Question

The solution of the differential equation $\mathrm{ydx}-\mathrm{xdy}+\left(\mathrm{logx}\right)\mathrm{dx}=0$ is

• A. $\mathrm{y}=\log \mathrm{x}+\mathrm{cx}$
• B. $\mathrm{y}+1+\mathrm{logx}=\mathrm{cx}$
• C. $\mathrm{y}+\mathrm{cx}=\log \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.\right)$
• D. None of these

Answer 1

The correct option is B

$\mathrm{y}+1+\mathrm{logx}=\mathrm{cx}$

The explanation for the correct answer.

Step 1: Simplify the given equation.

Given: $\mathrm{ydx}-\mathrm{xdy}+\left(\mathrm{logx}\right)\mathrm{dx}=0$

Divide the given equation by $\mathrm{xdx}$

$$\begin{gathered} \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}-\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{logx}=0 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}+\mathrm{logx}=0 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=-\mathrm{logx} \end{gathered}$$

This equation is now a linear differential equation.

Step 2: Find the Integrating Factor.

Integrating factor $={\mathrm{e}}^{\int \mathrm{Pdx}}$

$$\begin{gathered} ={\mathrm{e}}^{\int \frac{-1}{\mathrm{x}}\mathrm{dx}} \\ ={\mathrm{e}}^{-\mathrm{lnx}}\left[\int \frac{1}{\mathrm{x}}\mathrm{dx}=\mathrm{lnx}\right] \\ ={\mathrm{e}}^{{\mathrm{lnx}}^{-1}}\left[{\mathrm{e}}^{-\mathrm{a}}={\mathrm{e}}^{{\mathrm{a}}^{-1}}\right] \\ =\frac{1}{\mathrm{x}}\left[{\mathrm{e}}^{{\mathrm{lnb}}^{-1}}=\frac{1}{\mathrm{b}}\right] \end{gathered}$$

Step 3: Find the solution.

The solution is given by $\mathrm{y}·\mathrm{IF}=\int \left(\mathrm{Q}.\mathrm{IF}\right)\mathrm{dx}+\mathrm{C}$

$\Rightarrow \mathrm{y}·\frac{1}{\mathrm{x}}=\int \left(\frac{1}{\mathrm{x}}·\frac{\mathrm{logx}}{\mathrm{x}}\right)\mathrm{dx}+\mathrm{C}$

Take $\mathrm{logx}=\mathrm{t}\Rightarrow \mathrm{x}={\mathrm{e}}^{\mathrm{t}}$

$\frac{1}{\mathrm{x}}\mathrm{dx}=\mathrm{dt}$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\int {\mathrm{te}}^{-\mathrm{t}}\mathrm{dt}+\mathrm{C}$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{t}\int {\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}-\int \left(\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{t}\right).\int {\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\right)\mathrm{dt}$ [ Integrating using by parts]

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=-{\mathrm{te}}^{-\mathrm{t}}-\int -{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=-{\mathrm{te}}^{-\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}+\mathrm{C}$ [ Put value of $t,e^{-t}$]

$$\begin{gathered} \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{-\mathrm{logx}}{\mathrm{x}}-\frac{1}{\mathrm{x}}+\mathrm{C} \\ \Rightarrow \mathrm{y}=\mathrm{x}\left(\frac{-\mathrm{logx}}{\mathrm{x}}-\frac{1}{\mathrm{x}}+\right)+\mathrm{Cx} \\ \Rightarrow \mathrm{y}=-\mathrm{logx}-1+\mathrm{Cx} \\ \Rightarrow \mathrm{Cx}=\mathrm{y}+\mathrm{logx}+1 \end{gathered}$$

Hence option $B$ is the correct answer.