Let y=y(x) be the solution curve of x(x^2+e^x),dy+(e^x(x-2)y-x^3),dx=0, x0, passing throu…

Question

Let $y=y(x)$ be the solution curve of $x(x^2+e^x)\,dy+(e^x(x-2)y-x^3)\,dx=0$, $x>0$, passing through $(1,0)$. Then $y(2)$ is:

Accepted Answer

$$\frac{dy}{dx}+\frac{e^x(x-2)}{x(x^2+e^x)}y = \frac{x^3}{x(x^2+e^x)}.$$

I.F. $= \dfrac{e^x+x^2}{x^2}$ (obtained by decomposing the exponent of integration).

$$y\left(\frac{e^x+x^2}{x^2}\right) = x+c.$$

Using $y(1)=0$: $c=-1$. So $y\left(\dfrac{e^x+x^2}{x^2}\right)=x-1$.

At $x=2$: $y = \dfrac{4}{e^2+4}$.

Solutions · 1