Question

If the integrating factor of the differential equation $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ is $x$, then $P\left(x\right)$ is

• A. $x$
• B. $\frac{x^2}{2}$
• C. $\frac{1}{x}$
• D. $\frac{1}{x^2}$
• E. $\frac{1}{2x}$

Answer 1

The correct option is C

$\frac{1}{x}$

Explanation for the correct option:

The given differential equation, $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$.

The integrating factor, $e^{\int P\left(x\right)dx}=x$.

$$\begin{gathered} \Rightarrow \int P\left(x\right)dx=\log \left(x\right) \\ \Rightarrow \frac{d}{dx}\left(\int P\left(x\right)dx\right)=\frac{d}{dx}\left(\log \left(x\right)\right) \\ \Rightarrow P\left(x\right)=\frac{1}{x} \end{gathered}$$

Therefore the value of $P\left(x\right)$ is $\frac{1}{x}$.

Hence, option (C) $\frac{1}{x}$ is the correct answer.