Show that if y= (sin^-1 x)^2 then (1 - x^2) (d^2y)/(dx^2) - xdy/dx - 2 = 0 ?

May 12, 2018

We seek to show that:

$\left(1 - {x}^{2}\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} - x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 = 0$ where $y = {\left({\sin}^{-} 1 x\right)}^{2}$

Using the result:

$\frac{d}{\mathrm{dx}} \left(\arcsin x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

In conjunction with the chain rule, then differentiating $y = {\left({\sin}^{-} 1 x\right)}^{2}$ wrt $x$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \arcsin x}{\sqrt{1 - {x}^{2}}}$

And differentiating a second time, in conjunction with the quotient rule, we have:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\left(\sqrt{1 - {x}^{2}}\right) \left(\frac{2}{\sqrt{1 - {x}^{2}}}\right) - \left(\frac{\frac{1}{2} \left(- 2 x\right)}{\sqrt{1 - {x}^{2}}}\right) \left(2 \arcsin x\right)}{\sqrt{1 - {x}^{2}}} ^ 2$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{2 + \frac{2 x \setminus \arcsin x}{\sqrt{1 - {x}^{2}}}}{1 - {x}^{2}}$

And so, considering the LHS of the given expression:

$L H S = \left(1 - {x}^{2}\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} - x \frac{\mathrm{dy}}{\mathrm{dx}} - 2$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(1 - {x}^{2}\right) \left\{\frac{2 + \frac{2 x \setminus \arcsin x}{\sqrt{1 - {x}^{2}}}}{1 - {x}^{2}}\right\} - x \left\{\frac{2 \arcsin x}{\sqrt{1 - {x}^{2}}}\right\} - 2$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 + \frac{2 x \setminus \arcsin x}{\sqrt{1 - {x}^{2}}} - \frac{2 x \arcsin x}{\sqrt{1 - {x}^{2}}} - 2$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0 \setminus \setminus \setminus$ QED