Question

x(x-1)dy/dx-(x-2)y=x^2(2x-1)​

Answer 1

Step-by-step explanation:

We have,

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

Integrating factor =

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

$= {e}^{ - (2 ln(x) - ln(x - 1)) }$

$= \frac{ x - 1 }{ {x}^{2} }$

Now, required solution =

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

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

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

$\implies\frac{x - 1}{ {x}^{2} } .y = 2x - ln(x) + c$

$\implies \: y = \frac{2 {x}^{3} }{x - 1} - \frac{x ^{2} ln(x) }{x - 1} + \frac{cx^{2} }{x - 1}$