I'm trying to solve this system $$ \begin{align} x'&=x-3y+3z\\ y'&=-2x-6y+13z\\ z'&=-x-4y+8z \end{align} $$ must be reduced to a single equation I tried to express the x 3 and substitute in the other two but then I have not reduced y or z I can not understand how to solve it
Euler method can not be solved because it is written in the job to reduce to a single equation
On
I would look at the problem this way:
consider the vector $v = (x,y,z)^T$ and $ v' = (x',y',z')^T$. Then you can rewrite the problem like such:
$v'(t) = Av(t)$
where A is the matrix containing the coefficients of your functions. In your case:
$A :=\left( \begin{matrix} 1 &-3 & 3 \\ -2 & -6 & 13 \\ -1 & -4 & 8 \end{matrix} \right)$
Now you know that the solution must have the form $ v(t) = e^{At} \cdot c$ where $c \in \mathbb{R}^3$ is the vector with the constant values you could determine with given information about the functions. Assuming you know how the Jordan normal form works, you can rewrite this matrix as $ A = SJS^{-1}$. Then we have $ v(t) = e^{SJS^{-1}t} \cdot c$ and by using the definition of the exp function we get $ v(t) = S\cdot e^{Jt}\cdot S^{-1}\cdot c$. I will leave the matrix multiplication to you but I always thought that the $e^{Jt}$ was kinda tricky:
In your case:
$e^{Jt} = \exp\left( \begin{matrix} t & t & 0 \\ 0 & t & t \\ 0 & 0 & t \end{matrix} \right)$
You can divide this matrix into two separate ones: $I_n := \left( \begin{matrix} t & 0 & 0 \\ 0 & t & 0 \\ 0 & 0 & t \end{matrix} \right)$ and $N :=\left( \begin{matrix} 0 & t & 0 \\ 0 & 0 & t \\ 0 & 0 & 0 \end{matrix} \right)$
Then using $\exp(x+y) = \exp(x) \cdot \exp(y)$ you can now rewrite as such:
$e^{Jt} = e^{I_n} \cdot e^{N}$
and by taking a close look see that
$e^{I_n} = \left( \begin{matrix} e^t & 0 & 0 \\ 0 & e^t & 0 \\ 0 & 0 & e^t \end{matrix} \right)$
Almost done! But you must see that $N^k = 0 \forall k \geq 3$
That means you can calculate $e^N$ quite quickly and see that (if I calculated it correctly that is :) )
$e^N = \left( \begin{matrix} 1 & t & 0.5\cdot t^2 \\ 0 & 1 & t \\ 0 & 0 & 1 \end{matrix} \right)$
And therefore
$e^{Jt} = e^{I_n} \cdot e^{N} = \left( \begin{matrix} e^t & t\cdot e^t & 0.5\cdot t^2 \cdot e^t \\ 0 & e^t & t \cdot e^t \\ 0 & 0 & e^t \end{matrix} \right)$
And then you can finally see what your solutions must be by 'simple' matrix multiplication
$v(t) = S \cdot \left( \begin{matrix} e^t & t\cdot e^t & 0.5\cdot t^2 \cdot e^t \\ 0 & e^t & t \cdot e^t \\ 0 & 0 & e^t \end{matrix} \right) \cdot S^{-1} \cdot c $
I hope this could help you! If you are familiar with the Jordan normal form I think its a cool way to solve problems like this!
Best wishes,
Gustav
Matrix approach: name the equations as (1), (2), (3). To reduce them into one, you have to do $$a\cdot (1) +b\cdot (2) +c\cdot (3)$$ such that the right hand side is exactly a multiple of the right hand side. That is $$(ax+by+cz)'=(a-2b-c)x+(-3a-6b-4c)y+(3a+13b+8c)z\\ =\lambda(ax+by+cz)$$
This makes it an eigenvalue problem. You need to solve for eigenvalue $\lambda$, then find the eigenvectors. The eigenvector will make a good set of $a,b,c$.
Another approach: (without using matrix)
$$2\cdot (3)-(2): 2z'-y'=-2y+3z\tag{4}\\ (1)+(3): x'+z'=-7y+11z\implies x'=-z'-7y+11z$$ Now equate this with (1): $$-z'-7y+11z=x-3y+3z\implies x=-z'-4y+8z \tag{5}$$
This equation can be used to eliminate the variable $x$. Taking the derivative of (5) gives $$x'=-z''-4y'+8z' \tag{6}$$ Plug (5) and (6) into (1): $$-z''+9z'-11z=4y'-7y\tag{7}$$ $$(7)-4\cdot (4): y=-z''+z'+z\tag{8}$$ Take derivative of (8): $$y'=-z'''+z''+z'\tag{9}$$
Now plug (8) and (9) into (7): $$2z'-3z=-z'''+z''+z'-2(-z''+z'+z)$$
This is the single ode for $z$.