# Finding the Particular solution to a Diff. Eq. System

1. Nov 30, 2015

### dumbdumNotSmart

Mentor note: moved from mathematics forums, therefore no homework template

Hello mates. I have a System of Differential Equations on my hands.

$$W' = \left(\begin{matrix} 4 & 2 \\ 2 & 4 \end{matrix} \right) W + \begin{pmatrix} 6 \\6 \end{pmatrix}$$
where transposed W is (x y z). I have found the homogenous solution to said system. Finding the Particular solution is what I am not sure about. So far I have tried acquiring the particular using the homogenous solution, which I found to be:
$$W_H = c_1 e^{6t} \left(\begin{matrix} 1 \\ 1 \end{matrix} \right) + c_2e^{2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$$
Particular solution I am looking for is :
$$W_P=\begin{pmatrix} e^{6t} & e^{2t} \\ e^{6t} & -e^{2t} \end{pmatrix} \begin{pmatrix} u_1 \\u_2 \end{pmatrix}$$
And we can find it with the following equation (u' would be the derivative of u):
$$\begin{pmatrix} 6\\6 \end{pmatrix} =\begin{pmatrix} e^{6t} & e^{2t} \\ e^{6t} & -e^{2t} \end{pmatrix} \begin{pmatrix} u_1' \\u_2' \end{pmatrix}$$

This gives us the following equations:

$$u_1' e^{6t} + u_2' e^{2t} =6 \\ u_1'e^{6t}-u_2'e^{2t}=6 \\ \Rightarrow u_1=-e^{-6t} \land u_2=0$$

So my question is, when we integrate u2', would we consider the constant of integration... or not because we are finding the particular? Is my process correct? Thank you peeps

Last edited by a moderator: Nov 30, 2015
2. Nov 30, 2015

### Krylov

I don't understand why you would proceed from here on in the way you do. The matrix defining the homogeneous part of the differential equation is invertible, so calculating a particular solution just amounts to taking an inverse.

And what is a "peep"?

3. Nov 30, 2015

### Staff: Mentor

Singular of "peeps" -- or "people".

4. Nov 30, 2015

### Krylov

It sounded odd to me, thank you for the clarification.

5. Nov 30, 2015

### Ray Vickson

There is an easier way: suppose
$$A = \pmatrix{4 & 2 \\ 2 & 4}, \;\text{and} \; r = \pmatrix{6\\6}.$$
Your DE is $W' = AW + r.$ If $k$ is the column-vector solution of $Ak = r$, your DE reads as $W' = A(W+k),$ so the vector $Z = W+k$ satisfies $Z' = AZ$, because $k$ is constant. So, $W = Z-k$, where $Z$ is a solution of the homogeneous equation.

6. Dec 1, 2015

### HallsofIvy

Equivalently- since the "non-homogeneous" part of the equation is the constant vector, $\begin{pmatrix}6 \\ 6 \end{pmatrix}$, the part of the solution due to that must be constant, say, $\begin{pmatrix}A \\ B \end{pmatrix}$. The derivative of a constant is 0 so we have the equation
$$\begin{pmatrix}4 & 2 \\ 2 & 4 \end{pmatrix}\begin{pmatrix}A\\ B\end{pmatrix}=\begin{pmatrix}4A+ 2B\\ 2A+ 4B \end{pmatrix}= \begin{pmatrix}6 \\ 6 \end{pmatrix}$$
so we have the two equation 4A+ 2B= 6 and 2A+ 4B= 6.

7. Dec 1, 2015

### Krylov

I think we now have the same hint three times. Also,
mind the minus sign: To obtain a (constant) particular solution $W_p$, solve
$$\begin{bmatrix} 4& 2\\ 2& 4 \end{bmatrix} W_p = \color{red}{-} \begin{bmatrix} 6\\ 6 \end{bmatrix}$$

8. Dec 1, 2015

### dumbdumNotSmart

Thanks guys. It makes sense that the non-homogenous solution would have constant elements. Although I don't understand what you are saying krylov by minding the minus sign, when did it come in play?

9. Dec 1, 2015

### Krylov

When you substitute an unknown constant vector $W_p$ into the inhomogeneous equation, indeed you end up with a linear system, but you have to be careful that there is a minus sign in from of $(6,6)$ in the right-hand side. That minus sign was accidentally missing in the post of @HallsofIvy .