I asked this question before but with a linear algebra angle, however, I need to solve it using the elimination method.
I have been able to solve for $y$, which is $y=\frac{-y''}{2}+\frac{3y'}{2}$, however, I'm not going about solving this in any logical way. I'm all over the place and feel like I found $y$ by pure luck.
I'm randomly taking derivatives and substiting results into random equations and feel like there is no method to my madness.
How does one approach a problem like this in a logical way?
$x'=x+2y-z$
$y'=x+z$
$z'=4x-4y+5z$
I need to find $x(t)$, $y(t)$, and $z(t)$
Not sure where you could have made your mistake, but the solution given above for $y$ is wrong. Start with $$\begin{align}x^{\prime} & =x+2y-z\\ y^{\prime} & =x+z\\ z^{\prime} & =4x-4y+5z\end{align}$$ Solve the simplest equation available for one of the variables $$\begin{align}x & =y^{\prime}-z \end{align}$$ Then substitute into the other equations $$\begin{align}y^{\prime\prime}-z^{\prime} & =y^{\prime}-z+2y-z\\ z^{\prime} & =4y^{\prime}-4z-4y+5z\end{align}$$ Simplify $$\begin{align}y^{\prime\prime}-y^{\prime}-2y & =z^{\prime}-2z\\ 4y^{\prime}-4y & =z^{\prime}-z\end{align}$$ Subtract the second equation from the first $$\begin{align}y^{\prime\prime}-5y^{\prime}+2y & =-z\\ \end{align}$$ We can differentiate and substitute into one of the equations $$\begin{align}4y^{\prime}-4y & =-y^{\prime\prime\prime}+5y^{\prime\prime}-2y^{\prime}+y^{\prime\prime}-5y^{\prime}+2y\\ \end{align}$$ Simplify our differential equation for $y$ $$\begin{align}y^{\prime\prime\prime}-6y^{\prime\prime}+11y^{\prime}-6y & =0\\ \end{align}$$ If we try the solution $y=e^{rt}$ we find $$r^3-6r^2+11r-6=(r-1)(r-2)(r-3)=0$$ So the general solution for $y$ is $$y=c_1e^t+c_2e^{2t}+c_3e^{3t}$$ We had an equation a while back the gave $z$ in terms of $y$ so now $$z=c_1(-1+5-2)e^t+c_2(-4+10-2)e^{2t}+c_3(-9+15-2)e^{3t}=2c_1e^t+4c_2e^{2t}+4c_3e^{3t}$$ And now go back to that equation where so long ago we had solved for $x$ in terms of $y$ and $z$ $$x=c_1(1-2)e^t+c_2(2-4)e^{2t}+c_3(3-4)e^{3t}=-c_1e^t-2c_2e^{2t}-c_3e^{3t}$$ Check by differentiating and substituting into the original differentialequation, and you're done.