Question

$y=log(x+\sqrt{x^2+a^2})Provethat:(x^2+a^2)y^{\prime \prime }+x{y}^{\prime }=0$

Answer 1

$y=log(x+\sqrt{x^2+a^2})$
$y=log(x+\sqrt{x^2+a^2}):(x^2+a^2)y+x{y}^{\prime }=0Letx+\sqrt{x^2+a^2}=t\Rightarrow y=logt\frac{dt}{dx}=t+\frac{2x}{2\sqrt{x^2+a^2}}=\frac{\sqrt{x^2+a^2}2x}{\sqrt{x^2+a^2}}\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{1}{\mathrm{t/}}.\frac{\sqrt{\text{/}{x}^2{\mathrm{a/}}^2+x}}{\sqrt{x^2+a^2}}=\frac{1}{\sqrt{x^2}+a^2}\therefore y^{\prime }=\frac{1}{\sqrt{x^2+a^2}}Let{x}^2+a^2=x\frac{du}{dx}=2xy"=\frac{d}{dx}\frac{1}{\sqrt{u}}.\frac{du}{dx}=-\frac{1}{\text{/}2\sqrt{x^2+a^2}}.\frac{\text{/}2x}{(x^2+a^2)}=-\frac{x.y^1}{x^2+a^2}\therefore (x^2+y^2)y"+x{y}^{\prime }=0$