Question

$x(x-1)\frac{dy}{dx}-(x-2)y=x^3(2x-1)$.The solution to the above given differential equation is $y(x+k)=x^m(x^n-x+c)$. Find $k+m+n$ ?

Answer 1

Given $x(x-1)\frac{dy}{dx}-(x-2)y=x^3(2x-1)$
On rearranging the terms we get
$\frac{dy}{dx}-\frac{(x-2)y}{x(x-1)}=\frac{x^3(2x-1)}{x(x-1)}$
This is in the form of $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ which can be solved using integration factor(IF) method
IF$=e^{\int P\left(x\right)dx}=e^{\int -\frac{(x-2)}{x(x-1)}dx}=e^{\int (\frac{1}{x-1}-\frac{2}{x})dx}=e^{ln(x-1)-2lnx}=e^{ln\frac{(x-1)}{x^2}}=\frac{(x-1)}{x^2}$
Solution is in the form of $y.e^{Pdx}=\left[Q{e}^{Pdx}\right]dx+C$
$\frac{(x-1)}{x^2}y=\int (\frac{x^3(2x-1)}{x(x-1)}\times \frac{(x-1)}{x^2})dx+c$
$\Rightarrow (x-1)y=x^2(\int (2x-1)dx+c)$
$\Rightarrow (x-1)y=x^2(x^2-x+c)$
$k=-1$, $m=2$, $n=2$
$\Rightarrow k+m+n=3$