Let y=f(x) solve y(x+1),dx-x^2,dy=0, y(1)=e.

Question

Let $y=f(x)$ solve $y(x+1)\,dx-x^2\,dy=0$, $y(1)=e$. Then $\lim_{x	o0^+}f(x)$ is:

Accepted Answer

$$\frac{x+1}{x^2}\,dx = \frac{dy}{y} \implies \int\left(\frac{1}{x}+\frac{1}{x^2}\right)dx = \ln y+c'.$$

$$\ln x - \frac1x = \ln y+c.$$

Using $(1,e)$: $c=-2$. So $y = e^{\ln x-1/x+2}$; as $x\to0^+$, $-1/x\to-\infty$, so $y\to0$.

Solutions · 1