Solve u'-a*t*u=sin(t).

- anonymous

Solve u'-a*t*u=sin(t).

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- anonymous

Assume a is a constant and u is a function of t.
I'm using an integrating factor (e^((-at^2)/2), which is causing me problems. Specifically, I end up needing to integrate and integral that involves the error function as a solution, which I feel must be wrong.

- myininaya

Integrating factor is
\[v=e^{\int\limits_{}^{}-a t dt }=e^{-a \frac{t^2}{2}} \text{ we didn't need the +constant here}\]
So we have
\[e^{-a \frac{t^2}{2}} u'-e^{\frac{-a t^2}{2}} a t u=e^{\frac{-a t^2}{2}} \sin(t)\]

- myininaya

so we can write
\[(u e^{-a \frac{t^2}{2}})'=e^{-\frac{a t^2}{2}} \sin(t)\]

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## More answers

- myininaya

Now integrate both sides

- anonymous

This makes sense so far, but I don't know how to integrate the right side. I've tried integrating by parts, but that's not working.

- anonymous

Well for an equation:
\[u'(t)+p(t)u(t)=g(t)\]
The solution is given by:
\[u(t)=e^{\int\limits p(t)dt} \int\limits e^{-\int\limits p(t)dt}g(t)dt+Ce^{-\int\limits p(t)dt}\]

- anonymous

If I remember correctly D:

- myininaya

\[u e^{\frac{-a t^2}{2}}=\int\limits_{}^{}e^{\frac{-a t^2}{2}} \sin(t) dt +C\]

- myininaya

so lets look at that part we haven't integrated

- myininaya

\[\int\limits_{}^{}e^{\frac{-a t^2}{2}} \sin(t) dt=e^{\frac{-a t^2}{2}}[-\cos(t)]-\int\limits_{}^{}[-at] e^{\frac{-a t^2}{2}} [-\cos(t)] dt\]

- myininaya

so far I have performed one round on integration by parts

- anonymous

Okay, so this was integration by parts once?

- anonymous

And you must integrate by parts again then on the remaining integral?

- myininaya

\[\int\limits_{}^{}e^{-\frac{at^2}{2}} \sin(t) dt=-\cos(t) e^{-\frac{at^2}{2}}-a \int\limits_{}^{} t e^{-\frac{a t^2}{2}} \cos(t) dt\]

- myininaya

i just wanted to clean some stuff up
yes we need to do it again lol

- myininaya

this is nasty looking but yes we need to do it again

- anonymous

Alright this is making sense, I think I gave up too easy after a single round of integration by parts was looking like it was going nowhere.

- myininaya

\[\int\limits\limits_{}^{}e^{-\frac{at^2}{2}} \sin(t) dt=-\cos(t) e^{-\frac{at^2}{2}}-a \int\limits\limits_{}^{} t e^{-\frac{a t^2}{2}} \cos(t) dt \]
\[=- \cos(t) e^{-\frac{at^2}{2}}-a[\sin(t) t e^{-\frac{a t^2}{2}}-\int\limits_{}^{}\sin(t)[e^{-\frac{a t^2}{2}}+t (-at )e^{\frac{-a t^2}{2}} ] dt]\]

- myininaya

I'm sorry this getting to messey to do for me on here.
I think I will do this on paper and scan it for you.

- anonymous

Thank you for going through all this work. I might be able to do it myself if it's only one more round of integration by parts.

- myininaya

no it is not just one more

- myininaya

this thing looks more beastly to me

- anonymous

Yeah, it's pretty bad. It's supposed to be ODE review in a PDE class.

- anonymous

I took ODE's over 2 years ago, so I'm a little rusty. I thought maybe some other method besides integration by parts may work better.

- myininaya

i'm starting to think this isn't gonna work

- anonymous

I think it should integrate by some means, since I'm also supposed to sketch several members of the family of solutions to the ODE.

- anonymous

I don't have a problem sketching solutions, but I think that the integral must go away for me to be able to sketch solutions.

- myininaya

oh you know what?

- myininaya

this isn't an elementary integral

- myininaya

so we can't use our elementary ways

- myininaya

and that is all i know

- anonymous

I'm not sure I know of any non-elementary ways to integrate.

- myininaya

http://www.wolframalpha.com/input/?i=integrate%28e%5E%28-a+t%5E2%2F2%29+sin%28t%29%2Ct%29

- anonymous

Okay, well thank you for all of your time and help. I'll check out your link.

- asnaseer

I think you need to use an integrating factor here of:\[e^{\int{atdt}}\]

- anonymous

Oh no, the error function again. This is what I originally was running into problems with.

- myininaya

hey asnaseer thats not where we were having the problem

- asnaseer

sorry - missed a minus sign

- myininaya

yes you need that lol

- myininaya

\[\int\limits\limits_{}^{}e^{\frac{-a t^2}{2}} \sin(t) dt= \]

- myininaya

this is the part we got stuck on

- myininaya

and ike robin says we are getting the error function

- anonymous

I think this method is typically used for second order ODEs, but could we find the homogeneous and particular solutions?

- anonymous

I know the result should not involve the error function, sice II need to graph a family of curves, and I don't think the error function can be graphed.

- asnaseer

ah - ok - I see what you mean now - I've followed through the discusion above. Let me see if there is another approach that can be used here...

- myininaya

by the way thanks for coming asanaseer

- asnaseer

np

- anonymous

Thank you both for helping.

- asnaseer

yw - I'm off to do some research on this interesting problem - I'll be back if I find a solution

- anonymous

Good luck, I've been working on this for hours!

- JamesJ

The integral on the RHS doesn't have an expression in terms of elementary functions. If you want to deal with it all and get it beyond the form
\[ \int e^{-at^2/2} \sin t \ dt \ \ \ \ --(*)\]
then write
\[ \sin t = \frac{1}{2i}( e^{it} - e^{-it} ) \]
complete the square, and you're left with error functions involving complex numbers.
I'm not convinced that's a gain. I would probably just leave the integral (*) as it is.

- asnaseer

yes - I tried that approach as well and ended up with the error function. I was just wondering if there are other approaches to solving equations like this?

- JamesJ

No, not really. This integral isn't too bad.
Part of dealing with ODEs and PDEs is learning what sorts of expressions can and cannot be integrated. This one can't.

- asnaseer

ok - thx for the clarification - really appreciated.

- JamesJ

For what it's worth, even if you tried a different technique here such as Laplace transform, when you went to invert the solution, you'd find you had an integral equivalent to this one.

- asnaseer

Yes - I tried that as well :)

- anonymous

For anyone who comes across a problem like this in the future, I found a way to solve the differential equation, without having to integrate ∫e−at2/2sint dt.
General solution of differential equation = homogenous solution + particular solution.
Find homogenous solution:
\[u'-a*t*u = 0\]
This is separable.
\[du/dt =a*t*u\]
\[(1/u) du = (a*t) dt\]
\[\int\limits_{}^{}(1/u) du = \int\limits_{}^{} (at) dt\]
\[\ln \left| u \right|+c = (1/2) a *t^{2}+c\]
\[\left| u \right|=\exp((1/2)a*t ^{2}) + c\]
\[u=\pm\exp((1/2)a*t ^{2}) + c\]
Find the particular solution, using undetermined coefficients:
\[p(t) = A \sin t + B \cos t\]
\[p'(t) = A \cos t - B \sin t\]
\[u'-a*t*u = \sin t\]
\[A \cos t - B \sin t - a * t* (A \sin t + B \cos t) = \sin t\]
Group like terms and factor to get system of equations:
\[A - a*t*B = 0\]
\[-B - a*t*A =1\]
Solve for A and B.
\[A = (at)/(1-a ^{2}t ^{2})\]
\[B = 1/ (1-a ^{2}t ^{2}\]
Plug back A and B into particular solution, add homogenous solution to get final solution.
\[u =\pm e^{(1/2)at ^{2}} +(at)/(1-a ^{2}t ^{2})(\sin t) + 1/(1-a ^{2}t ^{2})(\cos t)+c\]
Please correct any mistakes I made.

- JamesJ

No, that's definitely not right because
(a) this solution doesn't satisfy the differential equation
(b) it falls apart when take the derivative. If the coefficients A and B are functions of t, then they must have non zero derivative, vs. the way you treat them above when you write down p'(t)

- anonymous

Ah, I wasn't sure if undetermined coefficients would work or not in this case. I see why it doesn't work now. Is there another way to find the particular solution then without using undetermined coefficients?

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