Proof/Intuition for Eigenvalues to Solve Linear Differential Equations To solve an equation of the form 
$$\frac{dx}{dt}=ax+by \\
\frac{dy}{dt}=cx+dy$$
Does anyone know the reasoning why you solve for eigenvalues and eigenvectors to determine the functions for x and y with e raised to the power of the eigenvalue times t? I can get an answer just curious as to why the method works out. I attached a link https://www.youtube.com/watch?v=1_EPFlwS7Kc with an example. So my question is to prove this method to solve linear differential equations.
 A: For your system of equations we can express them in matrix form like this:
$$\begin{pmatrix}
x'\\
y'
\end{pmatrix} =\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}$$ 
Assuming a solution for the column matrix for $x,y$ is in the form
$$\begin{pmatrix}
a_1\\
a_2
\end{pmatrix}e^{\lambda t}$$
Then we can substitute the column matix for $x,y$ and derive the following equation
$$\lambda\begin{pmatrix}
a_1\\
a_2
\end{pmatrix}e^{\lambda t} = \begin{pmatrix}
a & b\\
c & d
\end{pmatrix}\begin{pmatrix}
a_1\\
a_2
\end{pmatrix}e^{\lambda t}$$
Writing this as a system of equations we get
$$aa_1 + ba_2 = \lambda a_1 $$
$$ca_1 + da_2 = \lambda a_2$$
$$$$
$$(a - \lambda)a_1 + ba_2 = 0$$
$$ca_1 + (d  -\lambda)a_2 = 0$$
To avoid trivial solutions the determinant of the matrix must be zero so
$$\begin{vmatrix}
a-\lambda & b\\
c & d - \lambda\\
\end{vmatrix}= 0$$
Therefore $\lambda$ must be the matrix's eigen values. So you get that the solution is the sum of the product of the eigenvectors with is corresponding eigenvalue raised to the $e$th power
A: The intuition is this:  Solving coupled simultaneous differential equations in more than one variable is "too hard." So you look for some linear transformation of the original variables into other variables, such that when you write out the system of equations in the other variables, it splits off into a bunch of ordinary differential equations, in one variable each.
For the system of equations shown, finding that linear transformation is precisely equivalent to the problem of finding the eigenvectors of the matrix of  coefficients on the right. And once you apply that transformation, you get a diagonal matrix; that is, the equation for the $n$-th other variable is just 
$dy/dy = wx$ where $w$ is the diagonal element in row $n$ of the transformed matrix.  
Finally, the diagonal entries in the transformed matrix are of course the eigenvalues of the original matrix, and that is why this method works the way it does.
A: The primary intuition, in my opinion, comes from the fact that $e^{kt}$ is an eigenvector of the operator $\frac{d}{dt}$, with corresponding eigenvalue $k$.
If we rewrite the system you have in matrix notation:
$$ \frac{d \boldsymbol{x}}{dt} = A \boldsymbol{x} $$
we can see that a solution can be found of the form $\boldsymbol{x} = \boldsymbol{c} e^{kt}$.  Can you see that, when $\boldsymbol{c}$ is an eigenvector of $A$ and $k$ is the corresponding eigenvalue, then $\boldsymbol{x} = \boldsymbol{c} e^{kt}$ is a solution of this differential equation?
Here, you can think of it as solving for the eigenvalues and eigenvectors of $\frac{d}{dt}$.  The differential equations you begin with simply provide a way for you to express $\frac{d}{dt}$ as a matrix (i.e. $A =$ [a b; c d]).
A: You can think about superposition, which might help to see why this method is related to linear equations. The point is, if $Ax_0=\lambda x_0$, then the equation $x'=Ax,x(0)=x_0$ stays on the line spanned by $x_0$*. You can see this by thinking about a discrete approximation: you should have $x(h) \approx x_0+\lambda h x_0$, which is still a multiple of $x_0$. As a result, you can think about the problem as being a one parameter problem, in that $x(t)=s(t) x_0$ where $s(t)$ is some scalar-valued function. $s(t)$ solves $y'=\lambda y$ which you know very well how to solve.
Now, suppose $Ax_1=\lambda_1 x_1$ and $Ax_2=\lambda_2 x_2$, and consider $x'=Ax,x(0)=c_1 x_1+c_2 x_2$. Now you can exploit linearity to find that $x=c_1 e^{\lambda_1 t} x_1 + c_2 e^{\lambda_2 t} x_2$ for free. So if we could write any vector in this way, then we would be done. Attempting to write every vector in this way is the usual diagonalization procedure that you've learned.
* If $\lambda$ is complex then this line is actually in $\mathbb{C}^n$, not $\mathbb{R}^n$, but that's OK.
