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

  • Thread starter ascheras
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In summary: The three equations are: u'et+ v'cos(t)+ w'sin(t)= 0 u'et- v'sin(t)+ w'cos(t)= 0 u'et- v'cos(t)- w'sin(t)= e-tsint
  • #1
ascheras
14
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Use the method of variation of parameters to determine the general solution of the given DE:

y''' - y" + y' -y= e^(-t)sint
 
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  • #2
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
 
  • #3
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.
 
  • #4
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
 
  • #5
The general solution

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

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  • #6
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.
 
  • #7
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

[tex]S = <e^t, \cos{t}, \sin{t}>[/tex]

The Wronskian is given by

[tex]\textrm{det}(\left| \begin{array}{cccc}
S_1 & S_2 & \cdots & S_n \\
S_1' & S_2' & \cdots & S_n' \\
\vdots & \vdots & \ddots & \vdots \\
S_1^{(n)} & S_2^{(n)} & \cdots & S_n^{(n)}
\end{array} \right|) [/tex]

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

cookiemonster
 
  • #8
I take it that the 3 equations you are referring to are

u'et+ v'cos(t)+ w'sin(t)= 0
u'et- v'sin(t)+ w'cos(t)= 0 and
u'et- 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.
 

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model various physical and natural phenomena in fields such as physics, engineering, and economics.

What is a variation parameter?

A variation parameter is a constant that is added to a particular solution of a differential equation in order to find a more general solution that satisfies the equation. It allows for a family of solutions instead of just one specific solution.

What is the process for solving a differential equation using variation parameters?

The process for solving a differential equation using variation parameters involves finding a particular solution to the equation, adding a variation parameter to this solution, and then solving for the variation parameter using initial conditions or boundary conditions.

What is the general solution to the differential equation "y''' - y" + y' -y= e^(-t)sint"?

The general solution to this differential equation is y = C1e^t + C2e^-t + C3cos(t) + C4sin(t) + e^(-t)sint, where C1, C2, C3, and C4 are arbitrary constants.

How can the general solution be verified?

The general solution can be verified by plugging it back into the original differential equation and confirming that it satisfies the equation. It can also be verified by checking that it satisfies any given initial or boundary conditions.

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