Limit in an integral involving a DE

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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:

  1. $$y_1(t)=u\tag3$$

Where $u$ is just a constant.

  1. $$y_2(t)=r\cos\left(\omega t+\phi\right)\tag4$$

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 $\left(5\right)$?

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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