Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

Now integrate both sides

Answer 5

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.

Answer 6

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

Answer 7

If I remember correctly D:

Answer 8

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

Answer 9

so lets look at that part we haven't integrated

Answer 10

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

Answer 11

so far I have performed one round on integration by parts

Answer 12

Okay, so this was integration by parts once?

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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.

Answer 18

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

Answer 19

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.

Answer 20

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.

Answer 21

no it is not just one more

Answer 22

this thing looks more beastly to me

Answer 23

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

Answer 24

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.

Answer 25

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

Answer 26

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.

Answer 27

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.

Answer 28

oh you know what?

Answer 29

this isn't an elementary integral

Answer 30

so we can't use our elementary ways

Answer 31

and that is all i know

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

hey asnaseer thats not where we were having the problem

Answer 38

sorry - missed a minus sign

Answer 39

yes you need that lol

Answer 40

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

Answer 41

this is the part we got stuck on

Answer 42

and ike robin says we are getting the error function

Answer 43

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

Answer 44

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.

Answer 45

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...

Answer 46

by the way thanks for coming asanaseer

Answer 47

np

Answer 48

Thank you both for helping.

Answer 49

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

Answer 50

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

Answer 51

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.

Answer 52

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?

Answer 53

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.

Answer 54

ok - thx for the clarification - really appreciated.

Answer 55

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.

Answer 56

Yes - I tried that as well :)

Answer 57

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.

Answer 58

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)

Answer 59

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?