Let y=y(x) solve e^x√1-y^2,dx+yx,dy=0, y(1)=-1.

Question

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

Accepted Answer

*(As transcribed, the ODE in the source appears to read $e^x\,dx+\dfrac{y}{x\sqrt{1-y^2}}\,dy=0$-type; carrying out the source's integration:)*
$$\sqrt{1-y^2} = e^x(x-1)+c.$$

Using $(1,-1)$: $c=0$, so $\sqrt{1-y^2}=e^x(x-1)$. At $x=3$: $1-y^2 = (2e^3)^2 = 4e^6 \implies y^2=1-4e^6$.

Solutions · 1