Question

if y=sinh [m log {x+ √x^2+ ]1)}], show that (x^2 + 1)d^2y/dx^2 +xdy/dx=m^2 y.

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{y=sinh[m\,log(x+\sqrt{x^2+1})]}$

$\underline{\textbf{To show:}}$

$\mathsf{(x^2+1)\dfrac{d^2y}{dx^2}+x\,\dfrac{dy}{dx}=m^2\,y}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{y=sin\,h[m\,log(x+\sqrt{x^2+1})]}$

$\textsf{Differentiate with respect to 'x'}$

$\mathsf{\dfrac{dy}{dx}=cos\,h[m\,log(x+\sqrt{x^2+1})]{\times}m{\times}\dfrac{1}{x+\sqrt{x^2+1}}{\times}1+\dfrac{2x}{2\sqrt{x^2+1}}}$

$\mathsf{\dfrac{dy}{dx}=cos\,h[m\,log(x+\sqrt{x^2+1})]{\times}m{\times}\dfrac{1}{x+\sqrt{x^2+1}}{\times}\dfrac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}}$

$\mathsf{\dfrac{dy}{dx}=cos\,h[m\,log(x+\sqrt{x^2+1})]{\times}m{\times}\dfrac{1}{\sqrt{x^2+1}}}$

$\mathsf{\dfrac{dy}{dx}=cos\,h[m\,log(x+\sqrt{x^2+1})]{\times}\dfrac{m}{\sqrt{x^2+1}}}$

$\textsf{Squaring on bothsides, we get}$

$\mathsf{\left(\dfrac{dy}{dx}\right)^2=cos^2h[m\,log(x+\sqrt{x^2+1})]{\times}\dfrac{m^2}{x^2+1}}$

$\mathsf{(x^2+1)\left(\dfrac{dy}{dx}\right)^2=m^2\;cos^2h[m\,log(x+\sqrt{x^2+1})]}$

$\mathsf{(x^2+1)\left(\dfrac{dy}{dx}\right)^2=m^2\{1+sin^2h[m\,log(x+\sqrt{x^2+1})]\}}$

$\mathsf{(x^2+1)\left(\dfrac{dy}{dx}\right)^2=m^2\{1+y^2\}}$

$\textsf{Differentiate once again with respect to 'x'}$

$\mathsf{(x^2+1)\,2\left(\dfrac{dy}{dx}\right)\left(\dfrac{d^2y}{dx^2}\right)+\left(\dfrac{dy}{dx}\right)\;2x=m^2\;2y\;\dfrac{dy}{dx}}$

$\mathsf{Divide\;bothsides\;by\;2\left(\dfrac{dy}{dx}\right)}$

$\boxed{\mathsf{(x^2+1)\,\left(\dfrac{d^2y}{dx^2}\right)+x\left(\dfrac{dy}{dx}\right)=m^2\,y}}$