Mathematics - Differential Equation Question with Solution | TestHub

$y={\left(x+\sqrt{1+x^2}\right)}^n$ $$\begin{array}{c}\frac{dy}{dx}=n{\left(x+\sqrt{1+x^2}\right)}^{n-1}\left(1+\frac{1}{2}{\left(1+x^2\right)}^{-1/2}\cdot 2x\right)\\ \begin{array}{c}\frac{dy}{dx}=n{\left(x+\sqrt{1+x^2}\right)}^{n-1}\frac{\left(\sqrt{1+x^2}+x\right)}{\sqrt{1+x^2}}\\ =\frac{n{\left(\sqrt{1+x^2}+x\right)}^n}{\sqrt{1+x^2}}\end{array}\end{array}$$ or $\sqrt{1+x^2}\frac{dy}{dx}=ny$ or $\sqrt{1+x^2}{y}_1=ny$ $$\left(y_1=\frac{dy}{dx}\right)$$ Squaring, $\left(1+x^2\right)y_1^2=n^2{y}^2$ Differentiating, $$\left(1+x^2\right)2y_1{y}_2+y_1^2\cdot 2x=n^2\cdot 2y{y}_1$$ or $\left(1+x^2\right)y_2+x{y}_1=n^{2}y$