Question

$x(x-1)\frac{dy}{dx}-(x-2)y=x^2(2x-1)$

Answer 1

Given,

$x(x-1)\frac{dy}{dx}-(x-2)y=x^2(2x-1)$

$\frac{dy}{dx}-\frac{x-2}{x(x-1)}y=\frac{x^2(2x-1)}{x(x-1)}$

$\frac{dy}{dx}+\left(\frac{2-x}{x(x-1)}\right)y=\frac{x^2(2x-1)}{x(x-1)}$

$\frac{dy}{dx}+Py=Q$

$P=\frac{2-x}{x(x-1)},Q=\frac{x^2(2x-1)}{x(x-1)}$

$I.F=e^{\int Pdx}$

$I.F=e^{\int \frac{2-x}{x(x-1)}dx}=e^{\log (x-1)-2\log x}=e^{\log \frac{x-1}{x^2}}=\frac{x-1}{x^2}$

$y\times I.F=\int Q\times I.F\times dx$

$y\times \frac{x-1}{x^2}=\int \frac{x^2(2x-1)}{x(x-1)}\times \frac{x-1}{x^2}dx$

$y\frac{x-1}{x^2}=\int \frac{2x-1}{x}dx$

$y\frac{x-1}{x^2}=2x-\log x+c$

$y(x-1)=2x^3-x^2\log x+c{x}^2$