Separation of Variables

So far you have checked solutions that someone handed you. Now you will find them yourself. Separation of variables solves a big family of differential equations by moving everything with $y$ to one side, everything with $x$ to the other, and integrating. It is the main solving method in AP Calculus AB, and a separation-of-variables question appears on the free-response section most years.

Key ideas

Separable equations

A differential equation is separable if the right side can be written as a function of $x$ times a function of $y$:

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

For example, $\dfrac{dy}{dx} = 2xy$, $\dfrac{dy}{dx} = \dfrac{x^2}{y^2}$, and $\dfrac{dy}{dx} = e^{x + y} = e^x e^y$ are separable. But $\dfrac{dy}{dx} = x + y$ is not: you can’t split a sum into a product.

The method

1. Separate: get all the $y$‘s (with $dy$) on one side and all the $x$‘s (with $dx$) on the other, giving $\dfrac{1}{h(y)}\,dy = g(x)\,dx$.
2. Integrate both sides. Write one constant of integration, $+ C$, on the $x$ side.
3. Use the initial condition (if there is one) to find $C$. It is usually easiest to do this right after integrating, before solving for $y$.
4. Solve for $y$ if you can.
5. Check by substituting back into the differential equation and the initial condition.

You will often need u-substitution for the integrals, and $\displaystyle\int \frac{1}{y}\,dy = \ln|y| + C$ comes up all the time.

Handling ln and the constant

When you get $\ln|y| = (\text{something}) + C$, exponentiate both sides:

$$|y| = e^{(\text{something}) + C} = e^C \cdot e^{(\text{something})}$$

Since $e^C$ is just a positive constant, and the absolute value allows either sign, you can write $y = Ae^{(\text{something})}$ for a constant $A$. Then use the initial condition to find $A$.

The domain of a particular solution

A particular solution must be differentiable on an open interval that contains the initial $x$-value. If your formula has a vertical asymptote, a square root of a negative number, or a $\ln$ of a non-positive number somewhere, the solution only lives on the piece that contains the starting point. AP questions sometimes ask for this domain.

Worked examples

Example 1: A general solution

Find the general solution of $\dfrac{dy}{dx} = \dfrac{x^2}{y^2}$.

Solution. Multiply both sides by $y^2\,dx$ to separate:

$$y^2\,dy = x^2\,dx$$

Integrate both sides, with one constant:

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

Multiply by $3$. Since $3C$ is still just an arbitrary constant, call it $C$ again:

$$y^3 = x^3 + C \quad\Rightarrow\quad y = \sqrt[3]{x^3 + C}$$

Example 2: A particular solution

Solve $\dfrac{dy}{dx} = 2xy$ with $y(0) = 3$.

Solution. Separate and integrate:

$$\begin{aligned} \frac{1}{y}\,dy &= 2x\,dx \\ \int \frac{1}{y}\,dy &= \int 2x\,dx \\ \ln|y| &= x^2 + C \end{aligned}$$

Use the initial condition now: $\ln 3 = 0 + C$, so $C = \ln 3$. Then

$$|y| = e^{x^2 + \ln 3} = e^{\ln 3}\,e^{x^2} = 3e^{x^2}$$

Since $y(0) = 3$ is positive, $y = 3e^{x^2}$.

Check: $\dfrac{dy}{dx} = 3e^{x^2} \cdot 2x = 2x\big(3e^{x^2}\big) = 2xy$ ✓, and $y(0) = 3e^0 = 3$ ✓.

Example 3: A solution with a restricted domain

Solve $\dfrac{dy}{dx} = y^2$ with $y(0) = 1$, and state the domain of the solution.

Solution. Separate and integrate:

$$y^{-2}\,dy = dx \quad\Rightarrow\quad -\frac{1}{y} = x + C$$

At $(0, 1)$: $-1 = 0 + C$, so $C = -1$. Then

$$-\frac{1}{y} = x - 1 \quad\Rightarrow\quad y = \frac{-1}{x - 1} = \frac{1}{1 - x}$$

This formula has a vertical asymptote at $x = 1$. The solution must live on an open interval containing the starting value $x = 0$, so its domain is $x \lt 1$.

Only the branch through $(0, 1)$, where $x \lt 1$, is the particular solution. The dashed branch is not part of it.

Notice that nothing in the equation $\dfrac{dy}{dx} = y^2$ hints at trouble at $x = 1$. The solution “blows up” there anyway, because it grows faster and faster.

Example 4: Using u-substitution

Solve $\dfrac{dy}{dx} = \dfrac{x\cos(x^2)}{y}$ with $y(0) = 2$ (radians).

Solution. Separate:

$$y\,dy = x\cos(x^2)\,dx$$

For the right side, let $u = x^2$, so $du = 2x\,dx$ and $x\,dx = \tfrac{1}{2}\,du$:

$$\int x\cos(x^2)\,dx = \frac{1}{2}\int \cos u\,du = \frac{1}{2}\sin(x^2) + C$$

So

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

At $(0, 2)$: $\dfrac{4}{2} = \dfrac{1}{2}\sin 0 + C$, so $C = 2$. Multiply by $2$:

$$y^2 = \sin(x^2) + 4$$

Since $y(0) = 2$ is positive, take the positive square root:

$$y = \sqrt{\sin(x^2) + 4}$$

Because $\sin(x^2) \ge -1$, the expression under the root is always at least $3$, so this solution is defined for all real $x$.

Common mistakes

Forgetting $+ C$, or adding it at the end. The constant must appear as soon as you integrate. Writing $\ln|y| = x^2$ and then $y = e^{x^2} + C$ gives the wrong family. On AP free-response questions, a missing constant of integration usually costs the remaining points for that part, so build the habit now.

Separating a sum. $\dfrac{dy}{dx} = x + y$ can’t be split into $dy$ terms and $dx$ terms by multiplying or dividing. Only products and quotients separate. On the AP exam, an incorrect separation usually means no credit for the rest of the problem.

Exponentiating term by term. From $\ln|y| = x^2 + C$, the next step is $|y| = e^{x^2 + C} = e^C e^{x^2}$, not $y = e^{x^2} + e^C$.

Picking the wrong square root. When you solve $y^2 = \dots$, use the initial condition to choose the sign. If $y(0) = -3$, the solution is the negative square root.

Ignoring the domain. A particular solution can’t jump across an asymptote or a gap. State the interval that contains the initial $x$-value.

Not checking. Differentiating your answer and substituting it back takes a minute and catches most algebra slips.

Practice

1. (Warm-up) Which of these differential equations are separable?

• (a) $\dfrac{dy}{dx} = xy^2$
• (b) $\dfrac{dy}{dx} = x + y$
• (c) $\dfrac{dy}{dx} = e^{x + y}$
• (d) $\dfrac{dy}{dx} = \dfrac{y + 1}{x^2}$

Solution

(a) Separable: $\dfrac{1}{y^2}\,dy = x\,dx$.

(b) Not separable: a sum of $x$ and $y$ doesn’t factor into a function of $x$ times a function of $y$.

(c) Separable: $e^{x + y} = e^x e^y$, so $e^{-y}\,dy = e^x\,dx$.

(d) Separable: $\dfrac{1}{y + 1}\,dy = \dfrac{1}{x^2}\,dx$.

2. (Warm-up) Find the general solution of $\dfrac{dy}{dx} = \dfrac{x}{y}$.

Solution

$$y\,dy = x\,dx \quad\Rightarrow\quad \frac{y^2}{2} = \frac{x^2}{2} + C \quad\Rightarrow\quad y^2 = x^2 + C$$

(The constant $2C$ was renamed $C$.)

3. (Core) Solve $\dfrac{dy}{dx} = 3y$ with $y(0) = 4$.

Solution

$$\frac{1}{y}\,dy = 3\,dx \quad\Rightarrow\quad \ln|y| = 3x + C$$

At $(0, 4)$: $\ln 4 = C$. So $|y| = e^{3x + \ln 4} = 4e^{3x}$, and since $y(0) \gt 0$,

$$y = 4e^{3x}$$

Check: $\dfrac{dy}{dx} = 12e^{3x} = 3(4e^{3x}) = 3y$ ✓.

4. (Core) Solve $\dfrac{dy}{dx} = \dfrac{2x + 1}{2y}$ with $y(0) = -3$.

Solution

$$2y\,dy = (2x + 1)\,dx \quad\Rightarrow\quad y^2 = x^2 + x + C$$

At $(0, -3)$: $9 = 0 + 0 + C$, so $C = 9$ and $y^2 = x^2 + x + 9$.

Since $y(0) = -3$ is negative, take the negative root:

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

The quadratic $x^2 + x + 9$ has discriminant $1 - 36 \lt 0$, so it is always positive and the solution is defined for all real $x$.

5. (Core) Solve $\dfrac{dy}{dx} = xy^2$ with $y(0) = 1$, and state the domain of the solution.

Solution

$$y^{-2}\,dy = x\,dx \quad\Rightarrow\quad -\frac{1}{y} = \frac{x^2}{2} + C$$

At $(0, 1)$: $-1 = C$. So

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

The formula is undefined at $x = \pm\sqrt{2}$. The interval containing $x = 0$ is

$$-\sqrt{2} \lt x \lt \sqrt{2}$$

6. (Core) Solve $\dfrac{dy}{dx} = y\cos x$ with $y(0) = 2$ (radians). Then find $y\left(\dfrac{\pi}{2}\right)$, exactly and to 3 decimal places.

Solution

$$\frac{1}{y}\,dy = \cos x\,dx \quad\Rightarrow\quad \ln|y| = \sin x + C$$

At $(0, 2)$: $\ln 2 = 0 + C$. So $|y| = e^{\sin x + \ln 2} = 2e^{\sin x}$, and since $y(0) \gt 0$,

$$y = 2e^{\sin x}$$

Then $y\left(\dfrac{\pi}{2}\right) = 2e^{1} = 2e \approx 5.437$.

7. (Core) A student solves $\dfrac{dy}{dx} = 2xy$ with $y(0) = 5$ like this:

$$\ln|y| = x^2 \quad\Rightarrow\quad y = e^{x^2} + C \quad\Rightarrow\quad 5 = 1 + C \quad\Rightarrow\quad y = e^{x^2} + 4$$

Explain the error and find the correct solution.

Solution

The constant of integration has to appear when you integrate, not after solving for $y$. The correct line is $\ln|y| = x^2 + C$. (Check the student’s answer: $y = e^{x^2} + 4$ gives $\dfrac{dy}{dx} = 2xe^{x^2}$, but $2xy = 2xe^{x^2} + 8x$. They don’t match.)

Correct work: at $(0, 5)$, $\ln 5 = 0 + C$, so $|y| = e^{x^2 + \ln 5} = 5e^{x^2}$, and

$$y = 5e^{x^2}$$

8. (Challenge) Solve $\dfrac{dy}{dx} = \dfrac{y^2}{x}$ with $y(1) = -1$, and state the domain of the solution.

Solution

$$y^{-2}\,dy = \frac{1}{x}\,dx \quad\Rightarrow\quad -\frac{1}{y} = \ln|x| + C$$

At $(1, -1)$: $1 = \ln 1 + C = C$. So

$$-\frac{1}{y} = \ln|x| + 1 \quad\Rightarrow\quad y = -\frac{1}{1 + \ln|x|}$$

The starting value $x = 1$ is positive, and the equation is undefined at $x = 0$, so we need $x \gt 0$ (and $|x| = x$). We also need $1 + \ln x \ne 0$, that is, $x \ne e^{-1}$. The interval containing $x = 1$ is

$$x \gt \frac{1}{e}$$

so the solution is $y = -\dfrac{1}{1 + \ln x}$ for $x \gt \dfrac{1}{e}$.

Check: $\dfrac{dy}{dx} = \dfrac{1/x}{(1 + \ln x)^2} = \dfrac{1}{x} \cdot y^2$ ✓.

9. (Challenge) Solve $\dfrac{dy}{dx} = \dfrac{1 + y^2}{1 + x^2}$ with $y(0) = 1$. Write $y$ as a rational function of $x$, and state its domain. (Hint: use $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$.)

Solution

$$\frac{1}{1 + y^2}\,dy = \frac{1}{1 + x^2}\,dx \quad\Rightarrow\quad \arctan y = \arctan x + C$$

At $(0, 1)$: $\arctan 1 = \arctan 0 + C$, so $C = \dfrac{\pi}{4}$. Take the tangent of both sides:

$$y = \tan\left(\arctan x + \frac{\pi}{4}\right) = \frac{x + 1}{1 - x \cdot 1} = \frac{1 + x}{1 - x}$$

The formula is undefined at $x = 1$, and the interval containing $x = 0$ is $x \lt 1$.

Check: by the quotient rule, $\dfrac{dy}{dx} = \dfrac{(1 - x) + (1 + x)}{(1 - x)^2} = \dfrac{2}{(1 - x)^2}$. Also $1 + y^2 = \dfrac{(1 - x)^2 + (1 + x)^2}{(1 - x)^2} = \dfrac{2(1 + x^2)}{(1 - x)^2}$, so $\dfrac{1 + y^2}{1 + x^2} = \dfrac{2}{(1 - x)^2}$ ✓.