# Help with particular integral on 2nd order differential equation

#### u-will-neva-no

Joined Mar 22, 2011
230
Hello everyone, my second order to solve is:

y'' - 2y' = 12e^2x -8e^-2.

I have found the complementary function, which is Ae^2x + B but im not sure what my trial solution should be to find the particular solution.

I tried y = axe^2x - be^-2x, for the trial solution and had a nice answer. I don't think this is correct however...please let me know! Thankyou.

#### steveb

Joined Jul 3, 2008
2,436
Hello everyone, my second order to solve is:

y'' - 2y' = 12e^2x -8e^-2
Can you verify your equation? Is it the following?

$$y'' - 2y' = 12{\rm e}^{2x} -8 {\rm e}^{-2}$$

Or is it this?

$$y'' - 2y' = 12{\rm e}^2 x -8 {\rm e}^{-2}$$

Or, is it this?

$$y'' - 2y' = 12{\rm e}^{2x} -8 {\rm e}^{-2x}$$

#### u-will-neva-no

Joined Mar 22, 2011
230
Sorry, i messed up. Its the last one you typed.

#### steveb

Joined Jul 3, 2008
2,436
Sorry, i messed up. Its the last one you typed.
The way I would solve this is to make a couple of substitutions.

First, define an input function u=12exp(2x) - 8exp(-2x).

Then, define a new state variable z=dy/dx.

With these substitutions, the state space form is.

dy/dx=z
dz/dx=2z+u

which in matrix notation is

dW/dx=AW+Bu

where W=[y; z] as a column vector, A=[0 1; 0 2] as a square matrix and B=[0; 1] as a column vector

From there you can solve using standard methods such as state transition matrix approach, or Laplace transform method and the response to your input function will give you a particular solution matched to given initial conditions.

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• u-will-neva-no

#### steveb

Joined Jul 3, 2008
2,436
Maybe you haven't had a chance to look at this, but I had a few minutes at coffee break and solved this via state transition matrix. The solution made it clear that a valid particular solution is.

$$y=6x{\rm e}^{2x}-{\rm e}^{-2x}$$

So, the form of your suggested solution is correct, but you needed to solve for the "a" and "b" coefficients.

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• u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230
Wow, I have never come across this method to solve it although I think I will be taught this sometime soon at university. I will keep hold of this page and refer to it once I get my head around this method. Thank you for helping and putting all that effort in!

#### steveb

Joined Jul 3, 2008
2,436
Wow, I have never come across this method to solve it although I think I will be taught this sometime soon at university. I will keep hold of this page and refer to it once I get my head around this method. Thank you for helping and putting all that effort in!
You are welcome, and yes you should be learning this method a little later. It is one of the corner-stones of signals and systems theory and used extensively in controls system design. It probably looks complex to you now, but you will eventually see that it is straightforward, and wasn't much effort at all. It was just a nice little problem to solve on coffee break while my coworkers were chatting about nonsense. Since you say you want to save the pages, I should point out that there is a little error in there that I realized later. The solution for z(x) is a little off. It should be the following.

$$z(x)=y'(0){\rm e}^{2x}+12x{\rm e}^{2x}+2{\rm e}^{-2x}-2{\rm e}^{2x}$$

and NOT the following

$$z(x)=y'(0){\rm e}^{2x}+24x{\rm e}^{2x}+4{\rm e}^{-2x}-2{\rm e}^{2x}$$

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