Solving a coupled system of linear ODEs (one second order, the other first order)

$$\frac{d^2 T(x)}{d x^2}-\beta (T(x)-t(x))+K=0 \tag 1$$

$$\frac{d t(x)}{dx}-\alpha(T(x)-t(x))=0 \tag 2$$

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.

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 $\frac{dt(x)}{dx}$.