Where did I go wrong when solving this system of differential equations? We have been given two differential equation,
$$
\begin{alignat}{1}
\frac{dy}{dx}&=2y+z,\\
\frac{dz}{dx}&=3z.
\end{alignat}
$$
I need to solve this differential equation simultaneously.
My approach
From the first differential equation, on integrating with respect to x:
$$y = y^2+zy.$$
Putting this equation of $y$ into the second differential equation yields:
$$\frac{dz}{dx}=3y^2+3zy,$$
which is a linear differential equation with respect to $x$. Now, it can be solved as usual, and after solving my answer turned out to be:
$$(z+y)e^{-3yx}=C$$
Now, this answer does not match with any of the following four solutions that were provided:


*

*$(z+y)e^{3x}=C$

*$(z+3y)e^{-3x}=C$

*$(y-z)e^{3x}=C$

*$(z+3y)e^{3x}=C$


What did I do wrong?
 A: Solve the second equation first
$$z'=3z \implies \int \frac {dz}{z}=3\int dx \implies \ln |z|=3x \implies z=Ke^{3x}$$
Now you can susbtitute $z$ in the first equation
$$y'=2y+z \implies y'-2y=Ke^{3x}$$
You can solve that last equation easily....
A: $z$ is a function of $x$, so the first step in your solution is wrong.
Solve the second first to find
$$
z(x)=Ce^{3x}
$$
now you can tackle the first by substituting $z$.
A: In the context of a multiple choice question, you don't necessarily have to solve the system to arrive at the answer. Note that subtracting the 2 equations give
$$ \frac{dy}{dx} - \frac{dz}{dx} = 2y - 2z $$
$$ \frac{d}{dx}(y-z) = 2(y-z) $$
This gives
$$ y-z = Ce^{2x} $$
or $(y-z)e^{-2x} = C$, which is unfortunately none of the options given.
A: For some reason the author of your exercise seems to have mixed up left and right eigenvectors. For the linear combinations that decouple the system you need the left eigenvectors, for the solutions you need the right eigenvectors. The left eigenvectors of $\pmatrix{2&1\\0&3}$ are $\pmatrix{-1&1}$ for eigenvalue $2$ and $\pmatrix{0&1}$ for eigenvalue $3$, where the first one gives the answer by Dylan.
The right eigenvectors in contrast are $\pmatrix{1\\0}$  for eigenvalue $2$ and $\pmatrix{1\\1}$  for eigenvalue $3$.
Now if one were to consider the ODE with the transposed matrix,
\begin{align}
y'&=2y\\
z'&=y+3z
\end{align}
the left and right eigenvectors would change position and one would get for $\pmatrix{1&1}$ the equation
$$
y'+z'=3(y+z)\implies y+z=Ce^{3x}.
$$
which would be a mix of the first and second variant, $(z+y)e^{-3x}=C$.
