Let the solution curve of xdydx-y=y^2+16x^2, y(1)=3, be y=y(x).

Question

Let the solution curve of $x\dfrac{dy}{dx}-y=\sqrt{y^2+16x^2}$, $y(1)=3$, be $y=y(x)$. Then $y(2)$ is:

Accepted Answer

With $y=vx$: $x\dfrac{dv}{dx}=\sqrt{v^2+16}$; separating and integrating:
$$\ln\left(v+\sqrt{v^2+16}\right) = \ln x+\ln c \implies v+\sqrt{v^2+16}=cx.$$

I.e. $y+\sqrt{y^2+16x^2}=cx^2$. Using $y(1)=3$: $3+5=c \Rightarrow c=8$.

At $x=2$: $y+\sqrt{y^2+64}=32 \implies y=15$.

Solutions · 1