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anonymous
 4 years ago
Solve u'a*t*u=sin(t).
anonymous
 4 years ago
Solve u'a*t*u=sin(t).

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Assume 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
 4 years ago
Best ResponseYou've already chosen the best response.1Integrating 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
 4 years ago
Best ResponseYou've already chosen the best response.1so we can write \[(u e^{a \frac{t^2}{2}})'=e^{\frac{a t^2}{2}} \sin(t)\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1Now integrate both sides

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This 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
 4 years ago
Best ResponseYou've already chosen the best response.0Well 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
 4 years ago
Best ResponseYou've already chosen the best response.0If I remember correctly D:

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[u e^{\frac{a t^2}{2}}=\int\limits_{}^{}e^{\frac{a t^2}{2}} \sin(t) dt +C\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so lets look at that part we haven't integrated

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}e^{\frac{a t^2}{2}} \sin(t) dt=e^{\frac{a t^2}{2}}[\cos(t)]\int\limits_{}^{}[at] e^{\frac{a t^2}{2}} [\cos(t)] dt\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so far I have performed one round on integration by parts

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, so this was integration by parts once?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And you must integrate by parts again then on the remaining integral?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}e^{\frac{at^2}{2}} \sin(t) dt=\cos(t) e^{\frac{at^2}{2}}a \int\limits_{}^{} t e^{\frac{a t^2}{2}} \cos(t) dt\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1i just wanted to clean some stuff up yes we need to do it again lol

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1this is nasty looking but yes we need to do it again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright 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
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits\limits_{}^{}e^{\frac{at^2}{2}} \sin(t) dt=\cos(t) e^{\frac{at^2}{2}}a \int\limits\limits_{}^{} t e^{\frac{a t^2}{2}} \cos(t) dt \] \[= \cos(t) e^{\frac{at^2}{2}}a[\sin(t) t e^{\frac{a t^2}{2}}\int\limits_{}^{}\sin(t)[e^{\frac{a t^2}{2}}+t (at )e^{\frac{a t^2}{2}} ] dt]\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1I'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
 4 years ago
Best ResponseYou've already chosen the best response.0Thank 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
 4 years ago
Best ResponseYou've already chosen the best response.1no it is not just one more

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1this thing looks more beastly to me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, it's pretty bad. It's supposed to be ODE review in a PDE class.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.1i'm starting to think this isn't gonna work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.1this isn't an elementary integral

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so we can't use our elementary ways

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1and that is all i know

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not sure I know of any nonelementary ways to integrate.

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=integrate%28e%5E%28a+t%5E2%2F2%29+sin%28t%29%2Ct%29

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, well thank you for all of your time and help. I'll check out your link.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0I think you need to use an integrating factor here of:\[e^{\int{atdt}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh no, the error function again. This is what I originally was running into problems with.

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1hey asnaseer thats not where we were having the problem

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0sorry  missed a minus sign

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1yes you need that lol

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits\limits_{}^{}e^{\frac{a t^2}{2}} \sin(t) dt= \]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1this is the part we got stuck on

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1and ike robin says we are getting the error function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think this method is typically used for second order ODEs, but could we find the homogeneous and particular solutions?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0ah  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
 4 years ago
Best ResponseYou've already chosen the best response.1by the way thanks for coming asanaseer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you both for helping.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0yw  I'm off to do some research on this interesting problem  I'll be back if I find a solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Good luck, I've been working on this for hours!

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3The 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
 4 years ago
Best ResponseYou've already chosen the best response.0yes  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
 4 years ago
Best ResponseYou've already chosen the best response.3No, 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok  thx for the clarification  really appreciated.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3For 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yes  I tried that as well :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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. \[A = (at)/(1a ^{2}t ^{2})\] \[B = 1/ (1a ^{2}t ^{2}\] Plug back A and B into particular solution, add homogenous solution to get final solution. \[u =\pm e^{(1/2)at ^{2}} +(at)/(1a ^{2}t ^{2})(\sin t) + 1/(1a ^{2}t ^{2})(\cos t)+c\] Please correct any mistakes I made.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3No, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, 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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