Variation of parameters for higher order linear eq

In summary, the protagonist attempted to solve a differential equation using the method of variation of parameters, but ran into trouble because the Wronskian equation ended up being zero. He is not sure what he is doing wrong.
  • #1
KDizzle
5
0

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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  • #2
ok i redid it and still got the wronskian to be zero. what am i doing wrong??
 
  • #3
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
[tex]y(t)= u_1(t)cos(t)+ u_2(t)sin(t)+ u_3(t)tcos(t)+ u(4)tsin(t)[/tex]
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
[tex]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)[/tex]
If you assume that
[tex]- u_1sin(t)+ u_2cos(t)+ u_3cos(t)- u_3tsin(t)+ u_4sin(t)+ u_4tcos(t)= 0[/tex]
that becomes just
[tex]y'= u_1'cos(t)+ u_2'sin(t)+ u_3'tcos(t)+ u_4'tsin(t)[/tex]
to differentiate again. The differential equation, with those inserted, together with the 3 "= 0" equations, gives you 4 (algebraic) equations to solve for [itex]u_1'[/itex], [itex]u_2'[/itex], [itex]u_3'[/itex], and [itex]u_4'[/itex] which you can then integrate.
 
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  • #4
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?
 

FAQ: Variation of parameters for higher order linear eq

1. What is the variation of parameters method?

The variation of parameters method is a technique used to find a particular solution for a higher order linear differential equation. It involves finding a set of functions that satisfy the given differential equation and using them to form a particular solution.

2. When is the variation of parameters method used?

The variation of parameters method is used when the coefficients of a higher order linear differential equation are not constant. It is also used when the right side of the equation does not have a simple form, making it difficult to use other methods such as the method of undetermined coefficients.

3. How is the variation of parameters method applied?

The variation of parameters method involves finding a set of functions that satisfy the homogeneous form of the given differential equation. These functions are then used to form a particular solution by replacing the coefficients in the homogeneous solution with unknown functions and solving for them using the given non-homogeneous term.

4. What are the advantages of using the variation of parameters method?

One advantage of using the variation of parameters method is that it can be applied to a wide range of higher order linear differential equations, including those with variable coefficients. It also allows for the inclusion of initial conditions, making it easier to find a unique particular solution.

5. Are there any limitations to the variation of parameters method?

One limitation of the variation of parameters method is that it can be time-consuming and tedious, especially for higher order differential equations. It also requires a good understanding of homogeneous solutions and their properties. Additionally, it may not always provide a closed-form solution, making it difficult to interpret the results.

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