Fraction of integrals involving a limit and a DE

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Find:

$$\frac{\overline{\text{P}}_1}{\overline{\text{P}}_2}:=\displaystyle\frac{\displaystyle\lim_{\text{n}\to\infty}\frac{1}{\text{n}}\int_0^\text{n}\text{f}_1\left(t\right)\cdot\text{y}_1\left(t\right)\space\text{d}t}{\displaystyle\lim_{\text{n}\to\infty}\frac{1}{\text{n}}\int_0^\text{n}\text{f}_2\left(t\right)\cdot\text{y}_2\left(t\right)\space\text{d}t}\tag1$$

The things we know:

  1. $$\text{f}_1\left(t\right)=\text{a}_1\cdot\sin\left(\omega_1\cdot t+\varphi_1\right)\tag2$$
  2. $$\text{f}_2\left(t\right)=\text{a}_2\cdot\sin\left(\omega_2\cdot t+\varphi_2\right)\tag3$$
  3. For $\text{y}_1\left(t\right)$ we need to solve: $$\text{f}_1\left(t\right)=\text{R}\cdot\text{y}_1'\left(t\right)+\text{L}\cdot\text{y}_1''\left(t\right)+\frac{1}{\text{C}}\cdot\text{y}_1\left(t\right)\tag4$$
  4. For $\text{y}_2\left(t\right)$ we need to solve: $$\text{f}_2\left(t\right)=\text{R}\cdot\text{y}_2'\left(t\right)+\text{L}\cdot\text{y}_2''\left(t\right)+\frac{1}{\text{C}}\cdot\text{y}_2\left(t\right)\tag5$$

All the variables are real and positive.


My work:

The work I did was wrong, thanks to @Fimpellizieri so I deleted my work.

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The corresponding homogenous ODE for $(4)$ is linear with constant coefficients and can be easily solved with classical tools. With respect to a particular solution for $(4)$, this kind of ODE is fairly standard. First, we write

$$f_1(t)=a_1\Big(\sin(\omega_1t)\cos(\varphi_1)+\cos(\omega_1t)\sin(\varphi_1)\Big).$$

Notice that $\cos(\varphi_1)$ and $\sin(\varphi_1)$ are constants. Now, we guess

$$y_1(t)=a_1\Big(x\sin(\omega_1t)+z\cos(\omega_1t)\Big),$$

where $x$ and $z$ are to be determined. Then

$$y_1'(t)=a_1\Big(x\omega_1\cos(\omega_1t)-z\omega_1\sin(\omega_1t)\Big)$$ $$y_1''(t)=a_1\Big(-x\omega_1^2\sin(\omega_1t)-z\omega_1^2\cos(\omega_1t)\Big)$$

Plugging everything into the ODE and cancelling out $a_1$, we find that

$$\sin(\omega_1t)\cos(\varphi_1)+\cos(\omega_1t)\sin(\varphi_1)=\\ \Big(-Rz\omega_1-Lx\omega_1^2+\frac1Cx\Big)\sin(\omega_1t) + \Big(-Rx\omega_1-Lz\omega_1^2+\frac1Cz\Big)\cos(\omega_1t) $$

Therefore, a solution comes from the system

$$\left\{\begin{array}{l} \cos(\varphi_1)=\left(\frac1C-L\omega_1^2\right)x-\left(R\omega_1\right)z\\ \sin(\varphi_1)=\left(\frac1C-L\omega_1^2\right)z-\left(R\omega_1\right)x \end{array}\right.$$

on the unknowns $x,z$. One can similarly solve $y_2(t)$.

Do you think you can take it from here?