Q:

System of equations : $1$) $(1+2x)A+(1+2y)B=0 $, $2$) $Ae^{x}+Be^{y}=0$, solving for $A,B$, $x \neq y$ with $x,y$ fixed

As part of some of my DE notes, there is a system of equations where $A,B,x,y \in \Bbb R$, $x \neq y$, $x,y$ fixed.
$1$) $(1+2x)A+(1+2y)B=0 $
$2$) $Ae^{x}+Be^{y}=0$
This is all coming from a solution to a differential equation $y(t)=c_1e^{xt}+c_2e^{yt}$, with conditions $y(0)+2y'(0)=0$, $y(1)=0$.
When solving for $A,B$,the author states without explanation that $A=B=0$. 
All I notice so far is that $B=-Ae^{x-y}$. Any tips on how to see this appreciated, (or if there is wrong, the teacher makes mistakes frequently).

A:

The existence of a non-trivial solution, i.e. not $A = B = 0$, depends on the determinant of the matrix
$$
\begin{pmatrix} 1+ 2x & 1+2y \\ e^x & e^y
\end{pmatrix},
$$
which is
$$
\Delta = (1+2x)e^y-(1+2y)e^x.
$$
If you know that $(1+2x)e^y-(1+2y)e^x \neq 0$, then $A=B=0$ is the only solution, otherwise there is an explicit non-zero solution.

