Question

prove the given differencial equation ​

Answer 1

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

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

${\large{\underline{\underline{\textbf{To Prove :}}}}}$

$(1+x^2)y_2+xy_1-m^2y=0$

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

$\implies (1+x^2)y_2+xy_1-m^2y=0$

Solve by inserting chain rule

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$

$t=x+\sqrt{1+x^2}\:\:\:,\:\: y=t^m$

$\frac{dy}{dt}=mt^{m-1}\:\:\: ;\:\:\: \frac{dt}{dx}=1+\frac{x}{(1+x)^{\frac{1}{2}}}$

$y_1 = \frac{dy}{dx}=m(x+\sqrt{1+x^2})^{m-1}(1+\frac{x}{\sqrt{1+x^2}})$

$= m(x+\sqrt{1+x^2})^{m-1}(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}$

$= \frac{m}{\sqrt{1+x^2}}\times (x+\sqrt{1+x^2})^{m-1+1}$

$\implies y_1=\frac{my}{\sqrt{1+x^2}}$

$\textbf{squaring on both sides}$,

${y_1}^2 (1+x^2)=m^2y^2$

$\textbf{differentiating both sides},$

$2y_1 \frac{dy_1}{dx}(1+x^2)+{y_1}^2(2x)=m^22y.\frac{dy}{dx}$

$2y_1\frac{d}{dx}(\frac{dy}{dx})(1+x^2)+{y_1}^22x =m^22yy_1$

$2y_1y_2(1+x^2)+{y_1}^22x=m^22yy_1=0$

$\textbf{div 2y by both sides}$,

$y_2(1+x^2)+xy_1-m^2y=0$

$\textbf{Hence proved}.$