If y=y(x) solves 2x^2dydx-2xy+3y^2=0 with y(e)= e3, then y(1) is:

Question

If $y=y(x)$ solves $2x^2\dfrac{dy}{dx}-2xy+3y^2=0$ with $y(e)=\dfrac e3$, then $y(1)$ is:

Accepted Answer

$$2x(x\,dy-y\,dx)+3y^3\,dx=0 \implies 2\frac{x\,dy-y\,dx}{y^2}+3\frac{dx}{x}=0 \implies -\frac{2x}{y}+3\ln x=C.$$

Using $y(e)=e/3$: $-6+3=C \Rightarrow C=-3$. At $x=1$: $-\dfrac2y=-3 \Rightarrow y=\dfrac23$.

Solutions · 1