If $x^{3} d y+x y d x=x^{2} d y+2 y d x ; y(2)=e$ and $x>1$, then $y(4)$ is equal to:

(C) $\frac{3}{2} \sqrt{\mathrm{e}}$

(A) $\frac{\sqrt{\mathrm{e}}}{2}$

(B) $\frac{1}{2}+\sqrt{\mathrm{e}}$

(D) $\frac{3}{2}+\sqrt{\mathrm{e}}$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Differential Equations

x^{3} d y + x y d x = x^{2} d y + 2 y d x; y (2) = e

and

x > 1

, then

y (4)

is equal to:

Answer: (C)

Grouping the terms,

x^{3} d y + x y d x = x^{2} d y + 2 y d x

becomes

(x^{3} - x^{2}) d y = (2 - x) y d x

, a separable equation

\frac{d y}{y} = \frac{2 - x}{x ^{2} ( x - 1 )} d x

. Partial fractions give

\frac{2 - x}{x ^{2} ( x - 1 )} = - \frac{1}{x} - \frac{2}{x ^{2}} + \frac{1}{x - 1}

, so integrating,

ln y = - ln x + \frac{2}{x} + ln (x - 1) + C

. The condition

y (2) = e

fixes

C = ln 2

. At

x = 4

ln y = - ln 4 + \frac{1}{2} + ln 3 + ln 2 = ln \frac{3}{2} + \frac{1}{2}

, so

y (4) = \frac{3}{2} e

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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