$$\frac{dx}{dt}=2x +3y$$ $$\frac{dy}{dt}=3x +2y$$ Find general solution.
I know there is a solution through eigenvalues. But I want to solve it with Laplace transformation.
I almost get the right answer. I am missing a negative sign.
I get with one-sided Laplace: $$x(t)=C_1e^{-t}+C_2e^{5t}$$ $$y(t)=C_1e^{-t}+C_2e^{5t}$$
Real answer: $$x(t)=C_1e^{-t}+C_2e^{5t}$$
$$y(t)=-C_1e^{-t}+C_2e^{5t}$$ Have I forgot something very basic?
Letting capital letters denote the laplace transform and taking the laplace transform of both equations, we get
$$ sX(s)-x(0) = 2X(s) + 3Y(s) $$ $$ sY(s)-y(0) = 3X(s) + 2Y(s) $$
Collecting like terms we get
$$(s-2)X(s)-3Y(s)-x(0)=0$$ $$(s-2)Y(s)-3X(s)-y(0)=0$$
Using elimination, we can get
$$-9Y(s)-3x(0)+(s-2)^2Y(s)-(s-2)y(0)=0$$
Moving things around a bit
$$Y(s) = \frac{(s-2)y(0)+3x(0)}{(s-5)(s+1)}$$
Using partial fraction decomposition,
$$Y(s) = \frac{C_1}{s+1} + \frac{C_2}{s-5} \Rightarrow y(t) = C_1 e^{-t} + C_2 e^{5t}$$
Plugging back into your original second equation,
$$ -C_1 e^{-t} + 5C_2 e^{5t} = 3x(t) + 2(C_1 e^{-t} + C_2 e^{5t}) $$
Solving for $x(t)$, we get
$$x(t) = -C_1 e^{-t} + C_2 e^{5t}$$
which is equivalent to the solution you're looking for.
Since you haven't posted your own work, I can't point to the exact mistake you made, but its likely you just made an algebraic error.
Also, contrary to what other commenters have said, the choice of $-C_1$ versus $+C_1$ does matter since this is a system. It is critical that the $e^{-t}$ term in the solution for $x(t)$ has the opposite sign of $y(t)$.