Let $y=y(x)$ be the solution of the differential equation $e^{x} \sqrt{1-y^{2}} d x+\left(\frac{y}{x}\right) d y=0, y(1)=-1$ Then the value of $(y(3))^{2}$ is equal to :

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

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

Then the value of

(y (3))^{2}

is equal to :

Answer: (D)

Separating the variables,

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

Both sides integrate in closed form, since

\int x e^{x} d x = (x - 1) e^{x}

and

\int \frac{- y d y}{1 - y ^{2}} = 1 - y^{2}

, so the general solution is

(x - 1) e^{x} = 1 - y^{2} + C

x = 1

with

y = - 1

both

(x - 1) e^{x}

and

1 - y^{2}

vanish, which forces

C = 0

. Putting

x = 3

then gives

1 - y^{2} = 2 e^{3}

, and squaring gives

(y (3))^{2} = 1 - 4 e^{6}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option

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