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Variation of parameters for higher order linear eq

  1. Mar 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Use the method of variation of parameters to determine the general solution of the given differential equation: y^(4) + 2y'' + y = sin(t)

    2. Relevant 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.

    3. 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.
  2. jcsd
  3. Mar 6, 2008 #2
    ok i redid it and still got the wronskian to be zero. what am i doing wrong??
  4. Mar 7, 2008 #3


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    Staff Emeritus
    Science Advisor

    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.
    Last edited: Mar 7, 2008
  5. Mar 7, 2008 #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 isnt supposed to happen. so is there another way of solving the part. solution?
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