Limit in an integral involving a DE 
When we have the following differential equation:
$$y'(t)=a\cdot x'(t)+\frac{1}{b}\cdot x(t)+c\cdot x''(t)\tag1$$
With the following initial conditons: $x(f)=g,x(h)=k$.
Question, find:
$$P:=a\lim_{n\to\infty}\frac{1}{n}\int_0^nx^2(t)\space\text{d}t\tag2$$
All the variables are real and positive.


My work:
We know that $y(t)$ can have two forms:

*

*$$y_1(t)=u\tag3$$
Where $u$ is just a constant.


*$$y_2(t)=r\cos\left(\omega t+\phi\right)\tag4$$
When we have the first situation we get:
$$y'(t)=y_1(t)=\frac{\text{d}}{\text{d}t}\left(u\right)=0=a\cdot x'(t)+\frac{1}{b}\cdot x(t)+c\cdot x''(t)\space\Longleftrightarrow\space$$
$$x(t)=\exp\left(-\alpha t\right)\cdot\left\{K_1+K_2\cdot\exp\left(\beta t\right)\right\}\tag5,$$
in which
$$\alpha = \frac{a+\frac{\sqrt{ba^2-4c}}{\sqrt{b}}}{2c}$$
$$\beta =\frac{\sqrt{ba^2-4c}}{c\cdot\sqrt{b}}.$$

Now, what will be $P$ for $\left(5\right)$?

 A: I substituted the coefficients in your exponentials by $\alpha$ and $\beta$
The integral  from $0$ to $k$ is given by (I used Maple)
$$a/2\,{\frac { \left( -4\,K_{1}\,K_{2}\,{{\rm e}^{{\it \beta}\,k}}{{\it 
\alpha}}^{2}+4\,K_{1}\,K_{2}\,{{\rm e}^{{\it \beta}\,k}}{\it \alpha}\,{
\it \beta}-2\,{K_{2}}^{2}{{\rm e}^{2\,{\it \beta}\,k}}{{\it \alpha}}^{2}+
{K_{2}}^{2}{{\rm e}^{2\,{\it \beta}\,k}}{\it \alpha}\,{\it \beta}+2\,{K_{
1}}^{2}{{\rm e}^{2\,{\it \alpha}\,k}}{{\it \alpha}}^{2}-3\,{K_{1}}^{2}{
{\rm e}^{2\,{\it \alpha}\,k}}{\it \alpha}\,{\it \beta}+{K_{1}}^{2}{
{\rm e}^{2\,{\it \alpha}\,k}}{{\it \beta}}^{2}+4\,K_{1}\,K_{2}\,{{\rm e}
^{2\,{\it \alpha}\,k}}{{\it \alpha}}^{2}-4\,K_{1}\,K_{2}\,{{\rm e}^{2\,{
\it \alpha}\,k}}{\it \alpha}\,{\it \beta}+2\,{K_{2}}^{2}{{\rm e}^{2\,{
\it \alpha}\,k}}{{\it \alpha}}^{2}-{K_{2}}^{2}{{\rm e}^{2\,{\it \alpha}\,
k}}{\it \alpha}\,{\it \beta}-2\,{K_{1}}^{2}{{\it \alpha}}^{2}+3\,{K_{1}}^
{2}{\it \alpha}\,{\it \beta}-{K_{1}}^{2}{{\it \beta}}^{2} \right) {
{\rm e}^{-2\,{\it \alpha}\,k}}}{{\it \alpha}\, \left( 2\,{\it \alpha}-{
\it \beta} \right)  \left( {\it \alpha}-{\it \beta} \right) }}
$$
The next step would be to determine the limit of this expression
