JEE Advanced 2025 Paper 1 · Mathematics · Differential Equations · medium

Let y=y(x) satisfy (1+x^2dy/dx+2xy=3x^2 with y(0)=1. Find ∫0^1y(x) dx.

Numerical answer

Let

y=y(x)

satisfy

(1+x^{2})\dfrac{dy}{dx}+2xy=3x^{2}

with

y(0)=1

. Find

\displaystyle\int_{0}^{1}y(x)\,dx

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 0.94

The left side is the derivative of

(1+x^2)y

, because

\frac{d}{dx}\big[(1+x^2)y\big]=(1+x^2)y'+2xy

. Hence

(1+x^2)y=x^3+C

. The condition

y(0)=1

gives

C=1

, so

y=\frac{x^3+1}{1+x^2}

. Then

\int_0^1 y\,dx=\int_0^1\frac{x^3}{1+x^2}dx+\int_0^1\frac{dx}{1+x^2}

. For the first,

\frac{x^3}{1+x^2}=x-\frac{x}{1+x^2}

, giving

\left[\frac{x^2}{2}-\frac12\ln(1+x^2)\right]_0^1=\frac12-\frac{\ln2}{2}

. The second is

\frac{\pi}{4}

. The total is

\frac12-\frac{\ln 2}{2}+\frac{\pi}{4}=0.5-0.34657+0.78540=0.93883\approx0.94

. Numerical quadrature of

y

gives

0.938825

, matching.

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