Transforming 3rd order D.EQ into 1st order

[hils0005]

[SOLVED] Transforming 3rd order D.EQ into 1st order

Homework Statement

Transform the following diff eq into a system of first order diff eqs.

2y''' - 6y'' - y' + 3y =t^4cos(2t)

The Attempt at a Solution

heres what I've done so far:
y =A A'=y'
y' =B B'=y''
y'' =C C'=y'''=6y'' + y' - 3y + t^4cos(2t)

C'=-3A + B + 6C +t^4cos(2t)
B'= C
A'= B

I don't know if this is correct because the initial D.Eq is not homogenous, don't know what to do with the t^4cos(2t)

 

[Defennder]

Well, of course the system of 1st order ODE won't be homogenous. You started out with an inhomogenous equation. Why should you expect the final answer to be homogenous? You don't need C at all. You only need to express as a system of 3 linear ODE with variables y,A,B. You don't have to let A=y. Just use y itself. Once you've done that you can simply just add in the t^4cos2t as a column vector to the right of the RHS.

Here's what I have:

y'=A
A'=B
\frac{d}{dt} \left(\begin{array}{cc}y\\A\\B\end{array}\right) = \left(\begin{array}{ccc}0&1&0\\0&0&1\\-\frac{3}{2}&\frac{1}{2}&-3\end{array}\right)\left(\begin{array}{cc}y\\A\\B\end{array}\right) + \left(\begin{array}{cc}0\\0\\t^4cos2t\end{array}\right)

Are you required to solve this?

 

[hils0005]

No just need to set up the equations
so the equation would look like:
y'=A
A'=B
B'=(-3y+t^4cos2t) + (A+t^4cos2t) - (6B+t^4cos2t)

not sure why you multiplied B' by 1/2, and also how did you come up with -3B not a positive 3B??

 

[hils0005]

I see I forgot the coefficent 2 before y''', still can't see how you get -3B and not positive 3B

 

[lzkelley]

your solution does look correct.
If you actually wanted to solve this, you would probably first solve the homogenous version of the equation, then using that result... essentially guess solutions to the inhomogenous equation (the actual equation).

 

[Defennder]

I see I forgot the coefficent 2 before y''', still can't see how you get -3B and not positive 3B

You're right about that; it should be 3B, not -3B.

\frac{d}{dt} \left(\begin{array}{cc}y\\A\\B\end{array}\right) = \left(\begin{array}{ccc}0&1&0\\0&0&1\\-\frac{3}{2}&\frac{1}{2}&3\end{array}\right)\left(\begin{array}{cc}y\\A\\B\end{array}\right) + \left(\begin{array}{cc}0\\0\\t^4cos2t\end{array}\right)