Question

If $\mathrm{y}={\left(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)}^{\mathrm{n}}$,then $\left(1+{\mathrm{x}}^2\right)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}$ is

• A. ${\mathrm{n}}^2\mathrm{y}$
• B. $-{\mathrm{n}}^2\mathrm{y}$
• C. $-\mathrm{y}$
• D. $2{\mathrm{x}}^2\mathrm{y}$

Answer 1

The correct option is A

${\mathrm{n}}^2\mathrm{y}$

Finding the value of $\left(1+{\mathrm{x}}^2\right)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}$:

The given differential equation is $\mathrm{y}={\left(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)}^{\mathrm{n}}$

Differentiate the above equation with respect to $x$

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{n}{\left(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)}^{\mathrm{n}-1}\times \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)\mathbf{[}\mathbf{\because }\mathbf{}\frac{{\mathbf{dx}}^{\mathbf{n}}}{\mathbf{dx}}\mathbf{=}{\mathbf{nx}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{]} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{n}{\left(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)}^{\mathrm{n}-1}\times \left(1+\frac{2\mathrm{x}}{2\sqrt{1+{\mathrm{x}}^2}}\right) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{n}{\left(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)}^{\mathrm{n}-1}\times \left(\frac{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}{\sqrt{1+{\mathrm{x}}^2}}\right) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{n}{\left(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)}^{\mathrm{n}}}{\sqrt{1+{\mathrm{x}}^2}} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{ny}}{\sqrt{1+{\mathrm{x}}^2}} \\ \Rightarrow \sqrt{1+{\mathrm{x}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{ny} \end{gathered}$$

When both sides are squared, the result is

$\left(1+{\mathrm{x}}^2\right){\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2={\mathrm{n}}^2{\mathrm{y}}^2$

Differentiate the above equation with respect to $x$.

$\begin{array}{l}\left(1+{\mathrm{x}}^2\right)\times \left(2\frac{\mathrm{dy}}{\mathrm{dx}}\times \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\times 2\mathrm{x}=2{\mathrm{n}}^2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{}\left[\because \frac{d\left(a·b\right)}{\mathrm{dx}}=a\frac{d}{\mathrm{dx}}\left(b\right)+b\frac{d}{\mathrm{dx}}\left(a\right),\frac{d\left(x^n\right)}{\mathrm{dx}}={\mathrm{nx}}^{n-1}\right]\\ \Rightarrow 2\frac{\mathrm{dy}}{\mathrm{dx}}\left[\left(1+{\mathrm{x}}^2\right)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\right]=2{\mathrm{n}}^2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}\\ \Rightarrow \left(1+{\mathrm{x}}^2\right)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{n}}^2\mathrm{y}\end{array}$

Hence, the correct option is A.