System of differential equations - linear system, shooting method I need a little help with solving of this system, please. 
\begin{align}
y'' &= -(a/b)\, y'-c \,(z-y/e)\\
z' &= -(c/f)\, (z-y/e),
\end{align}
 where $a,b,c,f$ are known constants


*

*$z(0) = 100$ I.C.

*$a \,y(5)-b\, y'(5) = 0$ B.C.

*$y'(0) = 0$ I.C.


Firstly, I have to solve it for $e = 100$ as a system of linear diff equations and find value for $z(2)$. Then I have to calculate using shooting method and then use proper Runge-Kutta approximation (using $e = 10*sqrt(y)$, y $=<$ use $y/e = 0$). 
Is there anybody who can push me a bit? Or give me a link for similar problem? Thanks. 
 A: Define $\boldsymbol{y} := (y,y',z)^{\top}$ to obtain a linear system of first-order ordinary differential equations for $\boldsymbol{y}$:
\begin{eqnarray}
\boldsymbol{y}' = \left( \begin{array}{c}
y'\\
-\frac{a}{b} y' - c \left( z - \frac{y}{e} \right)\\
-\frac{c}{f} \left( z - \frac{y}{e} \right)
\end{array}
\right) =: \boldsymbol{f}(x,\boldsymbol{y}),
\end{eqnarray}
with conditions
\begin{equation}
z(0) = 100, \quad a y(5) - b y'(5) = 0, \quad y'(0) = 0.
\end{equation}
This is not an initial-value problem for $\boldsymbol{y}$ because the second condition is given at $x > 0$.
Introducing the condition $y(0) = \alpha$ with some unknown parameter $\alpha \in \mathbb{R}$, we may now solve the initial-value problem
\begin{equation}
\boldsymbol{y}' = \boldsymbol{f}(x,\boldsymbol{y}), \quad \boldsymbol{y}(0) = \left( \begin{array}{c}
\alpha\\
0\\
100
\end{array}
\right) =: \boldsymbol{y}_0(\alpha),
\end{equation}
whose solution $\boldsymbol{y}(x;\alpha) = \left(y(x;\alpha),y'(x;\alpha),z(x;\alpha)\right)^{\top}$ depends on the unknown parameter $\alpha$. Now we have to determine the value of $\alpha$ such that the nonlinear equation
\begin{equation}
F(\alpha) := a y(5;\alpha) - b y'(5;\alpha) = 0
\end{equation}
is satisfied. This will require multiple evaluations of $F$, and each evaluation of $F$ requires the solution of an initial-value problem with a different initial value.
Edit: All of this was assuming that $e$ is a constant as well. Of course with $e = 10 \sqrt{y}$ we obtain a nonlinear system which seems not easier but more difficult to solve!
A: Your boundary conditions on the left side, $t=0$, miss only one entry. Complementarily, on the right side you have one condition to satisfy. Selecting some input value for the missing left condition, you can integrate to obtain a value for the condition on the right side as output. This now is a continuous and, on a medium scale, differentiable scalar function. Now you can apply you preferred root seeking method, for instance the secant method or some bracketing method, to find the root.
In the first instance, this function will be linear up to the accuracy of the ODE integration, so that the secant method should converge in very few steps. In the second, non-linear case, you should observe the behavior of the root-seeking method on a typical non-linear function.
