Solve the separable differential equation dy/dx=2x/(1+y^2) with the initial condition y(0)=1. State the interval on which the solution is valid.

y=tan(x^2+π/4), valid for x∈(-π/4)^(1/2),(π/4)^(1/2)

Solve the separable differential equation $\frac{d y}{d x} = \frac{2 x}{1 + y ^{2}}$ with the initial condition $y (0) = 1$ . State the interval on which the solution is valid.

medium 2 marks Class 12 · Mathematics · Differential Equations

Answer

y = tan (x^{2} + \frac{π}{4})

, valid for

x \in (- \frac{π}{4})^{1/2}, (\frac{π}{4})^{1/2}

The correct answer is $y = tan (x^{2} + \frac{π}{4})$ , valid for $x \in (- \frac{π}{4})^{1/2}, (\frac{π}{4})^{1/2}$ — The tangent function is defined while its argument stays away from odd multiples of $\frac{π}{2}$ . The nearest singularities occur at $x^{2} + \frac{π}{4} = \pm \frac{π}{2}$ . Solving gives $∣ x ∣ < \frac{π}{4}$ .

Step-by-step solution

Separate 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 $arctan y = x^{2} + C$ because $\frac{d}{d y} arctan y = \frac{1}{1 + y ^{2}}$ .
Apply the initial condition $y (0) = 1$ : $arctan 1 = C \Rightarrow C = \frac{π}{4}$ .
Thus $arctan y = x^{2} + \frac{π}{4}$ .
Take tangent: $y = tan (x^{2} + \frac{π}{4})$ .
The tangent function is defined while its argument stays away from odd multiples of $\frac{π}{2}$ . The nearest singularities occur at $x^{2} + \frac{π}{4} = \pm \frac{π}{2}$ . Solving gives $∣ x ∣ < \frac{π}{4}$ .

Common mistake: Integrating

1/ (1 + y^{2})

y

instead of

arctan y

Concept tested: Separable ODE and arctan integration

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

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Step-by-step solution

Solve the separable differential equation $\frac{d y}{d x} = \frac{2 x}{1 + y ^{2}}$ with the initial condition $y (0) = 1$ . State the interval on which the solution is valid.