ODE Solution (Integrating Factor method)

[Angry Citizen]

Homework Statement

$$y^{'}+3y=t+e^{-2t}$$

Homework Equations

$$(μy)'=μy'+yμ'$$

The Attempt at a Solution

First I found the integrating factor: $$\frac{dμ}{dt}=3μ$$ which becomes $$μ=e^{3t}$$

I multiplied through by $$e^{3t}$$, yielding $$y'e^{3t}+3e^{3t}y=te^{3t}+e^{t}$$

I combined the LHS into $$(ye^{3t})'$$, then integrated both sides. The LHS became $$ye^{3t}$$ and the RHS required integration by parts, with $$u=t$$ $$du=1$$ $$dv=e^{3t}$$ $$v=\frac{e^{3t}}{3}$$

The part requiring integration by parts wound up as: $$\frac{te^{3t}}{3}+\frac{e^{3t}}{9}$$

The entire RHS becomes: $$\frac{te^{3t}}{3}+\frac{e^{3t}}{9}+e^{t}+c$$

The equation is now: $$ye^{3t}=\frac{te^{3t}}{3}+\frac{e^{3t}}{9}+e^{t}+c$$

Dividing through by $$e^{3t}$$ SHOULD yield $$y=\frac{t}{3}+\frac{1}{9}+e^{-2t}+ce^{-3t}$$

However, the solution's manual doesn't have the $$e^{-2t}$$ term, and I'm afraid I may have made a mistake. Can anyone help? This is my first differential equation, and I'm determined to get it right :)

 

[tiny-tim]

Hi Angry Citizen!

Must be a misprint … your e^-2t is definitely correct.

(But your 1/9 isn't. )

However, an integrating factor is not the way we'd normally do this, it's quicker easier and safer to use the standard method of finding the homgeneous solution and a particular solution.

 

[Angry Citizen]

But your 1/9 isn't.

That's interesting. The manual has that one. Could you tell me what's wrong with it? Should it not be there at all? Did I mess up the integration by parts, perhaps?

However, an integrating factor is not the way we'd normally do this, it's quicker easier and safer to use the standard method of finding the homgeneous solution and a particular solution.

Hmm. Well, like I said, this is my very first differential equation :) I started at chapter 2 of the Boyce-Diprima book, and am working my way through it sequentially. First section was integrating factors, so my repertoire of solution-finding is limited :)

 

[tiny-tim]

hi Angry Citizen!

That's interesting. The manual has that one. Could you tell me what's wrong with it? Should it not be there at all? Did I mess up the integration by parts, perhaps?

What's wrong is that it doesn't work …

(t/3 + 1/9)' + 3(t/3 + 1/9) = 1/3 + t + 1/3 ≠ t

Hmm. Well, like I said, this is my very first differential equation :) I started at chapter 2 of the Boyce-Diprima book, and am working my way through it sequentially. First section was integrating factors, so my repertoire of solution-finding is limited :)

Fair enough!

 

[Herr Malus]

I did my solution via the homgeneous + particular solution as well, so I'm not sure where your mistake is, but I got a -1/9 in my solution.

EDIT: Also, your mistake is in the integration by parts, the result you need is -integral(v*du), you have integral(v*du). So you should get t/3*exp(3t)-1/9*exp(3t).

 

[Angry Citizen]

Also, your mistake is in the integration by parts, the result you need is -integral(v*du), you have integral(v*du). So you should get t/3*exp(3t)-1/9*exp(3t).

Aha! Well, I guess I need to bone up on my integration techniques. I was thinking it was uv+int(vdu). Bloody sign changes!