Ordinary Differential Equation with 3 unknowns 
Solve the IVP system $\displaystyle{\begin{cases}x_1'=3x_1-4x_2+4x_3\\x_2'=4x_1-5x_2+4x_3\\x_3'=4x_1-4x_2+3x_3\\x_1(0)=2,\ x_2(0)=1,\ x_3(0)=-1\end{cases}}$

I am having trouble solving this. I know one method involves finding the eigenvalues and eigenvectors but is there not a method without using linear algebra and eigenvalues? 
 A: Hint
Solve this first then ...Substract 3) from 2)
$$x'_2-x'_3=-(x_2-x_3)$$
$$(x_2-x_3)'=-(x_2-x_3)$$
Substitute $z=x_2-x_3$ 
Then the equation becomes a simple first ODE easy to solve
$$z'=-z$$
$$\ln(z)=-t+K$$
$$z=Ke^{-t}$$
$$x_2(0)-x_3(0)=K \implies K=2 \implies x_2-x_3=2e^{-t}$$
First equation becomes ... 
$$
\begin{align}
1)x'_1&=3x_1-4x_2+4x_3\\
1)x'_1&=3x_1-4z\\
1)x'_1&=3x_1-8e^{-t}\\
\end{align}
$$
Which is easy to solve ..

First equation ($x_1$) 
$$x'_1=3x_1-8e^{-t}$$
$$x'_1-3x_1=-8e^{-t}$$
$e^{-3t}$ as integrating factor
$$x'_1e^{-3t}-3x_1e^{-3t}=-8e^{-t}e^{-3t}$$
$$(x_1e^{-3t})'=-8e^{-4t}$$
Simply integrate now
$$x_1e^{-3t}=-8\int e^{-4t}dx =2e^{-4t}+C$$
$$x_1=2e^{-t}+Ce^{3t}$$
We need to evaluate the constant C  for $t=0$
$$x_1(0)=2+C  \implies C=0 \implies x_1=2e^{-t}$$
$$\boxed{x_1=2e^{-t}}$$
Second Equation for $x_2$
$$x'_2=4x_1-5x_2+4x_3$$
We know the value of $x_1$ and we have a relation between $x_2$ and $x_3$
Because $x_1=2e^{-t}$
$$x'_2=8e^{-t}-5x_2+4x_3$$
Because $x_2-x_3=2e^{-t} \implies x_3= x_2-2e^{-t}$
$$x'_2=8e^{-t}-5x_2+4(x_2-2e^{-t})$$
$$x'_2=-x_2$$
$$x_2=Re^{-t} $$
Evaluate the constant R for $t=0$
$$x_2(0)=1 \implies R=1$$
$$\boxed {x_2=e^{-t} }$$
Third equation $x_3$
You don't need to solve any differential equation since you have a relation between $x_2$ and $x_3$
$$x_3=x_2-2e^{-t} =e^{-t}-2^{-t}$$
Therefore
$$\boxed{x_3=-e^{-t}}$$
You have finished...
