Integrating Factor for First Order Linear Differential Equation

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SUMMARY

The discussion focuses on finding an integrating factor for the first order linear differential equation given by \(\frac{dy}{dx} - \frac{y}{x} = xe^{2x}\). The integrating factor identified is \(e^{-\ln x} = x^{-1}\). By multiplying the equation by this integrating factor, the left-hand side becomes the exact derivative of \(\frac{y}{x}\), leading to the simplified equation \(\frac{d(y/x)}{dx} = e^{2x}\). The next step involves integrating both sides with respect to \(x\) to find the general solution.

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Homework Statement


Find an integrating factor for the first order linear differential equation
[itex]\frac{dy}{dx} - \frac{y}{x} = xe^{2x}[/itex]
and hence find its general solution

Homework Equations





The Attempt at a Solution


I found the integrating factor which is [itex]e^{-lnx} = x^{-1}[/itex]

and multiplying the equation with the integrating factor, will result in:
[itex]\frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = e^{2x}[/itex]

how do I go on from here?
 
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hi wowmaths! :smile:

the LHS is the exact derivative of … ? :wink:
 
Do you know why you found the "integrating factor"?

The whole point of an integrating factor for [itex]dy/dx+ a(x)y= f(x)[/itex] is that, with integrating factor [itex]\mu(x)[/itex], we will have
[tex]\mu(x)\frac{dy}{dx}+ \mu(x)a(x)y= \frac{d(\mu(x)y}{dx}= \mu(x)f(x)[/tex]

If [itex]\mu(x)= 1/x[/itex] here (I have not checked that) then your equation should reduce to
[tex]\frac{d(y/x)}{dx}= e^{2x}[/tex]
Integrate both sides of that with respect to x.
 

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