November 5th, 2016, 06:24 AM  #1 
Newbie Joined: Nov 2016 From: Ireland Posts: 3 Thanks: 0  2nd Order ODE Problem
I find 2nd Order ODEs quite hard and I've come across this question in my text book but am unsure of how to do it. Can anybody explain step by step how this is done? It would be a huge help considering I've an exam in a few weeks' time on this stuff. Last edited by skipjack; November 5th, 2016 at 11:24 AM. 
November 5th, 2016, 06:46 AM  #2 
Member Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2  The solution y = yh + yp (y homogen + y particular) First you find the homogen solution by converting the equation to mr^2 + cr + k = 0 It's simple: x'' = r^2, x' = r^1, kx = k Solve for r using basic algebra gives you r = something + sqrt(somethingelse) so you get r1 and r2. If r1 and r2 are different the yh (y homogen) is: yh = C1 * e^(r1*x) + C2 * e^(r2*x) If r1 = r2 are same you get: yh = (C1*x + C2) * e^(r1*x) C1 and C2 can only be solved for if you get conditions such as x(0) = 1 etc. Which you did get but you solve this after getting yp. To get yp you put the equation in standard form: y'' + k*y' + k2*y = k3 where k just being constants Now find the Integrating Factor, there's a lot of good tutorials on this on youtube. I suggest you learn it there. IF = e^integral(whatever was in front of y') Multiply both sides with IF and you will see that the left side can be written as: (y*IF)' = IF*right side integrate both sides and then divide both sides with IF to get y alone. Don't forget the constant when integrating right side. Now you got yp and yh, y = yp + yh Inserting x(0) (aka, y(0) = 1) in your equation should show clearly what constant should be what for the solution to work. I suggest youtube for help as well, since I probably did something wrong here. Gl! Last edited by skipjack; November 5th, 2016 at 10:57 AM. 
November 5th, 2016, 07:01 AM  #3  
Newbie Joined: Nov 2016 From: Ireland Posts: 3 Thanks: 0  Quote:
So for mx'' + cx' +kx = 0 I write it as mr^2 + cr + k = 0 Because that is a quadratic equation I can use the quadratic formula b +/ sqrt((b^24ac)/2a) And that will determine if I have real roots, one real root or a complex root right? Which is what you mentioned when you said If r1 and r2 are different the yh (y homogen) is: yh = C1 * e^(r1*x) + C2 * e^(r2*x) If r1 = r2 are same you get: yh = (C1*x + C2) * e^(r1*x) The integrating factor is e^(p(x)dx where p(x) is whatever is in front of the y' (first differential). I understand that part. I'm told the nat.frequency is Wn and the damping ratio is C, I need to show that mx'' + cx' + kx = 0 can be written as x'' + 2c,Wn x' +W^2nx = 0. I'm pretty confused as to how I sub in the Natural frequency and damping ratio into the above equation. Thanks for responding by the way! It's helped understand a good bit. Last edited by skipjack; November 5th, 2016 at 10:55 AM.  
November 5th, 2016, 07:45 AM  #4 
Member Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 
The way I see it, 2c*Wn is just a constant and W^2n is just another constant. Could be "shown" by substituting them with a single constant? Not sure either so much what they meant.. :/ Last edited by skipjack; November 5th, 2016 at 10:52 AM. 
November 5th, 2016, 07:51 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
The fact that there exist a "natural frequency" and a "damping ratio" tells you that your solution must be of the form $\displaystyle y(x)= e^{kx}(A \cos(\omega x)+ B \sin(\omega x))$ where $\displaystyle \omega$ is the "natural frequency" and k is the "damping ratio". That, in turn, means that the two roots of the characteristic equation are $\displaystyle k\pm i\omega$. And that, then, means that the characteristic equation is $\displaystyle (r k+ i\omega)(r k i\omega)= (r k)^2+ \omega^2= r^2 2k r+ k^2+ \omega^2= 0$ and, finally, that the differential equation is $\displaystyle \frac{d^2y}{dx^2} 2k\frac{dy}{dx}+ (k^2 \omega^2)y= 0$.
Last edited by skipjack; November 5th, 2016 at 10:51 AM. 
November 5th, 2016, 07:54 AM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra  Quote:
Matt, for the first part, you may well have some definition in your notes or text book. I would encourage you to look there. Alternatively, you could divide the first equation by $m$ (which cannot be zero) and then equate coefficients of $y$ and $y^2$. But that approach uses the answer to determine the definitions rather than using the definitions to determine the answer. The answer to the last part is, again, essentially book work. It should come straight out of the book or your notes. It may involve writing down and solving the characteristic polynomial for the equation and using the discriminant to determine the form that the solution of the equation takes, using the roots of the polynomial. Last edited by skipjack; November 5th, 2016 at 10:51 AM.  
November 5th, 2016, 12:10 PM  #7 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
This is a classic vibration problem and a classic ODE example. Differential Equations  Mechanical Vibrations 
November 5th, 2016, 01:22 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,309 Thanks: 1978 
1. One can define the natural frequency, $\omega_n$, as $\sqrt{k/m}$, and the damping ratio, $\zeta$, as $\dfrac{c}{2\sqrt{mk}}$. These seem to be what the question wants, though a textbook might explain why the terms "natural frequency" and "damping ratio" are used. It's easy to see how to use these definitions to finish this part of the question. 2. The dependent variable (for this problem) is $x$ and the independent variable is $t$ (time). The coefficients of the equation are constants, with $k$ and $m$ being positive. Solving it is bookwork, so please look up how to do that and make an attempt. 
November 7th, 2016, 08:55 PM  #9 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
Homogeneous Linear Ordinary Differential Equation with Constant Coefficients L(y) = $\displaystyle a_{0}y^{(n)}+...+a_{n1}y'+a_{n}y$ = 0 By substitution, $\displaystyle e^{rx}$ is a solution if r satisfies the polynomial P(r) = $\displaystyle a_{0}r^{n}+...+a_{n1}r+a_{n}$ = 0, which has n roots with complex roots occurring in pairs. Real root r, y = $\displaystyle e^{rx}$ Imaginary root, r = a ± ib If z complex satisfies L(y), so do real and imaginary parts. $\displaystyle e^{ibx}=\cos bx+i\sin bx$ $\displaystyle e^{bx}$ 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 = $\displaystyle e^{rx}, xe^{rx},..,x^{k1}e^{rx}$ Complex root of multiplicity k y = $\displaystyle e^{ax}\cos bx,e^{ax}\sin bx,xe^{ax}\cos bx,xe^{ax}\sin bx,..,x^{k1}e^{ax} \cos bx,x^{k1}e^{ax}\sin bx$ 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: OP 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$\displaystyle e^{i\omega_{f}t}$ Last edited by skipjack; November 8th, 2016 at 12:06 AM. 
November 10th, 2016, 04:37 AM  #10 
Newbie Joined: Nov 2016 From: Ireland Posts: 3 Thanks: 0 
I've done the 1st part and part (i) of the 2nd question. I am taking the initial values as x(0) =Xo and x'(0) = Xo' These are derivations. I know that when the damping ratio = 1 it's critically damped and therefore only has 1 real root. I've got as far as the line "Wn(zeta+/ (sqrt(zeta^2 1)) Last edited by skipjack; November 10th, 2016 at 05:27 AM. 

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