# Solve this Non exact differential equations(x^4-x+y)dx-xdy=0

Abhishek Singh
105 Points
one year ago
$(x^4 - x+y)dx -xdy = 0$
$\Rightarrow (x^4-x)dx + (ydx-xdy) = 0$
Dividing by x^2 we get
$\Rightarrow (x^2 - \frac{1}{x})dx + \frac{(ydx-xdy)}{x^2} = 0$
This is in exact form, Integrate it to get the answer
$\inline \Rightarrow \frac{x^3}{3}-ln(x) - y/x = C$

Alternatively, This can be solved as Linear Differential Equation(LDE)
Rearrange the terms to get equation as
$\frac{dy}{dx} -\frac{y}{x}= x^3 -1$
$\Rightarrow Integrating Factor(I.F) = e^{-\int \frac{1}{x}dx} = \frac{1}{x}$
Thus Solution is ==> y (IF ) = ∫Q*IF dx
$y*\frac{1}{x} = \int (x^3-1)*(\frac{1}{x})$
$\inline \Rightarrow \frac{y}{x} = \frac{x^3}{3} -lnx +C$