Consider y^2,dx+(x-1y)dy=0.

Question

Consider $y^2\,dx+\left(x-\dfrac1y\right)dy=0$. If $y=1$ at $x=1$, the value of $x$ for which $y=2$ is:

Accepted Answer

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

I.F.$=e^{-1/y}$. Solving via $u=-1/y$: $xe^{-1/y}=e^{-1/y}\left(\dfrac1y+1\right)+c$.

Using $y=1,x=1$: $c=-1/e$. At $y=2$: $x=\dfrac32-\dfrac{1}{\sqrt e}$.

Solutions · 1