# Transforming 3rd order D.EQ into 1st order

• hils0005
In summary, the conversation discusses transforming a 3rd order differential equation into a system of 1st order differential equations. The attempt at a solution involves expressing the equations in terms of variables y, A, and B, and adding the non-homogenous term t^4cos(2t) as a column vector. The correct system of equations is determined to be y'=A, A'=B, and B'=3B+(-3y+t^4cos(2t)). The conversation also mentions the possibility of solving the equation, which would involve solving the homogenous version first and then guessing solutions to the inhomogenous equation.

#### 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)

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?

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??

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

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