Question

If $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{xy}}{{\mathrm{x}}^2+{\mathrm{y}}^2}$,$\mathrm{y}\left(1\right)=1$; then a value of x satisfying $\mathrm{y}\left(\mathrm{x}\right)=\mathrm{e}$ is

• A. $\sqrt{3}\mathrm{e}$
• B. $\frac{1}{2}\sqrt{3}\mathrm{e}$
• C. $\sqrt{2}\mathrm{e}$
• D. $\frac{\mathrm{e}}{\sqrt{2}}$

Answer 1

The correct option is A

$\sqrt{3}\mathrm{e}$

Explanation for the correct answer:

Step 1: Differentiating with respect to $dx,$

Given, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{xy}}{{\mathrm{x}}^2+{\mathrm{y}}^2}...\left(1\right)$

Differentiating with respect to $dx,$

$$\begin{gathered} \therefore \mathrm{y}=\mathrm{vx} \\ \Rightarrow \frac{dy}{dx}=v\times 1+x\times \frac{dv}{dx}\left[\because \frac{d}{dx}\left(uv\right)=u\left(\frac{dv}{dx}\right)+v\left(\frac{du}{dx}\right)\right] \end{gathered}$$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}...\left(2\right)$

Step 2: Equate the equation (1) and (2)

$$\begin{gathered} \Rightarrow \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{{\mathrm{vx}}^2}{{\mathrm{x}}^2+{\mathrm{v}}^2{\mathrm{x}}^2}\left[\because y=vx\right] \\ \Rightarrow \frac{\mathrm{vdx}+\mathrm{xdv}}{\mathrm{dx}}=\frac{{\mathrm{vx}}^2}{{\mathrm{x}}^2+{\mathrm{v}}^2{\mathrm{x}}^2} \\ \Rightarrow \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{{\mathrm{vx}}^2}{{\mathrm{x}}^2+{\mathrm{v}}^2{\mathrm{x}}^2}-\mathrm{v} \\ \Rightarrow \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{{\mathrm{vx}}^2-{\mathrm{vx}}^2-{\mathrm{v}}^3{\mathrm{x}}^2}{{\mathrm{x}}^2+{\mathrm{v}}^2{\mathrm{x}}^2} \\ \Rightarrow \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-{\mathrm{v}}^3{\mathrm{x}}^2}{{\mathrm{x}}^2+{\mathrm{v}}^2{\mathrm{x}}^2} \\ \Rightarrow \frac{\mathrm{dv}}{-{\mathrm{v}}^3{\mathrm{x}}^2}=\frac{\mathrm{dx}}{\mathrm{x}({\mathrm{x}}^2+{\mathrm{v}}^2{\mathrm{x}}^2)} \\ \Rightarrow \frac{\mathrm{dv}}{-{\mathrm{v}}^3}=\frac{{\mathrm{x}}^2\mathrm{dx}}{{\mathrm{x}}^3(1+{\mathrm{v}}^2)} \\ \Rightarrow \frac{\mathrm{dv}}{-{\mathrm{v}}^3}=\frac{\mathrm{dx}}{\mathrm{x}(1+{\mathrm{v}}^2)} \\ \Rightarrow \frac{(1+{\mathrm{v}}^2)}{-{\mathrm{v}}^3}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}} \end{gathered}$$

Step 3: Integrating on both sides,

$$\begin{gathered} \Rightarrow \int \frac{\left(1+{\mathrm{v}}^2\right)}{-{\mathrm{v}}^3}\mathrm{dv}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\ \Rightarrow \int \left(\frac{-1}{{\mathrm{v}}^3}-\frac{1}{\mathrm{v}}\right)\mathrm{dv}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\ \Rightarrow \frac{-1}{2}\frac{1}{{\mathrm{v}}^2}+\ln \mathrm{v}=-\ln \mathrm{x}+\mathrm{c} \\ \Rightarrow \frac{-{\mathrm{x}}^2}{2{\mathrm{y}}^2}=-\ln \mathrm{y}+\mathrm{c}\left[\because \mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}}\right] \end{gathered}$$

when x=1 , y=1 then,

$$\begin{gathered} \Rightarrow \frac{-1}{2\left(1\right)}=\ln \left(1\right)+\mathrm{c} \\ \Rightarrow \frac{-1}{2}=\mathrm{c} \\ \therefore \frac{-{\mathrm{x}}^2}{2{\mathrm{y}}^2}=-\ln \mathrm{y}-\frac{1}{2} \\ \Rightarrow {\mathrm{x}}^2={\mathrm{y}}^2(1+2\ln \mathrm{y}) \end{gathered}$$

At,

$$\begin{gathered} \Rightarrow \mathrm{y}\left(\mathrm{x}\right)=\mathrm{e} \\ \Rightarrow {\mathrm{x}}^2={\mathrm{e}}^2\left(3\right) \\ \Rightarrow \mathrm{x}=\pm \sqrt{3}\mathrm{e} \\ \therefore \mathrm{x}=\sqrt{3}\mathrm{e} \end{gathered}$$

Hence, the correct answer is option (A).