JEE Main 2022 · Mathematics · Differential Equations · medium

If y(x) satisfies x dy/dx+y=x^3 and y(1)=1, then the value of y(2) is

Single correct option

y(x)

satisfies

x\,\dfrac{dy}{dx}+y=x^3

and

y(1)=1

, then the value of

y(2)

Show answer and solution

Answer: (D) $\frac{19}{8}$

Write it as

\frac{dy}{dx}+\frac yx=x^2

, linear with integrating factor

e^{\int dx/x}=x

. Then

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

, so

xy=\frac{x^4}{4}+c

. From

y(1)=1

1=\frac14+c

, so

c=\frac34

. Thus

y=\frac{x^3}{4}+\frac{3}{4x}

and

y(2)=2+\frac38=\frac{19}{8}

Original question written in the JEE Main 2022 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake