Question

The integrating factor of the differential equation $x\log \left(x\right)\frac{dy}{dx}+y=2\log \left(x\right)$ is given by

• A. $e^x$
• B. $\log \left(x\right)$
• C. $\log \left(\log x\right)$
• D. $x$

Answer 1

The correct option is B

$\log \left(x\right)$

Find the integrating factor of the given differential equation

Given the differential equation, $x\log \left(x\right)\frac{dy}{dx}+y=2\log \left(x\right)$.

$\Rightarrow \frac{dy}{dx}+\frac{1}{x\log \left(x\right)}y=\frac{2}{x}$.

Compare the differential equation with the general form of the linear differential equation $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{P}\mathit{y}\mathbf{=}\mathit{Q}$.

Here, $P$ and $Q$ are functions of $x$.

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

So, the integrating factor of the given differential equation can be provided by, $R=e^{\int Pdx}$.

$\Rightarrow R=e^{\int \left(\frac{1}{x\log \left(x\right)}\right)dx}$.

Let us assume that, $\log \left(x\right)=z$.

Differentiate both sides of the equation.

$\frac{dx}{x}=dz$.

Hence, $R=e^{\int \left(\frac{dz}{z}\right)}$

$$\begin{gathered} \Rightarrow R=e^{\log \left(z\right)} \\ \Rightarrow R=z \\ \Rightarrow R=\log \left(x\right)\left[\because \log \left(x\right)=z\right] \end{gathered}$$

Therefore, the integrating factor of the differential equation $x\log \left(x\right)\frac{dy}{dx}+y=2\log \left(x\right)$ is $\log \left(x\right)$, so, the correct answer is option (B).