Answer to Question #256783 in Differential Equations for Kvc

Question

(y^¹)^2= 1-y^2/1-x^2 ; y=1/2 when, x =1

Answer 1

Let us solve the differential equation $(y')^2= \frac{1-y^2}{1-x^2} ;$ $y=\frac{1}{2}$ when $x =1.$

This equation is eqivalent to the set of equations $y'= \sqrt{\frac{1-y^2}{1-x^2}}$, $y'= -\sqrt{\frac{1-y^2}{1-x^2}} .$ It follows that $\frac{ dy}{\sqrt{1-y^2} }= \frac{dx}{\sqrt{1-x^2}}$ or $\frac{ dy}{\sqrt{1-y^2} }= -\frac{dx}{\sqrt{1-x^2}}$. Therefore, $\int\frac{ dy}{\sqrt{1-y^2} }= \int\frac{dx}{\sqrt{1-x^2}}$ or $\int\frac{ dy}{\sqrt{1-y^2} }= -\int\frac{dx}{\sqrt{1-x^2}},$ and hence $\arcsin y=\arcsin x+C_1$ or $\arcsin y=-\arcsin x+C_2.$ Since $y=\frac{1}{2}$ when $x =1,$ we get that $\arcsin \frac{1}2=\arcsin 1+C_1$ or $\arcsin \frac{1}2=-\arcsin 1+C_2.$

It follows that $\frac{\pi}6=\frac{\pi}2+C_1,\ \frac{\pi}6=-\frac{\pi}2+C_1,$ and hence $C_1=-\frac{\pi}{3},\ C_2=\frac{2\pi}{3}.$

We conclude that the solutions are the following:

$\arcsin y=\arcsin x-\frac{\pi}{3},\ \arcsin y=-\arcsin x+\frac{2\pi}{3}.$