JEE Advanced 2016 Paper 1 · Mathematics · Differential Equations · medium

Let y(x) be the solution of (1+x^2dy/dx+2xy=2x with y(0)=0.

One or more options correct

Let

y(x)

be the solution of

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

with

y(0)=0

. Which of the following is/are correct?

Show answer and solution

Answer: A, C

The left side is

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

, so

(1+x^2)y=x^2+c

. With

y(0)=0

we get

c=0

and

y=\dfrac{x^2}{1+x^2}=1-\dfrac1{1+x^2}

. Then

y(1)=\tfrac12

, so (A) is true. Also

\int_0^1y\,dx=1-\tfrac{\pi}{4}\approx0.2146

, so (B) is false. Next

y'=\dfrac{2x}{(1+x^2)^2}

and

y''=\dfrac{2-6x^2}{(1+x^2)^3}

, which changes sign at

x=\pm1/\sqrt3

, so (C) is true. Finally

y'<0

for

x<0

and

y'>0

for

x>0

, so

x=0

is a local minimum, not a maximum, and (D) is false.

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