solve for $y$ :

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

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

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

Now evaluate at $t = 1$ :

$y = 2 (1)^{2} + 2 (1) + 4 = 2 + 2 + 4 = 22$

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

Separation of variables: Group variables and integrate to find a general solution.
Particular solutions: Use an initial condition to solve for the constant $C$ .

So far, you’ve seen differential equations like $\frac{d y}{d x} = 2 x$ , which can be solved by integrating both sides with respect to $x$ .

When the derivative 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

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

and rearranged into

$\frac{1}{g ( y )} d y = f (x) d x$

If the equation can’t be rewritten as a product of a function of $x$ and a function of $y$ , then separation of variables won’t apply.

General vs. particular solutions

To analyze a differential equation, check if you are given an initial starting point.

General solution: No initial condition is given. Your goal is to isolate $y$ , and the final answer will always contain $+ C$ .
Particular Solution: An initial condition $(x, y)$ is given. Use this point to solve for the exact value of $C$ .

AP tip:

Always substitute your initial condition to find $C$ immediately after integrating. Finding $C$ early keeps the algebra cleaner.

Case 1: General solutions (keeping $+ C$ )

Find the general solution to each differential equation:

1a) $\frac{d y}{d x} = 2 x y$

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

Solutions

1a) $\frac{d y}{d x} = 2 x y$

(spoiler)

1. Separate the variables:

Multiply by $d x$ and divide by $y$ :

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

2. Integrate both sides:

$\int \frac{1}{y} d y ln ∣ y ∣ = \int 2 x d x = x^{2} + C$

Because the integral is indefinite, add the constant of integration $C$ .

Conventionally, it’s added to the side with the independent variable (in this case, $x$ ), or the right hand-side.

3. Isolate $y$ :

Exponentiate both sides to undo the natural log:

$∣ y ∣ = e^{x^{2} + C}$

Using the product rule of exponents ( $e^{A + B} = e^{A} \cdot e^{B}$ ), rewrite the right side:

$∣ y ∣ = e^{C} \cdot e^{x^{2}}$

Since $e^{C}$ is just a constant, we can absorb the absolute value bars and replace $e^{C}$ with a new arbitrary constant, $C$ :

$y = C e^{x^{2}}$

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

(spoiler)

1. Separate the variables:

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

2. Integrate both sides:

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

3. Isolate $y$ :

Multiply by $- 1$ and take the reciprocal of both sides. The constant stays locked in the denominator:

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

Case 2: Particular solutions (solving for $C$ )

Find the particular solution to each differential equation given the initial condition.

2a) $\frac{d y}{d x} = \frac{x}{y}$ with initial condition $y (0) = - 3$

2b) $\frac{d y}{d x} = 1 - y$ with initial condition $(2, 0)$ .

Solutions

2a) $\frac{d y}{d x} = \frac{x}{y}$ with initial condition $y (0) = - 3$

(spoiler)

1. Separate the variables:

$y d y = x d x$

2. Integrate both sides:

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

3. Plug in the initial condition $y (0) = - 3$ to find $C$ :

$\frac{1}{2} (- 3)^{2} \frac{9}{2} = \frac{1}{2} (0)^{2} + C = C$

4. Substitute $C$ back into the equation and solve for $y$ :

$\frac{1}{2} y^{2} y^{2} y = \frac{1}{2} x^{2} + \frac{9}{2} = x^{2} + 9 = \pm x^{2} + 9$

Note: The initial condition states that $y = - 3$ (a negative value) when $x = 0$ . Therefore, select the negative root branch:

$y = - x^{2} + 9$

2b) $\frac{d y}{d x} = 1 - y$ with initial condition $(2, 0)$ .

(spoiler)

1. Separate the variables:

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

2. Integrate both sides:

On the left, $u$ -substitution with $u = 1 - y$ is used.

$\int \frac{1}{1 - y} d y - ln ∣1 - y ∣ = \int 1 d x = x + C$

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

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

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

$- ln ∣1 - y ∣ ln ∣1 - y ∣ e^{l n ∣1 - y ∣} 1 - y = x - 2 = 2 - x = e^{2 - x} = e^{2 - x}$

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

Method: Pre-separation algebra

You may have to factor or apply exponent rules before the variables can be separated.

Find the general solution to each differential equation:

3a) $\frac{d y}{d x} = x + x y^{2}$

3b) $\frac{d y}{d x} = e^{x - y}$

Solutions

3a) $\frac{d y}{d x} = x + x y^{2}$

(spoiler)

1. Factor the right-hand side:

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

2. Separate the variables:

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

3. Integrate both sides:

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

4. Isolate $y$ :

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

3b) $\frac{d y}{d x} = e^{x - y}$

(spoiler)

1. Rewrite the expression

By applying exponent rules,

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

2. Separate the variables:

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

3. Integrate both sides:

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

4. Isolate $y$ :

Take the natural log of both sides.

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

Finding a value from a particular solution

If

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

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

Solution

(spoiler)

In this case, $t$ is the independent variable (instead of $x$ ).

1. Separate the variables:

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

2. Integrate both sides:

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

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

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

*4. Solve for $y$ :

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

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

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

Now evaluate at $t = 1$ :

$y = 2 (1)^{2} + 2 (1) + 4 = 2 + 2 + 4 = 22$

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.