Question

The solution of $ydx-xdy+\log xdx=0$ is

• A. $y-\log x-1=Cx$
• B. $x-\log y+1=Cx$
• C. $y+\log x+1=Cx$
• D. $y+\log x-1=Cx$

Answer 1

The correct option is C

$y+\log x+1=Cx$

The explanation of the correct option:

Step-1: Converting equation into linear differential form:

The given differential equation: $ydx-xdy+\log xdx=0$.

$$\begin{gathered} \Rightarrow ydx+\log xdx=xdy \\ \Rightarrow \left(y+\log x\right)dx=xdy \\ \Rightarrow \left(y+\log x\right)=x\frac{dy}{dx} \\ \Rightarrow \frac{y}{x}+\frac{\log x}{x}=\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}-\frac{1}{x}y=\frac{\log x}{x} \end{gathered}$$

Step-2 Solution of linear differential equation:

Compare the derived equation with $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$.

Thus, $P\left(x\right)=-\frac{1}{x}$ and $Q\left(x\right)=\frac{\log x}{x}$.

Thus, the integrating factor can be given by, $R=e^{\int P\left(x\right)dx}$

$\begin{array}{rcl}& \Rightarrow & R=e^{\int -\frac{1}{x}dx}\\ & \Rightarrow & R=e^{-\log x}\\ & \Rightarrow & R=\frac{1}{e^{\log x}}\\ & \Rightarrow & R=\frac{1}{x}\end{array}$

Consider the equation: $\frac{dy}{dx}-\frac{1}{x}y=\frac{\log x}{x}$

Multiply both sides of the equation by the integrating factor.

$$\begin{gathered} \Rightarrow \frac{1}{x}\frac{dy}{dx}-\left(\frac{1}{x}\times \frac{1}{x}y\right)=\frac{\log x}{x}\times \frac{1}{x} \\ \Rightarrow \frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}=\frac{\log x}{x^2} \\ \Rightarrow \frac{d}{dx}\left(\frac{y}{x}\right)=\frac{\log x}{x^2} \\ \Rightarrow d\left(\frac{y}{x}\right)=\frac{\log x}{x^2}dx \end{gathered}$$

Integrate both sides of the equation.

$$\begin{gathered} \Rightarrow \int d\left(\frac{y}{x}\right)=\int \frac{\log x}{x^2}dx \\ \Rightarrow \frac{y}{x}=\log x\int \frac{dx}{x^2}-\int \left[\frac{d}{dx}\left(\log x\right)\int \frac{dx}{x^2}\right]dx\mathbf{[}\mathbf{\because }\mathbf{\int }\mathbf{u}\mathbf{·}\mathbf{v}\mathbf{d}\mathbf{x}\mathbf{=}\mathbf{u}\mathbf{\int }\mathbf{v}\mathbf{d}\mathbf{x}\mathbf{-}\mathbf{\int }\left(\frac{du}{dx}\int vdx\right)\mathbf{d}\mathbf{x}\mathbf{]} \\ \Rightarrow \frac{y}{x}=\log x\left(-\frac{1}{x}\right)-\int \left[\frac{1}{x}\left(-\frac{1}{x}\right)\right]dx \\ \Rightarrow \frac{y}{x}=-\frac{\log x}{x}+\int \frac{dx}{x^2} \\ \Rightarrow \frac{y}{x}=-\frac{\log x}{x}-\frac{1}{x}+C\mathbf{}\left[C=\mathrm{Integrating}\mathrm{Constant}\right] \\ \Rightarrow y=-\log x-1+Cx \\ \Rightarrow y+\log x+1=Cx \end{gathered}$$

Therefore, the solution of the differential equation $ydx-xdy+\log xdx=0$ is $y+\log x+1=Cx$.

Hence, option C is correct.