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 :

Correct answer: D. 4 4+e^2 aligned & x(x^2+e^x) d y+(e^x(x-2) y-x^3) d x=0 & x(x^2+e^x) (d y)/(d x)+e^x(x-2) y=x^3 & (d y)/(d x)+(e^x(x-2))/(x(x^2+e^x)) y=(x^2)/(x^2+e^x) & I.F. =e^ ∫ (e^x(x-2))/(x(x^2+e^x)) d x =e^ ∫ (e^x((1)/(x^2)-(2)/(x^2)) d x)/((1+(e^x)/(x^2))) d x aligned Let 1+(e^x)/(x^2)=t (x^2 e^x-e^x 2 x)/(x^4) d x=d t I.F. e^ (1+ e^2 x^2 ) =1+ e^ x x^2 Now y(1+(e^x)/(x^2))=∫ (x^2)/(x^2+e^x) · (x^2+e^x)/(x^2) d x+C y(1+(e^x)/(x^2))=x+C Passing through (1,0) aligned & C=-1 & y=(x-1)/(1+(e^x)/(x^2)) & y(2)=(1)/(1+(e^2)/(4))=(4)/(4+e^2) aligned

JEE Main 20257 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Differential Equations

Let y=y(x) be the solution curve of the differential equation...

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 \gt 0

passing through the point

(1,0)

. Then

y(2)

is equal to :

Options

A $\frac{4}{4-\mathrm{e}^2}$
B $\frac{2}{2+\mathrm{e}^2}$
C $\frac{2}{2-\mathrm{e}^2}$
D $\frac{4}{4+\mathrm{e}^2}$

Correct answer

D. $\frac{4}{4+\mathrm{e}^2}$

Step-by-step solution

\begin{aligned} & x\left(x^2+e^x\right) d y+\left(e^x(x-2) y-x^3\right) d x=0 \\ & x\left(x^2+e^x\right) \frac{d y}{d x}+e^x(x-2) y=x^3 \\ & \frac{d y}{d x}+\frac{e^x(x-2)}{x\left(x^2+e^x\right)} y=\frac{x^2}{x^2+e^x} \\ & \text{ I.F. }=e^{\int \frac{e^x(x-2)}{x\left(x^2+e^x\right)} d x}=e^{\int \frac{e^x\left(\frac{1}{x^2}-\frac{2}{x^2}\right) d x}{\left(1+\frac{e^x}{x^2}\right)} d x}\end{aligned}

Let

1+\frac{e^x}{x^2}=t \Rightarrow \frac{x^2 e^x-e^x 2 x}{x^4} d x=d t \Rightarrow \text{ I.F. } \mathrm{e}^{\ln

\left(1+\frac{\mathrm{e}^2}{\mathrm{x}^2}\right)}=1+\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}^2}

Now

y\left(1+\frac{e^x}{x^2}\right)=\int \frac{x^2}{x^2+e^x} \cdot \frac{x^2+e^x}{x^2} d x+C y\left(1+\frac{e^x}{x^2}\right)=x+C

Passing through

(1,0)

\begin{aligned}& \Rightarrow C=-1 \\ & y=\frac{x-1}{1+\frac{e^x}{x^2}} \\ & y(2)=\frac{1}{1+\frac{e^2}{4}}=\frac{4}{4+e^2}\end{aligned}

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