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

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

Grade:12th pass

1 Answers

Abhishek Singh
105 Points
2 years 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
\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})
\Rightarrow \frac{y}{x} = \frac{x^3}{3} -lnx +C
 

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