JEE Main 2024 · Mathematics · Differential Equations · medium

Given y(1)=1, the solution of dy/dx+y/x=x^2, evaluated at x=2, is

Single correct option

Given

y(1)=1

, the solution of

\dfrac{dy}{dx}+\dfrac{y}{x}=x^2

, evaluated at

x=2

, is

Show answer and solution

Answer: (A) $\dfrac{19}{8}$

The integrating factor is

e^{\int dx/x}=x

. So

\dfrac{d}{dx}(xy)=x^3

, giving

xy=\dfrac{x^4}{4}+C

. Using

y(1)=1

1=\tfrac14+C

C=\tfrac34

. At

x=2

2y=4+\tfrac34=\tfrac{19}{4}

, so

y=\tfrac{19}{8}

Original question written in the JEE Main 2024 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake