JEE Advanced 2017 Paper 1 · Mathematics · Differential Equations · hard

Let y=y(x) be the solution of xdy/dx+y=x^2y^2 with y(1)=1/2, defined on the largest interval containing x=1 on which it is a solution.

One or more options correct

Let

y=y(x)

be the solution of

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

with

y(1)=\dfrac12

, defined on the largest interval containing

x=1

on which it is a solution. Which of the following is (are) correct?

Show answer and solution

Answer: A, B, C

This is a Bernoulli equation. Divide by

y^2

and put

u=1/y

, so

u'=-y'/y^2

: the equation becomes

-xu'+u=x^2

, i.e.

u'-\frac{u}{x}=-x

. The integrating factor is

\frac1x

, giving

\left(\frac ux\right)'=-1

, so

u=x(C-x)

. Hence

y=\frac1{x(C-x)}

, and

y(1)=\frac12

gives

C=3

y=\frac{1}{x(3-x)}

(0,3)

. (A)

y(2)=\frac1{2\cdot1}=\frac12

; correct. (B)

x(3-x)

is maximal at

x=\frac32

with value

\frac94

, so the minimum of

y

\frac49

; correct. (C)

\int\frac{dx}{x(3-x)}=\frac13\ln\frac{x}{3-x}

, which evaluated from

1

2

\frac13(\ln2+\ln2)=\frac23\ln2

; correct. (D)

y\to\infty

x\to3^-

, so the solution exists only on

(0,3)

; incorrect. Answer: A, B, C.

Original question written in the JEE Advanced 2017 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