Let $y=y(x)$ be the solution curve of the differential equation $x\left(x^{2}+e^{x}\right) d y+\left(e^{x}(x-2) y-x^{3}\right) d x=0, x>0$, passing through the point $(1,0)$. Then $\mathrm{y}(2)$ is equal to :

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

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution curve of the differential equation

x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0

, passing through the point

(1, 0)

. Then

y (2)

is equal to :

Answer: (D)

In linear form the equation reads

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

. The integrating factor is

\frac{x ^{2} + e ^{x}}{x ^{2}} = 1 + \frac{e ^{x}}{x ^{2}}

, since its logarithmic derivative equals the coefficient of

y

. Multiplying through collapses the left side to

\frac{d}{d x} [y (1 + \frac{e ^{x}}{x ^{2}})] = 1

, so

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

. The point

(1, 0)

gives

C = - 1

, hence

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

. At

x = 2

this gives

y (2) = \frac{4}{4 + e ^{2}}

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option