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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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  • $\begingroup$ +1 Good question: If you already have the solution why not plug it into the integral and evaluate the integral limit first. If it converges then $P=0$. Have you tried that? I am too lazy to type this into Maple :D. Only if the integral is $O(n)$ then we can have a convergent non-zero answer. So maybe asymptotics should be added as a tag. $\endgroup$ – MrYouMath Oct 25 '17 at 14:15
  • $\begingroup$ @MrYouMath Thanks for giving the +1. Mathematica gives no answer. $\endgroup$ – Jan Oct 25 '17 at 14:22
  • $\begingroup$ I edited your question such that it is better to comprehend. $\endgroup$ – MrYouMath Oct 25 '17 at 14:34
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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

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  • $\begingroup$ That is what the question is about, not about finding the integral $\endgroup$ – Jan Oct 25 '17 at 16:03
  • $\begingroup$ If you know the coefficients $\alpha$ and $\beta$ it should be possible to see what happens with the limit. If the limit exists then you can spilt all components apart and see if for what conditions the limits exist. It is easy to see from the previous integral that if $\beta-2\alpha>0$ then the limit will not exist. On the other hand if $ $\endgroup$ – MrYouMath Oct 25 '17 at 16:14
  • $\begingroup$ It is clear that the question is not only about the integral. But the explicit solution of the integral can help you on the path to a solution. $\endgroup$ – MrYouMath Oct 25 '17 at 16:16

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