JEE Advanced 2020 Paper 1 · Mathematics · Differential Equations · medium

Let y=y(x), x>0, be the solution of xdy/dx+2y=x^2 with y(1)=1.

One or more options correct

Let

y=y(x)

x>0

, be the solution of

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

with

y(1)=1

. Which of the following is/are correct?

Show answer and solution

Answer: B, D

Multiplying by

x

gives

x^2y'+2xy=x^3

, i.e.

\dfrac{d}{dx}(x^2y)=x^3

. Hence

x^2y=\tfrac{x^4}{4}+C

, and

y(1)=1

gives

C=\tfrac34

, so

y=\dfrac{x^2}{4}+\dfrac{3}{4x^2}

. (A)

y(2)=1+\tfrac{3}{16}=\tfrac{19}{16}

, so (A) is false. (B)

y'=\tfrac x2-\tfrac{3}{2x^3}=0

gives

x^4=3

, and

y=\tfrac{\sqrt3}{4}+\tfrac{3}{4\sqrt3}=\tfrac{\sqrt3}{2}

, which is the minimum since

y\to\infty

at both ends; true. (C)

\int_1^2y\,dx=\left[\tfrac{x^3}{12}-\tfrac{3}{4x}\right]_1^2=\tfrac{23}{24}\ne\tfrac56

, false. (D)

y/x^2=\tfrac14+\tfrac{3}{4x^4}\to\tfrac14

, true.

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