ODEs dx/dt+xy=1 dy/dt+27x9y=(6e)^4t9 Find x(t) and y(t) subject to the initial conditions x(0)=3, y(0)=18 
Your complementary solution to $$\mathbf{x}'=\mathbf{Ax} \quad \text{where} \quad \mathbf{A}=\begin{pmatrix}1 & 1 \\ 27 & 9 \end{pmatrix}$$ will be augmented by a particular solution which will depend on the eigenvalues of $\mathbf{A}$, but may well be $$\mathbf{x}_p=\mathbf{c}_1 e^{4t} + \mathbf{c}_2 $$ where the vectors $\mathbf{c}_1$ and $\mathbf{c}_2$ can be determined by the method of undetermined coefficients. 
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$\displaystyle \begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} + x  y &= 1 \\ \frac{\mathrm{d}y}{\mathrm{d}t} + 27\,x  9\,y &= \left( 6\,\mathrm{e} \right) ^{4\,t}  9 \end{align*}$ 
Was your second equation intended to be dy/dt + 27x  9y = (6e)^(4t9), chris9598? Also, where did you get this problem from? What were the characters enclosing 6e that I've replaced with parentheses? 
hi, yeah thats the correct one. i got this from a problem sheet in college 
$$(6e)^{4t9}=e^{9} e^{(1+\ln 6)4t}$$ There's nothing too unusual here, we just have awkward values for constants. 
Did you mean $(6e)^{9} e^{(1+\ln 6)4t}$, v8archie? Quote:

The righthand side should be 6e^(4t)9 
Oh, yeah. I missed a bit. It's still $ae^{kt}$, just with a very messy $k$. That $a$ will get swallowed up in constants that we find using the method of undetermined coefficients. 
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