Question

If
`\mathrm {y = (x + \sqrt{1+x^(2)})^(m)}`, then prove that
`\mathrm {(x^(2) +1)y_(2) +x y_(1) - m^(2)y = 0}`.

Note : y₁ and y₂ refer to the first and second derivatives.

Answer 1

See below for proof.

Explanation:

Given:

`y=\left(x+√(1+x^2)\right)^m`

First derivative

`\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\frac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}`

`\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\frac{\text{d}y}{\text{d}x}=xn^(n-1)$\\\end{minipage}}`

`\begin{aligned} y_1=\frac{\text{d}y}{\text{d}x} & =m\left(x+√(1+x^2)\right)^(m-1) \cdot \left(1+(2x)/(2√(1+x^2)) \right)\\\\ & =m\left(x+√(1+x^2)\right)^(m-1) \cdot \left(1+(x)/(√(1+x^2)) \right) \\\\ & =m\left(x+√(1+x^2)\right)^(m-1) \cdot \left((x+√(1+x^2))/(√(1+x^2)) \right)\\\\ & = (m)/(√(1+x^2)) \cdot \left(x+√(1+x^2)\right)^(m-1) \cdot \left(x+√(1+x^2)\right)\\\\ & = (m)/(√(1+x^2))\left(x+√(1+x^2)\right)^m\end{aligned}`

Second derivative

`\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\frac{\text{d}y}{\text{d}x}=u\frac{\text{d}v}{\text{d}x}+v\frac{\text{d}u}{\text{d}x}$\\\end{minipage}}`

`\textsf{Let }u=(m)/(√(1+x^2))`

`\implies \frac{\text{d}u}{\text{d}x}=-(mx)/(\left(1+x^2\right)^(3)/(2))`

`\textsf{Let }v=\left(x+√(1+x^2)\right)^m`

`\implies \frac{\text{d}v}{\text{d}x}=(m)/(√(1+x^2)) \cdot \left(x+√(1+x^2)\right)^m`

`\begin{aligned}y_2=\frac{\text{d}^2y}{\text{d}x^2}&=(m)/(√(1+x^2))\cdot(m)/(√(1+x^2))\cdot\left(x+√(1+x^2)\right)^m+\left(x+√(1+x^2)\right)^m\cdot-(mx)/(\left(1+x^2\right)^(3)/(2))\\\\&=(m^2)/(1+x^2)\cdot\left(x+√(1+x^2)\right)^m+\left(x+√(1+x^2)\right)^m\cdot-(mx)/(\left(1+x^2\right)√(1+x^2))\\\\ &=\left(x+√(1+x^2)\right)^m\left((m^2)/(1+x^2)-(mx)/(\left(1+x^2\right)√(1+x^2))\right)\\\\\end{aligned}`

`= (\left(x+√(1+x^2)\right)^m)/(1+x^2)\right)\left(m^2-(mx)/(√(1+x^2))\right)`

Proof

`(x^2+1)y_2+xy_1-m^2y`

`= (x^2+1) (\left(x+√(1+x^2)\right)^m)/(1+x^2)\left(m^2-(mx)/(√(1+x^2))\right)+(mx)/(√(1+x^2))\left(x+√(1+x^2)\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m`

`= \left(x+√(1+x^2)\right)^m\left(m^2-(mx)/(√(1+x^2))\right)+(mx)/(√(1+x^2))\left(x+√(1+x^2)\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m`

`= \left(x+√(1+x^2)\right)^m\left[m^2-(mx)/(√(1+x^2))+(mx)/(√(1+x^2))-m^2\right]`

`= \left(x+√(1+x^2)\right)^m\left[0]`

`= 0`