Question

If $\mathrm{y}=\mathrm{xlog}\left(\frac{\mathrm{x}}{\mathrm{a}+\mathrm{bx}}\right)$ then ${\mathrm{x}}^3\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=?$

• A. $\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}$
• B. ${\left(\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}\right)}^2$
• C. $\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x}$
• D. ${\left(\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x}\right)}^2$

Answer 1

The correct option is B

${\left(\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}\right)}^2$

Explanation for the correct option:

Finding the value of ${\mathrm{x}}^3\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$:

The given differential equation is $\mathrm{y}=\mathrm{xlog}\left(\frac{\mathrm{x}}{\mathrm{a}+\mathrm{bx}}\right)$

$\frac{\mathrm{y}}{\mathrm{x}}=\log \mathrm{x}-\log (\mathrm{a}+\mathrm{bx})$

Differentiate the above equation with respect to $x$

$\begin{array}{rcl}\frac{\mathrm{x}{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}-\mathrm{y}}{{\mathrm{x}}^2}& =& \frac{1}{\mathrm{x}}-\frac{\mathrm{b}}{\mathrm{a}+\mathrm{bx}}\left[\because \frac{d\left(\mathrm{logx}\right)}{\mathrm{dx}}=\frac{1}{x},\frac{d}{\mathrm{dx}}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{\mathrm{du}}{\mathrm{dx}}}-u{\displaystyle \frac{\mathrm{dv}}{\mathrm{dx}}}}{v^2}\right]\\ \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}& =& \mathrm{x}-\frac{{\mathrm{bx}}^2}{\mathrm{a}+\mathrm{bx}}\\ & =& \frac{\mathrm{ax}+{\mathrm{bx}}^2-{\mathrm{bx}}^2}{\mathrm{a}+\mathrm{bx}}\\ & =& \frac{\mathrm{ax}}{\mathrm{a}+\mathrm{bx}}...\left(1\right)\\ & & \end{array}$

Again, differentiate with respect to $x$.

$\begin{array}{l}\text{x}\frac{{\text{d}}^2\text{y}}{{\mathrm{dx}}^2}+\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{dy}}{\mathrm{dx}}=1-\frac{\left[\left(\text{a}+\mathrm{bx}\right)2\mathrm{bx}-{\mathrm{bx}}^2\left(\text{b}\right)\right]}{{\left(\mathrm{a}+\mathrm{bx}\right)}^2}\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}\mathbf{(}\mathbf{a}\mathbf{·}\mathbf{b}\mathbf{)}}{\mathbf{dx}}\mathbf{=}\mathbf{a}\frac{\mathbf{db}}{\mathbf{dx}}\mathbf{+}\mathbf{b}\frac{\mathbf{da}}{\mathbf{dx}}\mathbf{,}\mathbf{}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}\frac{\mathbf{a}}{\mathbf{b}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{b}{\displaystyle \frac{\mathrm{da}}{\mathrm{dx}}}\mathbf{-}\mathbf{a}{\displaystyle \frac{\mathrm{db}}{\mathrm{dx}}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}\\ \text{⇒ x}\frac{{\text{d}}^2\text{y}}{{\mathrm{dx}}^2}=\frac{{\left(\mathrm{a}+\mathrm{bx}\right)}^2-2\mathrm{bx}\left(\text{a}+\mathrm{bx}\right)+{\text{b}}^2{\text{x}}^2}{{\left(\mathrm{a}+\mathrm{bx}\right)}^2}\\ =\frac{\left(\text{a}+\mathrm{bx}\right)\left[\text{a}+\mathrm{bx}-2\mathrm{bx}\right]+{\text{b}}^2{\text{x}}^2}{{(\text{a}+\mathrm{bx})}^2}\\ =\frac{\left(\text{a}+\mathrm{bx}\right)\left(\text{a}-\mathrm{bx}\right)+{\text{b}}^2{\text{x}}^2}{{(\text{a}+\mathrm{bx})}^2}\\ =\frac{{\text{a}}^2-{\text{b}}^2{\text{x}}^2+{\text{b}}^2{\text{x}}^2}{{(\text{a}+\mathrm{bx})}^2}\\ =\frac{{\text{a}}^2}{{(\text{a}+\mathrm{bx})}^2}\\ ={\left(\frac{\text{a}}{\text{a}+\mathrm{bx}}\right)}^2\end{array}$

$\mathrm{x}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\left(\frac{\mathrm{a}}{\mathrm{a}+\mathrm{bx}}\right)}^2...\left(2\right)$

Multiplying by $x^2$ in equation $\left(2\right)$,

,${\mathrm{x}}^3\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\mathrm{x}}^2{\left(\frac{\mathrm{a}}{\mathrm{a}+\mathrm{bx}}\right)}^2...\left(3\right)$

By squaring the equation $\left(1\right)$, we get

$\begin{array}{rcl}{\left(\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}\right)}^2& =& {\left(\frac{\mathrm{ax}}{\mathrm{a}+\mathrm{bx}}\right)}^2\\ & =& {\mathrm{x}}^3\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\left[\mathrm{from}3\right]\end{array}$

Hence, the correct option is B.