Question

The integrating factor of the differential equation $\left(y\log y\right)dx=\left[\log \left(y\right)-x\right]dy$ is

• A. $\frac{1}{\log y}$
• B. $\log \left(\log y\right)$
• C. $1+\log \left(y\right)$
• D. $\frac{1}{\log \left(\log y\right)}$
• E. $\log \left(y\right)$

Answer 1

The correct option is E

$\log \left(y\right)$

Explanation for the correct options:

Find the integrating factor of the given differential equation

Given the differential equation, $\left(y\log y\right)dx=\left[\log \left(y\right)-x\right]dy$.

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

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

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

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

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

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

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

Differentiate both sides of the equation.

$\frac{dy}{y}=dz$.

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

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

Therefore, the integrating factor of the differential equation $\left(y\log y\right)dx=\left[\log \left(y\right)-x\right]dy$ is $\log \left(y\right)$.