Question

If $\mathit{C}\mathbf{}$is an arbitrary constant, then the general solution of the differential equation $\mathit{y}\mathit{d}\mathit{x}\mathbf{-}\mathit{x}\mathit{d}\mathit{y}\mathbf{=}\mathit{x}\mathit{y}\mathit{d}\mathit{x}$ is given by

• A. $\mathit{y}\mathbf{=}\mathit{C}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$
• B. $\mathit{y}\mathbf{=}\mathit{C}\mathit{y}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$
• C. $\mathit{y}\mathbf{+}{\mathit{e}}^{\mathbf{x}}\mathbf{=}\mathit{C}\mathit{x}$
• D. $\mathbf{}\mathit{y}{\mathit{e}}^{\mathbf{x}}\mathbf{=}\mathit{C}\mathit{x}$

Answer 1

The correct option is A

$\mathit{y}\mathbf{=}\mathit{C}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$

Explanation for the correct options:

Step1 : Expressing the given data:

$\mathit{y}\mathit{d}\mathit{x}\mathbf{-}\mathit{x}\mathit{d}\mathit{y}\mathbf{=}\mathit{x}\mathit{y}\mathit{d}\mathit{x}$

$$\begin{gathered} ydx-xydx=xdy \\ y(1-x)dx=xdy \\ \frac{1-x}{x}dx=\frac{dy}{y} \end{gathered}$$

Step2 : Integrating Both Sides

$\begin{array}{rcl}\int \left(\frac{1-x}{x}\right)dx& =& \int \frac{dy}{y}\\ & \Rightarrow & {\log }_{e}x-x={\log }_{e}y+c\\ \Rightarrow {\log }_e\frac{x}{y}& =& x+c\\ \Rightarrow \frac{x}{y}& =& e^c{e}^x\end{array}$

$\Rightarrow y=e^{c}x{e}^{-x}$

Suppose $c=e^c$

then,

$\mathit{y}\mathbf{=}\mathit{c}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$

Hence, option (A) is correct answer.