Question

# Form the differential equation for y=(sin−1x)2+Acos−1x+B where A and B are arbitary constants, as its general solution.

Solution

## Given that, y=(sin−1x)2+Acos−1x+B On differentiating w.r.t. x, we get dydx=2sin−1x√1−x2+(−A)√1−x2⇒√1−x2dydx=2sin−1x−A Again, differentiating w.r.t.x we get √1−x2d2ydx2+dydx.−2x2√1+x2=2√1−x2⇒(1−x2)d2ydx2−x√1−x2.√1−x2dydx=2⇒(1−x2)d2ydx2−xdydx=2⇒(1−x2)d2ydx2−xdydx−2=0 which is the required differential equation.

Suggest corrections