JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Differential Equations

Let slope of the tangent line to a curve at any point

P (x, y)

be given by

\frac{x y ^{2} + y}{x}

. If the curve intersects the line

x + 2 y = 4

x = - 2

, then the value of

y

, for which the point (

3, y

) lies on the curve, is :

Answer: (C)

Solution

The curve satisfies

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

Dividing by

y^{2}

and setting

v = \frac{1}{y}

, so that

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

, gives the linear equation

\frac{d v}{d x} + \frac{v}{x} = - 1

whose integrating factor is

x

, so

xv = - \frac{x ^{2}}{2} + C

, that is

\frac{x}{y} = C - \frac{x ^{2}}{2}

x = - 2

the line

x + 2 y = 4

gives

y = 3

, so

- \frac{2}{3} = C - 2

and

C = \frac{4}{3}

. At

x = 3

\frac{3}{y} = \frac{4}{3} - \frac{9}{2} = - \frac{19}{6} \Rightarrow y = - \frac{18}{19}

The point of the curve with abscissa

3

has

y = - \frac{18}{19}

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option