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For a particular solution of $$ y^{(4)} - y''' - y'' + y' = t^2 + 4 + te^t$$

The solution to the homogeneous solution is represented by the characteristic equation:

$$ r(r-1)^2 (r+1)$$

so the solutions of the homogeneous solution are: $ 0,-1$ and $1$ with multiplicity 2.

So the homogeneous solution will have the form: $$ y_h = Ae^t + Bte^t + Ce^{-t} + D$$ As for the particular solution, I think it will have the form of

$$ y_p = A + Bt + Ct^2 + D + Ee^t + Ete^t$$ and then modify it so that it doesn't overlap with $y_h$.

I am not sure how to go from here. (The question is merely looking for the "form" of the particular solution)

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Are you familiar with the method of undetermined coefficients? –  AWertheim May 1 '13 at 20:29
    
Yes, but I the question here is just asking for the "form" of the particular solution. –  40Plot May 1 '13 at 20:29
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2 Answers

$y_p$ can have the form

$$ y_p = A_1 t + A_2 t^2 + A_3 t^3 + B_1 t^2 e^{t} + B_2 t^3 e^{t}. $$

Here is a general technique tells you how to find $y_p$.

Added: We will apply the general technique to find $y_p$. To annihilate the right hand side of the ode, we apply the operator $D^3(D-1)^2$, where $D=\frac{d}{dt}$, to both sides of the ode which results in

$$ D^4(D-1)^4(D+1) y(t)=0, $$

since $D^3 (t^4+4)=0$ and $(D-1)^2 te^{t} = 0 $.

Now, the above is a new homogenous ode with the the solution,

$$ y_h = c_0 + c_1 t + c_2 t^2 + c_3 t^3 + c_5 e^{t}+ c_6 te^{t}+ c_7 t^2 e^{t} + c_8 t^3e^{t} + c_9 e^{-t} $$

$$ = (c_0 + c_5e^{t}+c_6te^{t} + c_9e^{-t})+ c_1 t + c_2 t^2 + c_3 t^3 + c_7 t^2 e^{t} + c_8 t^3e^{t}.$$

You can see in the above equation what is between the brackets corresponds to $y_h$ of the old ode

$$ y_{h_{o}} = Ae^t + Bte^t + Ce^{-t} + D. $$

So, we can conclude our $y_p$ to be

$$ y_p = A_1 t + A_2 t^2 + A_3 t^3 + B_1 t^2 e^{t} + B_2 t^3 e^{t}.$$

Note that, the above form will do, but still you can choose the more general form which is based on the above form

$$ y_p = A_0+A_1 t + A_2 t^2 + A_3 t^3 + (B_0+ B_1 t + B_2 t^2 + B_3 t^3) e^{t} $$

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For each source term of the form $P(t)\,e^{\lambda t}$, where $P$ is a polynomial, the form of the particular solution will include $t^kQ(t)\,e^{\lambda t}$ where

  • $Q$ is a generic polynomial (with undetermined coefficients) of the same degree as $P$
  • $k$ is the multiplicity of $\lambda$ as a root of the characteristic equation. In particular, $k=0$ if $\lambda$ is not a root of the characteristic equation.

You have (second degree polynomial) $e^{0\cdot t}$ + (first degree polynomial)$e^{1\cdot t}$. And you already noticed that

  • $0$ is a root of multiplicity $k=1$
  • $1$ is a root of multiplicity $k=2$

Is the path clear now?

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