Question

Integrating factor of $\frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2}{x^2}$ is:

• A. $e^x$
• B. $\log x$
• C. $\frac{1}{x}$
• D. $e^{-x}$

Answer 1

Answer: (B)

$\log x$

Comparing given differential equation with standard linear first order linear differential equation $\frac{dy}{dx}+Py=Q$

We get $P=\frac{1}{x\ln x}$

Thus integrating factor $=e^{\int Pdx}=e^{\int \frac{1}{x\ln x}dx}=e^{\int \frac{d\left(\ln x\right)}{\ln x}}=e^{\ln \left(\ln x\right)}=\ln x$

Note: $a^{{\log }_{a}x}=x$