Question

Show that $y=e^{m{\sin }^{-1}x}$ is a solution of the differential equation $(1-x^2)y_2-x{y}_1-m^{2}y=0$.

Answer 1

$y=e^{msi{n}^{1}x}$

$y^{\prime }=msi{n}^{1}x$ $e^{msi{n}^{-1}x}$ $-\frac{1}{\sqrt{1-x^2}}$

$y^{\prime \prime }=\frac{\left(\sqrt{1-x^2}\right)\left(\frac{m^2{e}^{msi{n}^{-1}x}}{\sqrt{1-x^2}}\right)-m{e}^{msi{n}^{-1}}\left(\frac{1}{2}\right)(1-x^2{)}^{-1/2}(-2x)}{(\sqrt{1-x^2}{)}^2}$

$=\frac{m^2{e}^{msi{n}^{-1}x}+\frac{mx{e}^{si{n}^{-1}x}}{\sqrt{1-x^2}}}{1-x^2}$

$=\frac{m^{2}y+n{y}^{\prime }}{1-x^2}$

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

$(1-x^2)y^{\prime \prime }-m^{2}y-xy=0$