I could not understand the textbook clearly.
When you are trying to find a particular solution of
x' = -2x + y + 2e^(-t)
y' = x -2y + 3t
I understand that 2e^(-t) would have a form of ate^-t + b*e^-t
in which a and b needs to be determined and
3t would have a form of c*t + d
However, the book all the sudden subbed it into differential equation
and just says, "we obtain the following algebraic equations for a,b,c and d:"
Aa = -a
Ab = a - b - (2 0)' (' is transpose)
Ac = - (0 3)'
Ad = c
in x' = Ax + g(t) where A is the constant matrix and g(t) is particular solution.
Anyhow can anyone please show me how to solve
undetermined coefficient for 2 first order differential equation with steps?
You are given:
$$x' = -2x + y + 2e^{-t} \\ y' = x -2y + 3t$$
This can be written as:
$$X'(t) = \begin{bmatrix}x'(t)\\y'(t)\end{bmatrix} = AX(t) + F(t) = \begin{bmatrix}-2 & 1\\1 & -2\end{bmatrix} \begin{bmatrix}x(t)\\y(t)\end{bmatrix} + \begin{bmatrix}2e^{-t}\\ 3t \end{bmatrix}$$
Do you know how to approach solving this?
For the matrix $A = \begin{bmatrix}-2 & 1\\1 & -2\end{bmatrix}$, we find the eigenvalues and eigenvectors and end up with the general solution of the homogenous system as:
$$X_h(t) = c_1 \begin{bmatrix} 1\\-1\end{bmatrix}e^{-3t} + c_2 \begin{bmatrix} 1\\1\end{bmatrix}e^{-t}$$
Now, we can decompose $F(t)$ as:
$$F(t) = \begin{bmatrix} 2\\0\end{bmatrix}e^{-t} + \begin{bmatrix} 0\\3\end{bmatrix}t$$
Do you see how we do that?
Since we have the term $e^{-t}$ in the homogeneous solution, we have to multiply what we have for the particular solution by $t$. Are you clear of this from the Method of Undetermined Coefficients?
So, we assume that:
$$X_p(t) = v(t) = a t e^{-t} + be^{-t} + c t + d$$
Where $a, b, c$ and $d$ are vectors to be determined.
I will get you started with the first one and you can do the others.
We have:
$$x' = ae^{-t} - ate^{-t} - be^{-t} + c = Ax + F = A(a t e^{-t} + be^{-t} + c t + d) + \begin{bmatrix} 2\\0\end{bmatrix}e^{-t} + \begin{bmatrix} 0\\3\end{bmatrix}t$$
If we equate the terms on the left and right (start with the term $te^{-t}$), we have:
$$Aa = -a$$
Now, you have three other terms to equate and derive the three other expressions shown.