Solve DE: Use Variation Parameters - y''' - y" + y' -y= e^(-t)sint

[ascheras]

Use the method of variation of parameters to determine the general solution of the given DE:

y''' - y" + y' -y= e^(-t)sint

 

[cookiemonster]

First you have to find the solution to the homogeneous differential equation

y''' - y'' + y' - y = 0

Then you learned some formulas in your class that involves these solutions.

If you still need help, do at least the first step and post back your progress.

cookiemonster

 

[ascheras]

Ha, i wish i was taught in class.

i get u1e^t + u2cost + u3sint = Y(t)

the i get the 3X3 system of equations with the final row equalling e^(-t) sint.

I just don't know what to do with it.

 

[cookiemonster]

I think we've got different notation here.

What's u1, u2, u3 and Y(t)?

No class? Then it's in either a book or on the internet.

cookiemonster

 

[Max0526]

The general solution

Hi;
This is the general solution for the equation (see LinEq.gif).
Good luck,
Max.

 

[ascheras]

i have the answers in the back of my book. the problem is that i don't understand how to get the wronskian at t=0.

this is a variation of parameters of higher order DE (cookiemonster). Y(t) is the general solution to the homogeneous part of the problem.

 

[cookiemonster]

It's very nice that you can demonstrate you know how to use Maple, Max, but it doesn't tell you a thing about how to solve the problem.

Okay, ascheras. u1, u2, and u3 are constants, correct?

So our homogeneous solution set is

S = <e^t, \cos{t}, \sin{t}>

The Wronskian is given by

\textrm{det}(\left| \begin{array}{cccc}<br /> S_1 &amp; S_2 &amp; \cdots &amp; S_n \\<br /> S_1&#039; &amp; S_2&#039; &amp; \cdots &amp; S_n&#039; \\<br /> \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\<br /> S_1^{(n)} &amp; S_2^{(n)} &amp; \cdots &amp; S_n^{(n)}<br /> \end{array} \right|)

Use this for n = 3 to calculate the Wronskian. Do you need me to go through more steps?

cookiemonster

 

[HallsofIvy]

I take it that the 3 equations you are referring to are

u'e^t+ v'cos(t)+ w'sin(t)= 0
u'e^t- v'sin(t)+ w'cos(t)= 0 and
u'e^t- v'cos(t)- w'sin(t)= e^-tsint

The whole point is that those are three linear equations for u', v' and w' and can be solved by, for example: multiply the first equation by sin(t), the second equation by cos(t) and add to eliminate v'. Then add the first and third equations to eliminate v' so that you have two equations in u' and w'. Now eliminate w', etc. That's just algebra. After you have found u', v', w' separately, integrate to get u, v, w.