Question

If the solution of differential equation $x\ln x\frac{dy}{dx}+y=2\ln x$ is $y\left(\ln x\right)=(\ln x{)}^n+C$ then $n=$ ___

Answer 1

Type of the differential equation is what you have to identify. Here if you divide the given equation by $x\ln x$, it becomes $\frac{dy}{dx}+\frac{1}{x\ln x}\times y=\frac{2}{x}$ which when compared to standard form of linear differential equation matches it. Here $P\left(x\right)=\frac{1}{x\ln x}$

$Q\left(x\right)=\frac{2}{x}$
Solution will be $y\times I.F=\int Q\times (I.F.)dx+C$
$I.F.=e^{\int P\left(x\right)dx}$
So $I.F=\ln x$.
So general solution will be
$y\times \ln x=\int \frac{2}{x}\ln xdx+C$
$y\ln x=(\ln x{)}^2+C$
So $n=2$.