Question

Differential equation, having $y=(si{n}^{-1}x{)}^2+A(co{s}^{-1}x)+B$ where A and B are arbitary constants is $(p-x^2)\frac{d^{2}y}{d{x}^2}-\frac{xdy}{dx}=q$; then $p+q=$ ___

Answer 1

$y=(si{n}^{-1}x{)}^2+Aco{s}^{-1}x+B$

$y^{\prime }=\frac{2si{n}^{-1}x}{\sqrt{1-x^2}}-\frac{A}{\sqrt{1-x^2}}$

$\therefore \sqrt{1-x^2}{y}^{\prime }=2si{n}^{-1}x-A$

$(1-x^2)(y^{\prime }{)}^2=(2si{n}^{-1}x-A{)}^2$ .....(1)

Now, $y=(si{n}^{-1}x{)}^2+Aco{s}^{-1}x+B$

$\Rightarrow y=(si{n}^{-1}x{)}^2+A(\frac{\pi }{2}-si{n}^{-1}x)+B$

$\Rightarrow y=(si{n}^{-1}x{)}^2-Asi{n}^{-1}x+\frac{\pi A}{2}+B$

$\Rightarrow 4y=4(si{n}^{-1}x{)}^2-4Asi{n}^{-1}x+A^2-A^2+2\pi A+4B$

$\Rightarrow (2si{n}^{-1}x-A{)}^2=4y+A^2-4B-2\pi A$ ....(2)

From (1) & (2)

$(1-x^2)(y^{\prime }{)}^2=4y+A^2-4B-2\pi A$

$(1-x^2)2y^{\prime }{y}^{\prime \prime }+\left(y^{\prime }{)}^2\right(-2x^2)=4y^{\prime }$

$(1-x^2)y^{\prime \prime }-x{y}^{\prime }=2$

$\therefore p=1;q=2$