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I am asked to find the general solution set of the following system of differential equations:

$$\begin{cases} x' = 3x -2y-2 \\ y' = 6x-4y-1 \end{cases} $$

I found the general solution set of the corresponding homogeneous system to be:

$$\{ \left[ \begin{array}{c} x\\ y \end{array} \right] = k_1\left[ \begin{array}{c} 2\\ 3 \end{array} \right] + k_2e^{-t}\left[ \begin{array}{c} 1\\ 2 \end{array} \right] : k_1, k_2 ∈ R\}$$

Now I have to find a particular solution. Since $\left[ \begin{array}{c} 2\\ 1 \end{array} \right]$ is a particular solution of the homogeneous equation I can't use the method of undetermined coefficients, correct? If so, how can I find a particular solution?

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A particular solution is not $\left[ \begin{array}{c} 2\\ 1 \end{array}\right]$ but is $\left[ \begin{array}{c} 2\\ 3 \end{array}\right]$ . The method of undetermined coefficients continue to be usable.

A particulat solution of the non-homogeneous equation is to be searched on the form : $$\left[ \begin{array}{c} x\\ y \end{array}\right]=\left[ \begin{array}{c} 2\\ 3 \end{array}\right]a\:t+ \left[ \begin{array}{c} b\\ c \end{array}\right] $$ Bring it back into the ODE system and identify the coefficients.( I obtained $a=-3$ , $b=0$ , $c=2$ , to be checked )

For more explanation, see http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients and example [2].

Alternatively, one could use the method of "variation of coefficients", but it should be more arduous. In this method, one remplace $k_1$ and $k_2$ by $f(t)$ and $g(t)$ respectively and solve the system for $f(t)$ and $g(t)$.

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  • $\begingroup$ Thanks, this already helps a lot. But after substituting the particular solution back into the ODE system I end up with a system with three unknowns and only two equations. How do I solve this? $$\begin{cases} 2a = 3b -2c-2 \\ 3a = 6b-4c-1 \end{cases} $$, $\endgroup$ Commented Jun 3, 2015 at 14:50
  • $\begingroup$ $\big\{ \begin{matrix} 2a=D-2 \\ 3a=2D-1 \end{matrix}$ where $D=3b-2c$. So there are two unknowns, leading to $a=-3$ and $D=-4$. Since you are looking for one particular solutions (any one among many) you can chose any value for $b$ (for example $b=0$) and the corresponding $c=(3b-D)/2$ . If you chose another value of $b$ and the corresponding $c$, you have another particular solution which is as well convenient in the formula for the general solution (this only changes the coefficient $k_1$ in your formula for the general solution). $\endgroup$ Commented Jun 3, 2015 at 15:36
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if you multiply the first by $2$ and subtract from the second, you get $$2x' - y' = -3$$ we can satisfy this by picking $$x = at+b, y = (2a+3)t +b $$ where $a, b$ are to fixed later. we will try to satisfy $$x' = 3x - 2y -2 \to a = 3(at+b) -2[(2a+3)t + b]-2=-t(a+6)+b \tag 1$$ we will choose $$a = b = -6 $$ to make $(1)$ an identity.

therefore, a particular solution is $$x = -6(t+1), \, y = -3(3t+2) $$

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