JEE Advanced 2018 Paper 1 · Mathematics · Differential Equations · medium

Let y=f(x) be the solution of dy/dx=y/x+(y^2/(x^2 on (1,e) with f(1)=1. Find the value of f(√ e)+f'(√ e).

Numerical answer

Let

y=f(x)

be the solution of

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

(1,e)

with

f(1)=1

. Find the value of

f(\sqrt e)+f'(\sqrt e)

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 9.30

The equation is homogeneous: put

y=vx

, so

\frac{dy}{dx}=v+x\frac{dv}{dx}

and the equation becomes

x\frac{dv}{dx}=v^2

. Separating,

-\frac1v=\ln x+c

. With

v(1)=1

we get

c=-1

, so

v=\frac1{1-\ln x}

and

f(x)=\frac{x}{1-\ln x}

, valid for

x<e

. At

x=\sqrt e

1-\ln x=\frac12

, so

f(\sqrt e)=2\sqrt e\approx3.2974

. Then

\frac yx=2

and

f'(\sqrt e)=\frac yx+\left(\frac yx\right)^2=2+4=6

. The required sum is

2\sqrt e+6\approx9.2974

, i.e.

9.30

Original question written in the JEE Advanced 2018 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake