Show the system of differential equations has the solution of the form $ \phi(x)=e^{kx}\alpha$ 
Consider the following linear system
  $$y_1 '=ay_1+by_2$$$$y_2 '=cy_1+dy_2$$
  where $a,b,c,d$ are constants. Then show that this system always has a solution $\phi(x)=e^{kx}\alpha,$ where $\alpha=(\alpha_1,\alpha_2)\ne(0,0)$ is a constant vector and $k$ is a constant. 

My approach was to suppose $\phi(x)$ to be a solution and define the function vector $\psi(x)=e^{-kx} \phi(x)$. And show that $\psi'(x)=(0,0)=\mathbf 0$ by using the given system and the fact that $\phi(x)$ is the solution. However, I did get a trouble in showing that $\psi'(x)=(0,0)=\mathbf 0$. 
Could anyone give me a hand on this part? OR does anyone have a simpler way to show this? 
Thanks in advance. 
 A: Hint: You aren't asked to show that EVERY solution is of the form $e^{kx}\alpha$; you're just asked to show that you can choose $\alpha=(\alpha_1,\alpha_2)\neq(0,0)$ and $k$ such that $\phi(x)=e^{kx}\alpha$ is a solution.
Here, you're trying to take $y_1=\alpha_1e^{kx}$ and $y_2=\alpha_2e^{kx}$, and you want
$$
\begin{align*}
k\alpha_1e^{kx}&=a\alpha_1e^{kx}+b\alpha_2e^{kx}=(a\alpha_1+b\alpha_2)e^{kx},\\
k\alpha_2e^{kx}&=c\alpha_1e^{kx}+d\alpha_2e^{kx}=(c\alpha_1+d\alpha_2)e^{kx}.
\end{align*}
$$
So, you want to show that you can choose $\alpha$ and $k$ such that
$$
k\alpha_1=a\alpha_1+b\alpha_2\qquad\text{and}\qquad k\alpha_2=c\alpha_1+d\alpha_2.
$$
Can you figure out how to do that?  Again, you don't need to find EVERY solution -- just a solution.
A: A different way to approach the problem: we can write this system as $\vec y' = A\vec y $, where $\vec y = (y_1\quad y_2)^T$ and
$$ 
A = \pmatrix{a&b\\c&d}
$$
Now, $A$ must have some eigenvalue and corresponding eigenvector.  Let $k$ be this eigenvalue and let $\alpha$ be this eigenvector.  Show that $\phi(x) = e^{kx}\alpha$ is a solution to this system of differential equations.
