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I am seeking a particular integral of the differential equation:

$u^{(4)}(t) - 5u''(t) + 4u(t) - t^3 = 0$

I simply interested in the technique, not just an answer (Mathematica suffices for that). How would one apply undetermined coefficients in this case?

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Make a guess: $u_p=At^3+Bt^2+Ct+D$ (this should seem a reasonable guess). Now compute the required derivatives and substitute into your equation. Then solve for the coefficients. You might find this page helpful towards understanding the "method of undetermined coefficients". – David Mitra Feb 5 '13 at 0:49
up vote 2 down vote accepted

We have $(D^4-5D+4)u=t^3$ where in $Du=u'$. So $$(D^2-1)(D^2-4)u=t^3$$ so $$D^4(D^2-1)(D^2-4)=0$$ So the general solution of the OE has a form of $$u=(At^3+Bt^2+Ct+E)+y_c$$ where in $y_c=C_1e^t+C_2e^{-t}+C_3e^{-2t}+C_4e^{2t}$.

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+1 nicely done... – amWhy Feb 6 '13 at 0:07

The standard general technique is variation of parameters. In this case, as the ODE is linear with constant coefficients and the RHS is a polynomial, just try a $3 + 4 = 7$ degree polynomial.

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The characteristic polynomial is $$z^4-5z^2+4=(z^2-1)(z^2-4)=(z-1)(z+1)(z-2)(z+2)$$ Since $z=0$ is not a root, the homogeneous equation has no polynomial solutions. Therefore, the idea David Mitra gives in the comments will work.

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