If y=y(x) is the solution of √4-x^2dydx = ((^-1(x/2))^2-y)^-1(x/2), -2 x2, y(2)=^24-2, the…

Question

If $y=y(x)$ is the solution of $\sqrt{4-x^2}\dfrac{dy}{dx} = \left((\sin^{-1}(x/2))^2-y\right)\sin^{-1}(x/2)$, $-2\le x\le2$, $y(2)=\dfrac{\pi^2}{4}-2$, then $y^2(0)$ is equal to \_\_\_\_\_\_.

Answer: 4

Accepted Answer

Using the substitution $t=\sin^{-1}(x/2)$ and solving the resulting linear ODE, one obtains
$$y=(\sin^{-1}(x/2))^2-2+c\,e^{-\frac12(\sin^{-1}(x/2))^2}.$$

Using $y(2)=\pi^2/4-2 \Rightarrow c=0$. Then $y(0)=-2$, so $y^2(0)=4$.

Solutions · 1