Separation of variables | Differential equations

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 are easy to solve by directly integrating. But if the right-hand side depends on both $x$ and $y$ , we use a technique called separation of variables. This method works by rearranging the equation so that all terms involving $y$ are on one side and all terms involving $x$ are on the other, allowing us to integrate with respect to a single variable. In order to separate the variables, the differential equation must be written in the form

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

If the equation can’t be written this way, then separation of variables won’t work.

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

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

Generally, the easiest way to separate the variables is to multiply both sides by $d x$ and then rearrange so that the portion with $y$ goes to the other side with $d y$ - in this case, by multiplying 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, then just solve the equation for $y$ to obtain the general solution.

In this case, there is an initial condition $(x, y) = (0, ln (5))$ so plug it in to find $C$ :

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

$5 = 1 + C$

$C = 4$

Then

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

Solving for $y$ ,

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

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

Examples

1. Find the general solution to

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

Solution

(spoiler)

First, factor so that the equation is written as a product:

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

Next, separate the variables by multiplying both sides by $d x$ and then dividing 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 $y$ :

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

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

Using the product law of exponents,

$y = e^{(x + x^{3} /3)} \cdot e^{C}$

Since $e^{C}$ is also a constant, we can change it to $C$ , which is any real number rather than a specific value. So the general solution to the differential equation is

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

2. 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:

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

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

Plug in initial condition $(x, y) = (0, 1)$ :

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

$C = - 1$

Lastly, solve for $y$ to find the particular solution after substituting $C = - 1$ :

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

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

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

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

3. 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 (use $u$ -substitution with $u = y + 1$ for the left-hand side):

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

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

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

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

$0 = 2 + C$

$C = - 2$

Solve for $y$ to find the particular solution after substituting $C = - 2$ :

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

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

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

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

4. If

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

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

Solution

(spoiler)

Although the variables this time are $y$ and $t$ , separation of variables works the same way:

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

Integrate both sides:

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

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

Solve for $C$ given initial condition $y = 2$ and $t = 0$ :

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

$2 = 0 + C$

$C = 2$

Then

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

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

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

Note: The initial condition $y (0) = 2$ tells us that $y$ is positive; therefore, the positive branch $y = 2 t^{2} + 2 t + 4$ is the only solution.

Solving for $y$ when $t = 1$ ,

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

$= 2 + 2 + 4$

$= 22$

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 a tangent line at a point $(a, f (a))$ is

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

$(a, f (a))$ was given as $(1, 4)$ . To find $f^{'} (a)$ , or the slope of the tangent line, evaluate the differential equation at the point:

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

$= - 2 π$

Then the equation of the tangent line is

$y - 4 = - 2 π (x - 1)$

The linear approximation of $f (0.9)$ can be found by plugging in $x = 0.9$ into the tangent line equation:

$y - 4 = - 2 π (0.9 - 1)$

$y = - 2 π (- 0.1) + 4$

$f (0.9) \approx 0.2 π + 4$

b) Underapproximate or overapproximate?

(spoiler)

Since $f^{''} (x) < 0$ , the graph of the function is concave down on that interval.

When a function is concave down, the tangent line lies above the curve, so the linear approximation from the tangent line is an overestimate of the actual value.

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:

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

$\int y^{- 1/2} d y = sin (π x) + C$

$2 y^{1/2} = sin (π x) + C$

Plug in initial condition $(1, 4)$ to solve for $C$ :

$24 = sin (π \cdot 1) + C$

$4 = 0 + C$

$C = 4$

Lastly, solve for $y$ :

$2 y^{1/2} = sin (π x) + 4$

$y^{1/2} = \frac{1}{2} sin (π x) + 2$

$y = (\frac{1}{2} sin (π x) + 2)^{2}$

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

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

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

If the equation can’t be written this way, then separation of variables won’t work.

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

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

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

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, then just solve the equation for $y$ to obtain the general solution.

In this case, there is an initial condition $(x, y) = (0, ln (5))$ so plug it in to find $C$ :

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

$5 = 1 + C$

$C = 4$

Then

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

Solving for $y$ ,

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

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

Examples

1. Find the general solution to

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

Solution

(spoiler)

First, factor so that the equation is written as a product:

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

Next, separate the variables by multiplying both sides by $d x$ and then dividing 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 $y$ :

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

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

Using the product law of exponents,

$y = e^{(x + x^{3} /3)} \cdot e^{C}$

Since $e^{C}$ is also a constant, we can change it to $C$ , which is any real number rather than a specific value. So the general solution to the differential equation is

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

2. 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:

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

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

Plug in initial condition $(x, y) = (0, 1)$ :

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

$C = - 1$

Lastly, solve for $y$ to find the particular solution after substituting $C = - 1$ :

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

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

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

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

3. 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 (use $u$ -substitution with $u = y + 1$ for the left-hand side):

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

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

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

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

$0 = 2 + C$

$C = - 2$

Solve for $y$ to find the particular solution after substituting $C = - 2$ :

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

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

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

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

4. If

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

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

Solution

(spoiler)

Although the variables this time are $y$ and $t$ , separation of variables works the same way:

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

Integrate both sides:

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

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

Solve for $C$ given initial condition $y = 2$ and $t = 0$ :

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

$2 = 0 + C$

$C = 2$

Then

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

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

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

Note: The initial condition $y (0) = 2$ tells us that $y$ is positive; therefore, the positive branch $y = 2 t^{2} + 2 t + 4$ is the only solution.

Solving for $y$ when $t = 1$ ,

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

$= 2 + 2 + 4$

$= 22$

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 a tangent line at a point $(a, f (a))$ is

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

$(a, f (a))$ was given as $(1, 4)$ . To find $f^{'} (a)$ , or the slope of the tangent line, evaluate the differential equation at the point:

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

$= - 2 π$

Then the equation of the tangent line is

$y - 4 = - 2 π (x - 1)$

The linear approximation of $f (0.9)$ can be found by plugging in $x = 0.9$ into the tangent line equation:

$y - 4 = - 2 π (0.9 - 1)$

$y = - 2 π (- 0.1) + 4$

$f (0.9) \approx 0.2 π + 4$

b) Underapproximate or overapproximate?

(spoiler)

Since $f^{''} (x) < 0$ , the graph of the function is concave down on that interval.

When a function is concave down, the tangent line lies above the curve, so the linear approximation from the tangent line is an overestimate of the actual value.

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:

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

$\int y^{- 1/2} d y = sin (π x) + C$

$2 y^{1/2} = sin (π x) + C$

Plug in initial condition $(1, 4)$ to solve for $C$ :

$24 = sin (π \cdot 1) + C$

$4 = 0 + C$

$C = 4$

Lastly, solve for $y$ :

$2 y^{1/2} = sin (π x) + 4$

$y^{1/2} = \frac{1}{2} sin (π x) + 2$

$y = (\frac{1}{2} sin (π x) + 2)^{2}$