Question

Form the differential equation having \[y = \left( \sin^{- 1} x \right)^2 + A \cos^{- 1} x + B\], where A and B are arbitrary constants, as its general solution.

Answer 1

We have two constnts in the solution, so we will differentiate both sides twice and eliminate the constants A and B.
\[y = \left( \sin^{- 1} x \right)^2 + A \cos^{- 1} x + B\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2 \sin^{- 1} x}{\sqrt{1 - x^2}} - \frac{A}{\sqrt{1 - x^2}}\]
\[ \Rightarrow \left( \sqrt{1 - x^2} \right)\frac{dy}{dx} = 2 \sin^{- 1} x - A\]
\[\Rightarrow \left( \sqrt{1 - x^2} \right)\frac{d^2 y}{d x^2} + \frac{\left( - 2x \right)}{2\sqrt{1 - x^2}}\frac{dy}{dx} = \frac{2}{\sqrt{1 - x^2}}\]
\[ \Rightarrow \left( \sqrt{1 - x^2} \right)\frac{d^2 y}{d x^2} - \frac{x}{\sqrt{1 - x^2}}\frac{dy}{dx} = \frac{2}{\sqrt{1 - x^2}}\]
\[ \Rightarrow \left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} = 2\]
\[ \Rightarrow \left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - 2 = 0\]