Solving system of ODE with initial value problem (IVP) I have a question about a system of ODE. 
If we have:
$\frac{dx}{dt}=x+2y$
$\frac{dy}{dt}=3x+2y$
with $x(0)=6$ and $y(0)=4$, 
how come the solution to the IVP is: 
$x(t)=4e^{4t}+2e^{-t}$
$y(t)=6e^{4t}-2e^{-t}$
I tried doing integral by separating the variable but I didn't get that solution. That example & solution are from my numerical method book, please see the attached image
example & solution
 A: take laplace transform on both sides of both differential equations to get 
$(s-1)X(s)=2Y(s)+6  \ \ $ ; 
$(s-2)Y(s)=3X(s)+4\\$  respectively 
solve for $X(s)$ and $Y(s)$ like algebric equations to give ;
$Y(s)=\dfrac{4s+14}{(s+1)(s-4)}\implies y(t)=-2e^{-t}+6e^{4t}$
$X(s)=\dfrac{6s-4}{(s+1)(s-4)}\implies x(t)=2e^{-t}+4e^{4t}$
A: Hint: With $$y=\frac{x'-x}{2}$$ we get
$$y'=3x+\frac{3}{2}(x'-x)$$ and we get $$y'=\frac{x''-x'}{2}$$ so we obtain
$$\frac{x''-x'}{2}=3x+\frac{3}{2}(x'-x)$$
Can you finish?
From here you will get
$$x''-4x'-3x=0$$ make the ansatz $$x=e^{\lambda t}$$
A: Hint:
One possibility is to eliminate one of the unknowns.
$$x'=x+2y\implies x''=x'+2y'$$ and $$y'=3x+2y=3x+(x'-x),$$
hence
$$x''-3x'-4x=0.$$
A: Let us consider your system of ODE:
$$\frac{dx}{dt}=x+2y\\\frac{dy}{dt}=3x+2y$$
It follows that
$$\frac{dx}{dt}+\frac{dy}{dt}=4x+4y\rightarrow \frac{d(x+y)}{dt}=4(x+y)\rightarrow \frac{d(x+y)}{x+y}=4dt$$
Integration on both sides yields
$$\ln(x+y)=4t+C\rightarrow x+y=Ce^{4t}\rightarrow y=Ce^{4t}-x$$
Substitution into $\frac{dx}{dt}$ gives
$$\frac{dx}{dt}=x+2(Ce^{4t}-x)\rightarrow \frac{dx}{dt}+x=2Ce^{4t}$$
Multiply by $e^t$ on both sides:
$$\frac{dx}{dt}e^{t}+xe^{t}=2Ce^{5t}\rightarrow\frac{d(xe^{t})}{dt}=2Ce^{5t}$$
Integration on both sides yields
$$xe^{t}=\frac{2C}{5}e^{5t}+K\rightarrow x=\frac{2C}{5}e^{4t}+Ke^{-t}$$
Substitution into $y$ gives
$$y=\frac{3C}{5}e^{4t}-Ke^{-t}$$
Apply conditions:
$$x(0)+y(0)=6+4=10\rightarrow Ce^{4(0)}=10\rightarrow C=10\\x(0)=6\rightarrow \frac{2(10)}{5}e^{4(0)}+Ke^{-(0)}=6\rightarrow K=2$$
Thus,
$$x(t)=4e^{4t}+2e^{-t}\\y(t)=6e^{4t}-2e^{-t}$$
A: Observe that you have 
$$  \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \cdot \begin{pmatrix}x \\ y \end{pmatrix}  \text{.}  $$
The eigenvalues of this matrix are $4, -1$, so both solutions are of the form $a \mathrm{e}^{4x} + b \mathrm{e}^{-1x}$.  As others have shown, you then match the coefficients to the initial value data.
