Variation of parameters for higher order linear eq

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Homework Help Overview

The discussion revolves around finding the general solution of a fourth-order linear differential equation using the method of variation of parameters. The equation in question is y^(4) + 2y'' + y = sin(t).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the Wronskian for the functions involved and express confusion over obtaining a value of zero, which raises questions about the correctness of their approach. There is also mention of alternative methods, such as undetermined coefficients, for finding the particular solution.

Discussion Status

The discussion is ongoing, with participants exploring different methods to find the particular solution. Some guidance has been offered regarding the use of variation of parameters and the simplification of the problem, but there is no consensus on the best approach yet.

Contextual Notes

Participants are grappling with the implications of a zero Wronskian and whether the use of variation of parameters is necessary for this problem. There is an indication of imposed homework rules that may limit the methods available for solving the equation.

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Homework Statement


Use the method of variation of parameters to determine the general solution of the given differential equation: y^(4) + 2y'' + y = sin(t)


Homework Equations


characteristic equation is factored down to (r^2 + 1)^2, so r = +/- i. this gives the general solution to be y(t) = c1*cos(t) + c2*sin(t) + c3*tcos(t) + c4*tsin(t) + Y(t) where Y(t) is the particular solution.


The Attempt at a Solution



ok. so i did the Wronskian W(cos(t), sin(t), tcos(t), tsin(t)) and after doing all the distribution, the terms ended up canceling out and it equaled 0. so i don't know what to do next since I'm pretty sure the Wronskian isn't supposed to equal 0.
 
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ok i redid it and still got the wronskian to be zero. what am i doing wrong??
 
I keep saying this: No one can tell you what you did wrong if you don't tell us what you did! I did not calculate the whole thing but I did do it for t= 0 and got a value of 4, not 0.

But my question is why calculate that rather complicated Wronskian?

Are you required to use "variation of parameters"? It is obvious that the particular solution must be of the form At2cos(t)+ Bt2sin(t) and it would be easy to use "undetermined coefficients".

To use variation of parameters, you would look for a solution of the form
y(t)= u_1(t)cos(t)+ u_2(t)sin(t)+ u_3(t)tcos(t)+ u(4)tsin(t)
Differentiate that 4 times to get the derivatives to put into the equation. Since you are only looking for a "specific" solution, you can simplify by assuming that any part not involving the derivative of each ui sums to 0.

For example, the derivative of y is
y'= u_1'cos(t)- u_1sin(t)+ u_2'sin(t)+ u_2cos(t)+ u_3'tcos(t)+ u_3cos(t)- u_3tsin(t)+ u_4'tsin(t)+ u_4sin(t)+ u_4tcos(t)
If you assume that
- u_1sin(t)+ u_2cos(t)+ u_3cos(t)- u_3tsin(t)+ u_4sin(t)+ u_4tcos(t)= 0
that becomes just
y'= u_1'cos(t)+ u_2'sin(t)+ u_3'tcos(t)+ u_4'tsin(t)
to differentiate again. The differential equation, with those inserted, together with the 3 "= 0" equations, gives you 4 (algebraic) equations to solve for u_1', u_2', u_3', and u_4' which you can then integrate.
 
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is there another way to find the particular solution without doing the wronskian? i did the wronskian of cos(t), sin(t), tcos(t), tsin(t) and it ended up being zero, which isn't supposed to happen. so is there another way of solving the part. solution?
 

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