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 November 29th, 2016, 10:12 AM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Linear Ordinary Differential Equation with Constant Coefficients Linear Ordinary Differential Equation with Constant Coefficients. This is a concise outline and summary of a solution paradigm for the topic, ie, just follow the steps, and do and interpret the resulting algebra. L(y) = $\displaystyle a_{0}y^{(n)}+...+a_{n-1}y'+a_{n}y$ = 0 By substitution, $\displaystyle e^{rx}$ is a solution if r satisfies the polynomial P(r) = $\displaystyle a_{0}r^{n}+...+a_{n-1}r+a_{n}$ = 0, which has n roots with complex roots occurring in pairs. Real root r, y = $\displaystyle e^{rx}$ Imaginary root, r = ± ib If z complex satisfies L(y)=0, so do real and imaginary parts. y = $\displaystyle e^{ibx}=\cos bx+i\sin bx$ $\displaystyle e^{-ibx}$ doesn't add anything new, so y = C$\displaystyle _{1}\cos bx$ + C$\displaystyle _{2}\sin bx$ Complex root, r = (a ± ib) y = $\displaystyle e^{(a\pm ib)x}=e^{ax}\cos bx\pm ie^{ax}\sin bx$ y = $\displaystyle C_{1}e^{ax}\cos bx+C_{2}e^{ax}\sin bx$ Note: C$\displaystyle _{1}\cos bx$ + C$\displaystyle _{2}\sin bx$ = A$\displaystyle \sin(bx+\phi$), A and $\displaystyle \phi$ arbitrary constants $\displaystyle \phi=\tan^{-1}\frac{C_{1}}{C_{2}}$, A = $\displaystyle \sqrt{C_{1}^{2}+C_{2}^{2}}$, from $\displaystyle \sin(\alpha +\beta)=\sin\alpha\cos\beta + \sin\beta\cos\alpha$ Real root of multiplicity k y = $\displaystyle e^{rx}, xe^{rx},..,x^{k-1}e^{rx}$ Complex root of multiplicity k y = $\displaystyle e^{ax}\cos bx,e^{ax}\sin bx,xe^{ax}\cos bx,xe^{ax}\sin bx,..,x^{k-1}e^{ax} \cos bx,x^{k-1}e^{ax}\sin bx$ In general, there will be n arbitrary constants corresponding to n roots of P(r), with a solution: $\displaystyle y_{c}=C_{1}y_{1}+C_{2}y_{2}+....+C_{n}y_{n}$ and n initial conditions to determine C$\displaystyle _{i}$'s. Example1: y' + py = 0 y = $\displaystyle e^{rx}$ r + p = 0 r = -p y = C$\displaystyle e^{-px}$ Example 2: 2nd order linear ODE with constant coefficients (google). Now suppose L(y) = F$\displaystyle \cos\omega_{f}t$ Consider L(y) = F$\displaystyle e^{i\omega_{f}t}$ If z complex satisfies L(z) = F$\displaystyle \cos\omega_{f}t$, then R(z) satisfies L(R(z)) = R(F$\displaystyle e^{i\omega_{f}t}$) Let z = A$\displaystyle e^{i\omega_{f}t}$ and sub into L(z) = F$\displaystyle e^{i\omega_{f}t}$. Then solve P(A) = F for A (complex) and then R(A$\displaystyle e^{i\omega_{f}t}$) satisfies L(y) = F$\displaystyle \cos\omega_{f}t$. If A = a + ib = B$\displaystyle e^{i\theta }$, R(A$\displaystyle e^{i\omega_{f}t}$) = B$\displaystyle \cos(\omega_{f}t+\theta)$ = y$\displaystyle _{p}$ y = y$\displaystyle _{c}$+y$\displaystyle _{p}$ Finally, if an arbitrary periodic function can be expressed as a Fourier series, the above applies. Ref: z = u + iv, $\displaystyle \bar{z}$= u - iv, R(z) = u $\displaystyle e^{z}=e^{u}e^{iv}=e^{u}(\cos v+i\sin v)$ z' = u' + iv' $\displaystyle (e^{z})$' = $\displaystyle e^{z}$z' L(u+iv) = L(u) + iL(v) = F$\displaystyle e^{i\omega_{f}t}$ ------------------------------------- Minor corrections added from the original post in: 2nd Order ODE Problem
 November 30th, 2016, 06:25 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 That covers only homogeneous linear ODEs with constant coefficients.
November 30th, 2016, 12:29 PM   #3
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Quote:
 Originally Posted by skipjack That covers only homogeneous linear ODEs with constant coefficients.
"Now suppose L(y) = Fcosω$\displaystyle _{f}$t"

I switched independent variable to t. Engineering force of habit.
This also takes care of case where forcing function is a Fourier series.

Last edited by zylo; November 30th, 2016 at 12:35 PM.

 November 30th, 2016, 12:36 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 What does "F" mean?
November 30th, 2016, 12:51 PM   #5
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Quote:
 Originally Posted by skipjack What does "F" mean?
What does x=Acoswt mean? F, like A, is a constant.

 November 30th, 2016, 06:56 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra There's no justification there for "converting" the imaginary parts of your solution into real parts. In fact, there's no real justification for most solutions, most especially the ones from repeated roots. You also fail to treat the majority of cases of non-homogeneous equations, most significantly in the context of your writing, the case of resonance. Last edited by skipjack; November 30th, 2016 at 10:05 PM.

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