Consider the differential equation, $y^{2} d x+\left(x-\frac{1}{y}\right) d y=0$. If value of $y$ is 1 when $x=1$, then the value of x for which $\mathrm{y}=2$, is:

(C) $\frac{3}{2}-\frac{1}{\sqrt{\mathrm{e}}}$

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

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

(D) $\frac{5}{2}+\frac{1}{\sqrt{\mathrm{e}}}$

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Differential Equations

Consider the differential equation,

y^{2} d x + (x - \frac{1}{y}) d y = 0

. If value of

y

is 1 when

x = 1

, then the value of x for which

y = 2

, is:

Answer: (C)

Treat

x

as a function of

y

: dividing

y^{2} d x + (x - \frac{1}{y}) d y = 0

y^{2} d y

gives

\frac{d x}{d y} + \frac{x}{y ^{2}} = \frac{1}{y ^{3}}

, a linear equation with integrating factor

e^{\int y^{- 2} d y} = e^{- 1/ y}

. Then

\frac{d}{d y} (x e^{- 1/ y}) = \frac{e ^{- 1/ y}}{y ^{3}}

, and integrating the right side (substitute

u = - 1/ y

) yields

x e^{- 1/ y} = e^{- 1/ y} (1 + \frac{1}{y}) + C

. Applying

x = 1

y = 1

gives

e^{- 1} = 2 e^{- 1} + C

, so

C = - e^{- 1}

. At

y = 2

x e^{- 1/2} = \frac{3}{2} e^{- 1/2} - e^{- 1}

, hence

x = \frac{3}{2} - \frac{1}{e}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option