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Particular solution

  1. Oct 20, 2015 #1
    how can I find the particular solution for something like this
    x = 120 cos 6t

    I know how to find a particular solution for equations like x''+x'+x=0, you can easily find the characteristic equation then find the particular solution, but I have no idea how to deal with an equation like that..
    Last edited: Oct 20, 2015
  2. jcsd
  3. Oct 20, 2015 #2


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    Are you sure you wrote the problem down correctly? The equation you've given does not have to be solved. It already gives x unambiguously in terms of t. What did you imagine a solution would look like?
  4. Oct 21, 2015 #3


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    If you mean, say, a differential equation of the form x''+ x'+ x= 120 cos(6t), there are a number of different ways to find a "specific solution" to the entire equation. Two methods are given in any differential equations textbook. You should start by finding the general solution to the associated homogeneous equation, x''+ x'+ x= 0. That has, as you say, characteristic equation [itex]r^2+ r+ 1= 0[/itex] which has solutions [itex]r= \frac{-1\pm i\sqrt{3}}{2}[/itex]. That, in turn, tells us that the general solution to the associated homogeneous equation is [itex]x(t)= e^{-t/2}\left(C_1\cos\left(\frac{\sqrt{3}}{2}t\right)+ C_2\sin\left(\frac{\sqrt{3}}{2}t\right)\right)[/itex].

    As I said, there are a number of ways of finding just one function that satisfies the entire equation. The two covered in text books are "undetermined coefficients" and "variation of parameters".

    The first requires that you use what you know of differentiation to guess the general form of the solution. Here, since we know that the derivative of sine and cosine always give sine and cosine again, we "guess" a solution of the form x(t)= A cos(6t)+ B sin(6t). Take the first and second derivatives of that, put them into the equation and determine the values of A and B that will satisfy the equation.

    The second does not require that you be able to guess the general form but is more complicated. Knowing the general solution to the associated homogeneous equation, we look for a solution of the form [itex]x(t)= e^{-t/2}(u(t) cos(\sqrt{3}t/2)+ v(t)sin(\sqrt{3}t/2))[/itex]. Finding the first and second derivatives of that and putting them into the equation will give two separate equations for u' and v'. Solve those for u' and v' and then integrate.
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