# Higher Order Linear Differential Equations - Solving for the particular solution

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?

• If you're doing differential equations like this, shouldn't you have learned how to take a derivative long ago? – Mike Nov 28 '12 at 19:24

## 3 Answers

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  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.

There is a way to get the complete solution directly. It helps that all our roots are real. There is a way to "factor" the left side like a polynomial. This method attempts to get the equation into the form

$$z'+az=f(x)$$

which can be solved through the use of an integrating factor.

You have already that the characteristic polynomial is $(r-5)(r+1)(r-1)$, which I will rewrite as $(r-5)(r^2-1)$. Now let $z=y''-y$, which you'll notice has the characteristic polynomial $r^2-1$. Now the original equation can be rewritten.

$$y'''-y'-5y''+5y=(y''-y)'-5(y''-y)=z'-5z=3e^{-x}$$

You'll notice that the characteristic polynomial for this equation in $z$ corresponds to the other factor $r-5$. This is of course solved my multiplying through by the integrating factor $e^{-5x}$

$$e^{-5x}z'-5e^{-5x}z=(e^{-5x}z)'=3e^{-6x}$$ $$e^{-5x}z=-\frac12e^{-6x}+k_1$$ $$z=-\frac12e^{-x}+k_1e^{-5x}$$

As long as we don't get our equation into the form

$$u'+u=f(x)$$

the $e^{-x}$ term will not cancel out prior to integration and introduce an $xe^{-x}$ term, which would make further integration more complicated. So now we have

$$z=y''-y=-\frac12e^{-x}+k_1e^{-5x}$$

Again, we choose a substitution based on a factor of the characteristic polynomial. To avoid the problem mentioned above, we make the substitution $u=y'+y$. This yields

$$y''-y=(y'+y)'-(y'+y)=u'-u=-\frac12e^{-x}+k_1e^{-5x}$$ $$e^{-x}u'-e^{-x}u=(e^{-x}u)'=-\frac12e^{-2x}+k_1e^{-6x}$$ $$e^{-x}u=\frac14e^{-2x}+k_2e^{-6x}+k_3,k_2=-\frac{k_1}6$$ $$u=y'+y=\frac14e^{-x}+k_2e^{-5x}+k_3e^x$$

And finally we can solve for $y$.

$$e^xy'+e^xy=(e^xy)'=\frac14+k_2e^{-4x}+k_3e^{2x}$$ $$e^xy=\frac14x+k_4e^{-4x}+k_5e^{2x}+k_6$$ $$y=\frac14xe^{-x}+k_4e^{-5x}+k_5e^x+k_6e^{-x}$$

Here is an easier approach (I will give only a recipe, but it can be easily justified).

You have the characteristic polynomial, in your case this is $$p(r)=r^3-5r^2-r+5$$ You need to find your particular solution to the problem with the right-hand side $3e^{-x}$ and you know that $-1$ is a root of the characteristic polynomial (multiplicity 1). A particular solution will have the form $$y_p(x)=\frac{3xe^{-x}}{p'(-1)}=\frac{1}{4}xe^{-x}.$$

Added: If you need to find a particular solution to the problem with the right hand side in the form $Ae^{ax}$ and you know that $a$ is not a root of the characteristic polynomial, then a particular solution can be taken as $$y_p=\frac{Ae^{ax}}{p(a)},$$ there is no need for any derivatives.