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I have the following fourth order differential equation and was asked to find the general solution for it by using the method of undetermined coefficients.

$y^{(4)} +2y'' +y = (t-1)^2$

So, solving for the characteristic equation,

$r^4+2r^2 +1 =0$

I got,

$r=±i$

and from there I obtained the particular solution

$ y_p=c_1 \cos t + c_2 \sin t + c_3t \cos t +c_4t\sin t $

Now, my question is: how do I use the forcing function to come up with a "guessed" equation to solve for the general solution?

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  • $\begingroup$ The notation $y^4$ is unfortunate, because it suggests $y\cdot y\cdot y\cdot y$, particularly when you use $y''$ in the same expression. $\endgroup$ – almagest Apr 20 '16 at 7:52
  • $\begingroup$ Oh, im sorry! I'll fix that! Thanks for pointing it out1 $\endgroup$ – Marco Neves Apr 20 '16 at 7:52
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For a right-hand side of the form $(t-1)^2=t^2-2t+1$, you propose a particular solution of the form: $$y_p = At^2+Bt+C$$ Substitution into the differential equation will give you a system of 3 linear equations in the undetermined coefficients $A$, $B$ and $C$.

This method only works for certain forms of the right-hand side: but this should be in your book, course notes or otherwise easily found online (e.g. see Method of undetermined coefficients).

Remark: when your proposed solution is already a part of the homogeneous solutions (which is not the case for your differential equation; but which would be the case if the RHS was, for example, $\sin t$), you will need to multiply the proposed solution with a sufficiently large power of $t$ so that it is no longer contained in the homogeneous solution.

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  • $\begingroup$ Does the proposed solution always follow the base format of the right-hand side? i.e, if it were a linear function I would propose Ax + B ? $\endgroup$ – Marco Neves Apr 20 '16 at 7:45
  • $\begingroup$ Correct: for a polynomial, you propose a polynomial of the same degree (careful: this means that if the right-hand side is just $t^3$, you still take $At^3+Bt^2+Ct+D$). $\endgroup$ – StackTD Apr 20 '16 at 7:46
  • $\begingroup$ Yeah i get it now! Thanks a bunch, will upvote and mark it as answered! $\endgroup$ – Marco Neves Apr 20 '16 at 7:48
  • $\begingroup$ You're welcome; good luck! $\endgroup$ – StackTD Apr 20 '16 at 7:49
  • $\begingroup$ But that is NOT true for general right hand side. For example if the equation were y''''+ 2y''+ y= ln(t) or = tan(t), we could not make a reasonable guess as to what to use. In that case, try "variation of parameters", You can make such a "guess" for functions that are of the type we expect as solutions to a homogeneous linear differential equation with constant coefficients: exponentials, sine and cosine, polynomials, and products of such functions. $\endgroup$ – user247327 May 8 '18 at 13:42

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