<s>[INST] Forward and Backward Euler. I want to consider this differential system:
$$
\ \frac{dx}{dt} = -y(t)\\
\frac{dy}{dt} = \ x(t)
$$
where $t>0$ with initial condition$ (x(0),y(0))=(1,0).$
First I want to show that this differential equation admits an invariant of $I = x(t)^2 + y(t)^2$ Also, Can someone help me to figure out if Forward Euler, backwards Euler, or implicit trapezoidal rule admit invariants similar to $I = x(t)^2 + y(t)^2$? 
 [/INST] I would like to explain this, using explicit Euler, because it is the most elementary.
We have:
\begin{align}
\begin{pmatrix}
x \\ y
\end{pmatrix}'
 = 
\begin{pmatrix}
-y \\ x 
\end{pmatrix}
= f(x,y)
\end{align}
Explicit euler looks like this: $z_{k+1} = z_k +h \cdot f(z_k)$, applied to our problem it becomes:
\begin{align}
\begin{pmatrix}
x_{k+1} \\ y_{k+1}
\end{pmatrix}
=
\begin{pmatrix}
x_{k} \\ y_{k}
\end{pmatrix}
+h 
\begin{pmatrix}
-y_{k} \\ x_{k}
\end{pmatrix}
=
\begin{pmatrix}
x_{k}-hy_k \\ y_{k}+h x_k
\end{pmatrix}
\end{align}
So if we want to know, whether $x(t)^2+y(t)^2$ are constant, we check
\begin{align}
x_{k+1}^2+y_{k+1}^2 = (x_k -h y_k)^2 +(y_k+ h x_k)^2 = (x_k^2+y_k^2)(1+h^2)
\end{align}
Give the initial values $x_0=1$ and $y_0=0$ and using recursion, we get
\begin{align}
x_{k+1}^2+y_{k+1}^2 &=(x_k^2+y_k^2)(1+h^2) =(x_{k-1}^2+y_{k-1}^2)(1+h^2)^2 =\  ... \ =(x_0^2+y_0^2)(1+h^2)^{k+1}\\ &= (1+h^2)^{k+1}
\end{align}
So it is no invariant.
Next implicit euler. The method reads  $z_{k+1} = z_k +h \cdot f(z_{k+1})$
So we get
\begin{align}
& \begin{pmatrix}
x_{k+1} \\ y_{k+1}
\end{pmatrix}
=
\begin{pmatrix}
x_{k} \\ y_{k}
\end{pmatrix}
+h 
\begin{pmatrix}
-y_{k+1} \\ x_{k+1}
\end{pmatrix}
=
\begin{pmatrix}
x_{k}-hy_{k+1} \\ y_{k}+h x_{k+1}
\end{pmatrix}
\\
 & \Leftrightarrow
\begin{pmatrix}
x_{k+1}+ h y_{k+1} \\ y_{k+1} -h x_{k+1} 
\end{pmatrix}
= 
\begin{pmatrix}
x_{k} \\ y_{k} 
\end{pmatrix}
\\
& \Leftrightarrow
\begin{pmatrix}
1 & h \\ 1 & -h
\end{pmatrix}
\begin{pmatrix}
x_{k+1} \\ y_{k+1}
\end{pmatrix}
=
\begin{pmatrix}
x_{k} \\ y_{k} 
\end{pmatrix}
\end{align}
So as you can see, you end up with a linear system you have to solve for $x_{k+1} $ and $y_{k+1}$. This is the price you pay for an implicit scheme. If you solve for $x_{k+1} $ and $y_{k+1}$ you can again check the invariant.
Good luck,
Thomas
</s>