Solve the separable differential equation dy/dx=x^2/y and state the domain of the solution.

y^2=2/3x^3+C (so y=±√(2/3x^3+C)), domain: any interval where 2/3x^3+C≥0

Solve the separable differential equation $\frac{d y}{d x} = \frac{x ^{2}}{y}$ and state the domain of the solution.

easy 2 marks Class 12 · Mathematics · Differential Equations

Answer

y^{2} = \frac{2}{3} x^{3} + C (so y = \pm \frac{2}{3} x^{3} + C), domain: any interval where \frac{2}{3} x^{3} + C \geq 0

The correct answer is $y^{2} = \frac{2}{3} x^{3} + C (so y = \pm \frac{2}{3} x^{3} + C), domain: any interval where \frac{2}{3} x^{3} + C \geq 0$ — Domain requires the expression under the root to be non‑negative, i.e., $\frac{2}{3} x^{3} + C \geq 0$ , so the solution is valid on any interval where this holds.

Step-by-step solution

Separate variables: $y d y = x^{2} d x$ .
Integrate both sides: $\int y d y = \int x^{2} d x$ .
Obtain $\frac{y ^{2}}{2} = \frac{x ^{3}}{3} + C_{1}$ , multiply by 2: $y^{2} = \frac{2}{3} x^{3} + C$ .
Take square root: $y = \pm \frac{2}{3} x^{3} + C$ .
Domain requires the expression under the root to be non‑negative, i.e., $\frac{2}{3} x^{3} + C \geq 0$ , so the solution is valid on any interval where this holds.

Common mistake: Multiplying by

y

after integrating instead of before, which loses the constant of integration.

Concept tested: Separable differential equation

Published 5 September 2026 · How this solution was prepared and checked

← All Differential Equations questions

Solve the separable differential equation dxdy​=yx2​ and state the domain of the solution.

Step-by-step solution

Solve the separable differential equation $\frac{d y}{d x} = \frac{x ^{2}}{y}$ and state the domain of the solution.