TITLE: Solving a coupled system of linear ODEs (one second order, the other first order)
QUESTION [5 upvotes]: I have two coupled ODEs for $T(x)$ and $t(x)$:
$$\frac{d^2 T(x)}{d x^2}-\beta (T(x)-t(x))+K=0 \tag 1$$
$$\frac{d t(x)}{dx}-\alpha(T(x)-t(x))=0 \tag 2$$
$\alpha, \beta$ and $K$ are constants $>0$. Also, it is known that $t(x=0)=t_i$. Additionally, for $(1)$ we know:
$$\frac{d T(x=0)}{d x}=\frac{d T(x=L)}{d x} = 0$$
I need to determine $T(x)$ and $t(x)$.  Can anyone suggest a way towards moving ahead with this problem ?
Probably this system of coupled equations can be solved using the matrix method, but I am not aware of it. I normally solve a single equation using either the method of integrating factor or using a characteristic equation and finding the roots.

REPLY [6 votes]: $$\frac{d^2 T(x)}{d x^2}-\beta (T(x)-t(x))+K=0 \tag 1$$
$$\frac{d t(x)}{dx}-\alpha(T(x)-t(x))=0 \tag 2$$
HINT :
From $(2) \qquad T=\frac{1}{\alpha}\frac{d t}{dx}+t$
$\frac{d^2T}{dx^2}=\frac{1}{\alpha}\frac{d^3t}{dx^3}+\frac{d^2t}{dx^2}$
Puting them into $(1)$ :
$\frac{1}{\alpha}\frac{d^3t}{dx^3}+\frac{d^2t}{dx^2}-\frac{\beta}{\alpha}\frac{d t}{dx} +K=0$
$$\frac{d^3t}{dx^3}+\alpha\frac{d^2t}{dx^2}-\beta\frac{d t}{dx} +\alpha K=0$$
This is a linear ODE with constant coefficients. I suppose that you can take it from here.

REPLY [3 votes]: Hint :
Substitute $T(x)=\frac{1}{\alpha}\frac{d t(x)}{dx}+t(x)$ in $(1)$ :
$$  \frac{1}{\alpha}\frac{d^3t(x)}{dx^3}+\frac{d^2t(x)}{dx^2}-\frac{\beta}{\alpha}\frac{d t(x)}{dx} +K = 0$$
Then solve for $\frac{dt(x)}{dx}$.