User:Egm4313.s12.team11.arrieta/Report 5

Problem R5.4
Solved by Francisco Arrieta

Problem Statement
Show that $$ y_p(x)=\sum_{i=0}^{n}y_{p,i}(x)\!$$ is indeed the overall particular solution of the L2-ODE-VC $$ y''+p(x)y'+q(x)y=r(x) \!$$ with the excitation $$ r(x)=r_1(x)+r_2(x)+...+r_n(x)=\sum_{i=0}^{n}r_i(x)\!$$.

Discuss the choice of $$ y_p(x)\!$$ in the above table e.g., for $$ r(x)=k \cos\omega x\!$$ why would you need to have both $$ \cos \omega x \!$$ and $$\sin \omega x \!$$ in $$ y_p(x) \!$$ ?

Solution
Because the ODE is a linear equation in y and its derivatives with respect to x, the superposition principle can be applied:

$$r_1(x) \!$$ is a specific excitation with known form of $$ y_{p1}(x)\!$$ and $$r_2(x) \!$$ is a specific excitation with known form of $$ y_{p2}(x)\!$$

$$+\begin{cases} & \text{ } y_{p1}''+p(x)y_{p1}'+q(x)y_{p1}=r_1(x) \\ & \text{ } y_{p2}''+p(x)y_{p2}'+q(x)y_{p2}=r_2(x) \end{cases}\!$$

becomes

$$(y_{p1}+y_{p2})''+p(x)(y_{p1}+y_{p2})'+q(x)(y_{p1}+y_{p2})=r_1(x)+r_2(x)\!$$

proving that

$$ y_p(x)=\sum_{i=0}^{n}y_{p,i}(x)\!$$ is indeed the overall particular solution of the L2-ODE-VC $$ y''+p(x)y'+q(x)y=r(x) \!$$ with the excitation $$ r(x)=\sum_{i=0}^{n}r_i(x)\!$$

According to Fourier Theorem periodic functions can be represented as infinite series in terms of cosines and sines:

$$f(x)=a_0+\sum_{n=1}^{\infty}[a_n\cos\omega x+b_n\sin\omega x] \!$$

where the coefficients $$ a_0, a_n , b_n \!$$ are the Fourier coefficients calculated using Euler formulas.

So even though the system is being excited by functions like $$ r(x)=k\cos \omega x \!$$ the particular solution would still include both $$ \sin \omega x \!$$ and $$ \cos \omega x \!$$ in $$ y_p(x) \!$$ because the excitation is a periodic function that can be represented as the Fourier infinite series in terms of both $$ \sin \omega x \!$$ and $$ \cos \omega x \!$$ times the Fourier coefficients