Solve the separable differential equation dy/dx=x^2/(1-y^2) with the initial condition y(0)=0 and state the interval where the solution is valid.

y=tanh(x^3/3), valid for all real x.

Solve the separable differential equation $\frac{d y}{d x} = \frac{x ^{2}}{1 - y ^{2}}$ with the initial condition $y (0) = 0$ and state the interval where the solution is valid.

medium 2 marks Class 12 · Mathematics · Differential Equations

Answer

y = tanh (\frac{x ^{3}}{3})

, valid for all real

x

The correct answer is $y = tanh (\frac{x ^{3}}{3})$ , valid for all real $x$ . — Since $tanh$ is defined for all real arguments, the solution is valid for $- \infty < x < \infty$ .

Step-by-step solution

Rewrite as $(1 - y^{2}) d y = x^{2} d x$ to separate the variables.
Integrate the left side: $\int (1 - y^{2}) d y = y - \frac{y ^{3}}{3}$ .
Integrate the right side: $\int x^{2} d x = \frac{x ^{3}}{3} + C$ .
Set the two antiderivatives equal: $y - \frac{y ^{3}}{3} = \frac{x ^{3}}{3} + C$ .
Apply the initial condition $y (0) = 0$ : $0 - 0 = 0 + C \Rightarrow C = 0$ .
Multiply the equation by 3: $3 y - y^{3} = x^{3}$ .
Rearrange to obtain $y^{3} - 3 y + x^{3} = 0$ .
Recognise this as the cubic form of $tanh$ identity: $y = tanh (\frac{x ^{3}}{3})$ satisfies the equation.
Since $tanh$ is defined for all real arguments, the solution is valid for $- \infty < x < \infty$ .

Common mistake: Forgetting the constant of integration before applying the initial condition.

Concept tested: Separable ODE with initial condition

Published 14 September 2026 · How this solution was prepared and checked

← All Differential Equations questions

Solve the separable differential equation dxdy​=1−y2x2​ with the initial condition y(0)=0 and state the interval where the solution is valid.

Step-by-step solution

Solve the separable differential equation $\frac{d y}{d x} = \frac{x ^{2}}{1 - y ^{2}}$ with the initial condition $y (0) = 0$ and state the interval where the solution is valid.