If y=y(x) solves x^2,dy+(y-1x)dx=0, x0, y(1)=1, then y(12) is:

Question

If $y=y(x)$ solves $x^2\,dy+\left(y-\dfrac1x\right)dx=0$, $x>0$, $y(1)=1$, then $y\left(\dfrac12\right)$ is:

Accepted Answer

$$\frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3}.$$

I.F.$=e^{-1/x}$; solving via $t=-1/x$ gives $y = \dfrac1x+1-\dfrac{e^{1/x}}{e}$.

Using $y(1)=1$ fixes the constant consistently; at $x=1/2$: $y=2+1-\dfrac{e^2}{e}=3-e$.

Solutions · 1