Let y=y(x) solve (x^2-3y^2),dx+3xy,dy=0, y(1)=1.

Question

Let $y=y(x)$ solve $(x^2-3y^2)\,dx+3xy\,dy=0$, $y(1)=1$. Then $6y^2(e)$ is:

Accepted Answer

$$\frac{dy}{dx} = \frac{3y^2-x^2}{3xy} = \frac{y}{x}-\frac{x}{3y}.$$

With $y=vx$: $x\dfrac{dv}{dx}=-\dfrac{1}{3v} \implies v\,dv = -\dfrac{dx}{3x}$.

Integrating: $\dfrac{v^2}{2}=-\dfrac13\ln x+c$. Using $y(1)=1 \Rightarrow v=1 \Rightarrow c=\dfrac12$.

$$\frac{y^2}{2x^2} = -\frac13\ln x+\frac12 \implies y^2 = -\frac23x^2\ln x+x^2.$$

At $x=e$: $y^2(e) = -\dfrac23e^2+e^2 = \dfrac{e^2}{3} \implies 6y^2(e)=2e^2$.

Solutions · 1