Let the solution curve of x,dy=(x+x^2+y^2+y)dx, x0, intersect x=1 at y=0 and x=2 at y=.

Question

Let the solution curve of $x\,dy=\left(x+\sqrt{x^2+y^2}+y\right)dx$, $x>0$, intersect $x=1$ at $y=0$ and $x=2$ at $y=\alpha$. Then $\alpha$ is:

Accepted Answer

With $y=vx$: $\dfrac{dv}{\sqrt{1+v^2}}=\dfrac{dx}{x} \implies \ln\left(v+\sqrt{1+v^2}\right)=\ln x+C$.

At $x=1,y=0$: $C=0$, so $\dfrac{y}{x}+\sqrt{1+\dfrac{y^2}{x^2}}=x$.

At $x=2$: $\dfrac{y}{2}+\sqrt{1+\dfrac{y^2}{4}}=2$; solving gives $y=\dfrac32$.

Solutions · 1