Question

Solution of the differential equation$\frac{dy}{dx}+\frac{y}{x}=\ln x$ is:
(where $C$ is integration constant)

• A. $y=\left(\ln x\right)-\frac{x}{4}+\frac{C}{x}$
• B. $y=\frac{x}{2}\left(\ln x\right)-x+\frac{C}{x}$
• C. $y=\frac{x}{2}\left(\ln x\right)-\frac{x}{4}+\frac{C}{x}$
• D. $y=\frac{x^2}{2}\left(\ln x\right)-\frac{x}{8}+\frac{C}{x}$

Answer 1

The correct option is C $y=\frac{x}{2}\left(\ln x\right)-\frac{x}{4}+\frac{C}{x}$

It is a linear differential equation of the form $\frac{dy}{dx}+Py=Q\left(x\right)$
Here, $P=\frac{1}{x},Q=\ln \left(x\right)$
The I.F.$=e^{\int Pdx}=e^{\int \frac{1}{x}dx}=e^{\ln |x|}=x$
Hence the general solution is
$y\left(\text{IF}\right)=\int Q\left(\text{IF}\right)dx+C$
$\Rightarrow yx=\int \left(\ln x\right)xdx+C$
On integrating By parts we get:
$\Rightarrow yx=\left(\ln x\right)\cdot \frac{x^2}{2}-\int \frac{1}{x}\cdot \frac{x^2}{2}dx+C$
$\Rightarrow yx=\frac{x^2}{2}\left(\ln x\right)-\frac{x^2}{4}+C$
$\Rightarrow y=\frac{x}{2}\left(\ln x\right)-\frac{x}{4}+\frac{C}{x}$