Separable Equations
Differential Equations · Section 2.2 · generated practice set
Separate, integrate, apply the initial condition early, pick the right branch, and state the interval of validity.
Practice this set → Fresh numbers on every attempt. No account needed.
Try one now
Solve $$\frac{dy}{dx} = \frac{3x}{y}, \qquad y(2) = 1,$$ then evaluate the solution at $x = 3$. Write the implicit solution as $y^2 = 3x^2 + K$.
Solution
- Separate: $y\,dy = 3x\,dx$.
- Integrate both sides: $\frac{y^2}{2} = \frac{3x^2}{2} + c$, i.e. $y^2 = 3x^2 + K$.
- Apply $y(2) = 1$: $1 = 12 + K \Rightarrow K = -11$.
- Since $y(2) = 1 > 0$, take the positive root: $y = \sqrt{3x^2 - 11}$.
- At $x = 3$: $y = \sqrt{16} \approx 4$.
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Method
Separable means you can write $\dfrac{dy}{dx} = \dfrac{M(x)}{N(y)}$, i.e. split the right side into an $x$-only factor and a $y$-only factor.
- Separate. $N(y)\,dy = M(x)\,dx$.
- Integrate both sides — one constant $C$, conventionally on the $x$ side.
- Apply the IC immediately, while the solution is still implicit. It is much easier than solving for $y$ first.
- Solve for $y$ if the problem asks for an explicit solution, and choose the branch (the sign of the square root) that matches the initial condition.
- Interval of validity: the largest interval containing $x_0$ where the explicit solution is defined and differentiable — watch for square roots going negative and denominators hitting zero.
Separable vs. linear
$\frac{dy}{dx} = \frac{ay}{x}$ is both. $\frac{dy}{dx} = \frac{ax}{y}$ is separable only. $y' + p(t)y = g(t)$ with $g \neq 0$ is linear only. When an equation is both, separate — it is fewer steps.
Definitions and theorems
Worked example
Solve $\dfrac{dy}{dx} = \dfrac{3x^2}{2y}$ with $y(0) = -2$, and give the interval of validity.
- Separate: $2y\,dy = 3x^2\,dx$.
- Integrate: $y^2 = x^3 + C$.
- Apply $y(0) = -2$: $4 = 0 + C \Rightarrow C = 4$, so $y^2 = x^3 + 4$.
- Solve for $y$, choosing the negative branch because $y(0) = -2 < 0$: $y = -\sqrt{x^3 + 4}$.
- Need $x^3 + 4 > 0$, i.e. $x > -4^{1/3} \approx -1.587$. Interval of validity: $\left(-\sqrt[3]{4},\ \infty\right)$.
Common mistakes
- Picking the wrong branch. $y = \pm\sqrt{\cdot}$ — the initial condition decides the sign, and it is not always positive.
- Strict inequality at the endpoint. The interval excludes the point where the radicand is 0, because $y'$ blows up there.
- Constants on both sides. One $C$ is enough; two just get merged.
- Dividing by $y$ without noting $y = 0$. State it, then dismiss it using the initial condition.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.