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 (a separation-of-variables question appears on the AP free-response section most years), and it is the standard method in IB HL too.

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. Some questions (including on the AP exam) 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 (and in most marking schemes), 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 a test, 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 $\dfrac{1 + y^2}{1 + x^2} = \dfrac{2}{(1 - x)^2}$ ✓.