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 :

Question

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 :

Aicharya · Accepted Answer

$\begin{aligned} & x\left(x^2+e^xight) d y+\left(e^x(x-2) y-x^3ight) d x=0 \ & x\left(x^2+e^xight) \frac{d y}{d x}+e^x(x-2) y=x^3 \ & \frac{d y}{d x}+\frac{e^x(x-2)}{x\left(x^2+e^xight)} y=\frac{x^2}{x^2+e^x} \ & 	ext { I.F. }=e^{\int \frac{e^x(x-2)}{x\left(x^2+e^xight)} d x}=e^{\int \frac{e^x\left(\frac{1}{x^2}-\frac{2}{x^2}ight) d x}{\left(1+\frac{e^x}{x^2}ight)} d x}\end{aligned}$ Let $1+\frac{e^x}{x^2}=t \Rightarrow \frac{x^2 e^x-e^x 2 x}{x^4} d x=d t$ $\Rightarrow 	ext { I.F. } \mathrm{e}^{\ln \left(1+\frac{\mathrm{e}^2}{\mathrm{x}^2}ight)}=1+\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}^2}$ Now $y\left(1+\frac{e^x}{x^2}ight)=\int \frac{x^2}{x^2+e^x} \cdot \frac{x^2+e^x}{x^2} d x+C$ $y\left(1+\frac{e^x}{x^2}ight)=x+C$ Passing through $(1,0)$ $\begin{aligned} & \Rightarrow C=-1 \ & y=\frac{x-1}{1+\frac{e^x}{x^2}} \ & y(2)=\frac{1}{1+\frac{e^2}{4}}=\frac{4}{4+e^2} \end{aligned}$

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)…

Solution