Solving the Differential Equation e2xy' + 2e2xy = (6x + 5)e2x

[Icebreaker]

y' + 2y = 6x + 5

Can I solve it by...

\frac{dy}{dx} + 2y = 6x + 5

dy = (6x + 5 - 2y)dx

y = 3x^2 + 5x - 2yx

y + 2yx = 3x^2 + 5x

y = \frac{3x^2 + 5x}{1 + 2x}

 

[arildno]

No you can't.
1) Why do you think this works?
2) Have you checked if it works?

 

[dextercioby]

Nope.It's a nonhomogenous I-st order linear equation with constant coefficients.Find an integrating factor.

Daniel.

 

[Icebreaker]

Damn...

 

[whozum]

I think it would be e^{2t} [/tex], right?

 

[arildno]

That would be the integrating factor, yes.

 

[HallsofIvy]

Well, e^2x, actually. \frac{d(e^{2x}y)}{dx}= 2e^{2x}y+ e^{2x}y' so multiplying the entire equation by e^2x gives e^2xy'+ 2e^2xy= (e^2xy)'= (6x+5)e^2x. Integrating the left side gives e^2xy and the right side can be integrated by parts.

By the way, the reason your first method fails is that the integral of -2ydx is not -2xy because y is itself a function of x.