Solve the separable differential equation $\frac{d y}{d x} = \frac{2 x}{1 - y ^{2}}$ and state the maximal interval on which the solution is defined, given the initial condition $y (0) = 0$ .

easy 2 marks Class 12 · Mathematics · Differential Equations

Answer

y = tanh (x^{2} + C)

with

C = 0

, so

y = tanh (x^{2})

; the solution is valid for all real

x

The correct answer is $y = tanh (x^{2} + C)$ with $C = 0$ , so $y = tanh (x^{2})$ ; the solution is valid for all real $x$ . — Since $tanh$ is defined for all real arguments, the solution exists for all $x \in R$ .

Step-by-step solution

Separate the variables: $(1 - y^{2}) d y = 2 x d x$ .
Integrate both sides: $\int (1 - y^{2}) d y = \int 2 x d x$ .
Left integral gives $y - \frac{y ^{3}}{3}$ ; right integral gives $x^{2} + C$ .
Rewrite as $y - \frac{y ^{3}}{3} = x^{2} + C$ .
Recognise the left side as the antiderivative of $\frac{d}{d y} (tanh^{- 1} y) = \frac{1}{1 - y ^{2}}$ , so solve for $y$ using hyperbolic tangent: $tanh^{- 1} y = x^{2} + C$ .
Apply the inverse hyperbolic tangent: $y = tanh (x^{2} + C)$ .
Use the initial condition $y (0) = 0$ to find $C$ : $0 = tanh C \Rightarrow C = 0$ .
Thus $y = tanh (x^{2})$ .
Since $tanh$ is defined for all real arguments, the solution exists for all $x \in R$ .

Common mistake: Treating

1 - y^{2}

as a constant while integrating.

Concept tested: Separable differential equation

Published 11 September 2026 · How this solution was prepared and checked

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Solve the separable differential equation dxdy​=1−y22x​ and state the maximal interval on which the solution is defined, given the initial condition y(0)=0.

Step-by-step solution

Solve the separable differential equation $\frac{d y}{d x} = \frac{2 x}{1 - y ^{2}}$ and state the maximal interval on which the solution is defined, given the initial condition $y (0) = 0$ .