[Math] Higher Order Linear Differential Equations – Solving for the particular solution

ordinary differential equations

Given $y''' – 5y'' – y' + 5y = 3e^{-x}$, find the general solution.

I found the roots for the homogeneous solution to be 5, 1, and -1:

$$(r – 5)(r + 1)(r – 1)=0$$

$$y_h(x) = c_1e^x + c_2e^{-x} + c_3e^{5x}$$

Setting up the particular solution, I have:

$$g(x) = 3e^{-x}$$

$$y_p(x) = Axe^{-x}$$

$$y_p'(x) = Ae^{-x} – Axe^{-x} = A(1 – x)e^{-x}$$

I know I need to use the product rule to continue differentiating $y'$, but is there an easier method to do so?

Best Answer

  1. How to differentiate $y'$ using the product rule? $y''=A(1-x)'e^{-x}+A(1-x)(e^{-x})'$, and so on.

  2. It is easier (IMO) to use annihilator method [1] which says:

    • If your ODE is $L[y]=f(x)$, where $f(x)=P_k(x)e^{\alpha x}\cos(\beta x)$ or $f(x)=P_k(x)e^{\alpha x}\sin(\beta x)$ (where $P_k(x)$ is a polynomial and: $k$, $\alpha$, $\beta$ can be $0$ in order to get rid of one of these elements).
    • Apply the following differential operator on both sides: $((D-\alpha)^2+\beta^2)^{k+1}$. Applying this on the right side will result in a zero.
    • This will result in $y=y_h+y_p$. Since you already found $y_h$, you can identify $y_p$.
    • Solve the original equation for $y_p$ in order to find the constant coefficients.
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