Question

Integrating factor of $x\frac{dy}{dx}-y=x^4-3x$ is

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

Answer 1

Answer: (C)

$\frac{1}{x}$

We have, $x\frac{dy}{dx}-y=x^4-3x$

$\frac{dy}{dx}-\frac{1}{x}y=x^3-3$

Above equation is of the form $\frac{dy}{dx}+Py=Q$

Hence Integrating factor $=e^{\int Pdx}=e^{\int (-\frac{1}{x})dx}=e^{-\ln x}=e^{\ln x^{-1}}=x^{-1}=\frac{1}{x}$

$I.F=\frac{1}{x}$

Since $a^{{\log }_{a}x}=x$

$\Rightarrow y\left(\frac{1}{x}\right)=\int \frac{1}{x}\times (x^4-3)dx$

$\Rightarrow \frac{y}{x}=\int (x^3-\frac{3}{x})dx$

$\Rightarrow \frac{y}{x}=(\frac{x^4}{4}-3\ln x)+c$

$\Rightarrow y=\frac{x^3}{4}-3x\ln x+cx$