# Integral Question

1. May 21, 2014

### rwooduk

d4y/dx4 - 4 d3y/dx3 - 5 d2y/dx2 + 36 dy/dx - 36y = -8e^x

Given that λ^4 - 4λ^3 - 5λ^2 + 36λ - 36 = 0 can be factored to obtain (λ-2)^2 (λ^2 - 9) = 0

find the general solution to this equation

I can answer the question no problem BUT where does the -8 on the RHS of the equation go??

If i set y=e^λx i get λ^4 - 4λ^3 - 5λ^2 + 36λ - 36 = -8 ??

so why hasnt he factorised λ^4 - 4λ^3 - 5λ^2 + 36λ - 28 = 0

??

Totally confused. Thanks for any help.

2. May 21, 2014

### CAF123

That is guiding you to the general solution of the corresponding homogenous equation only. To obtain the general solution to your original equation, you also need a solution solving the equation with the term on the RHS present.

3. May 21, 2014

### Fightfish

The differential equation you have there is an inhomogeneous differential equation.
The characteristic polynomial that is provided in the hint provides the solution to the homogenous differential equation
$$y^{(4)} - 4y^{(3)} -5y''+36y'-36y = 0$$
The final solution will be the sum of the homogenous solution and the particular solution for the imhomogenous part.

4. May 21, 2014

### rwooduk

the particular solution is given as e^x

so the complete solution will be

λ = 2, 2, 3, -3

-> y = Ae^2x + Bxe^2x + Ce^3x + De^-3x + e^x

are you saying that the particular solution given takes care of the RHS, so I dont need to worry about it? Or do I have to include something else in the complete solution above?

thanks for the help.

EDIT

oh, i see, it doesnt want a solution to the original equation. thanks again for the help!!

Just out of interest how would the solution to the original question differ from my answer?

Last edited: May 21, 2014
5. May 21, 2014

### Fightfish

Yes, that is the role of the particular solution. You can think of it this way:
We have a differential equation in the form
$\hat{D} y = f(x)$
The general solution can be written in terms of the homogenous and particular solutions:
$y = y_{h} + y_{p}$
which satisfy
$\hat{D} y_{h} = 0$
$\hat{D} y_{p} = f(x)$
So, clearly,
$\hat{D} (y_{h}+y_{p}) = f(x)$

6. May 21, 2014

### rwooduk

excellent! got it! Many thanks!!