# Second-order linear homogeneous ODEs **URL:** \ **Category:** Factual Questions\ **Created:** [September 27, 2003, 8:45pm UTC](https://boards.straightdope.com/t/second-order-linear-homogeneous-odes/204203 "2003-09-27T20:45:03Z")\ **Posts on this page:** 4\ **Page:** 1
**Author:** ![barking\_frog](https://avatars.discourse-cdn.com/v4/letter/b/96bed5/32.png) [@barking\_frog](https://boards.straightdope.com/u/barking_frog)\ **Post date:** [September 27, 2003, 8:45pm UTC](https://boards.straightdope.com/t/second-order-linear-homogeneous-odes/204203/1 "2003-09-27T20:45:03Z")
Could someone please outline the method (or provide a link) on how to solve such equations when they have an initial condition? Without an IC the problem is quite trivial, but we never covered such problems in class, nor is it in the book (and yes, I tried using google, but all I could find was it listed in course syllabi and not actually explained). ---
**Author:** ![missing\_link](https://avatars.discourse-cdn.com/v4/letter/m/d2c977/32.png) [@missing\_link](https://boards.straightdope.com/u/missing_link)\ **Post date:** [September 27, 2003, 9:09pm UTC](https://boards.straightdope.com/t/second-order-linear-homogeneous-odes/204203/2 "2003-09-27T21:09:05Z")
I’m sorry, but I don’t quite see the problem here. Maybe I am misunderstanding the question. If you solve a linear homogenous 2nd degree ODE (y’’ + ay’ + by = 0) you should get a function y(t) with two coefficients C1 and C2. I assume that you are given something like y(t=0) = A and y’(t=0) = B as initial conditions. Then you just put t=0 into your solution and set it equal to A. Then you take the first derivative of your solution, set t=0 and equal it to B. from this you can calculate C1 and C2. Thats all. ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [September 27, 2003, 9:21pm UTC](https://boards.straightdope.com/t/second-order-linear-homogeneous-odes/204203/3 "2003-09-27T21:21:58Z")
Have you used matrices to solve systems of linear equations before? That will do it, but it’s not absolutely necessary, since this only involves two equations with two unknowns, which is pretty straightforward. Say you have: y’’ - 5y’ + 6y = 0 with initial conditions y(0) = 1, y’(0) = 3 You know how to get the general solution: y(t) = Ae[sup]2t[/sup] + Be[sup]3t[/sup] From here, first plug in your IC information: y(0) = **A + B = 1** , and (since y’(t) = 2Ae[sup]2t[/sup] + 3Be[sup]3t[/sup]): y’(0) = **2A + 3B = 3**. Solve the two bolded equations for A and B: A = 1 - B 2(1 - B) + 3B = 3 2 + B = 3 B = 1 A = 1 - 1 = 0 And so our final solution is: y(t) = e[sup]3t[/sup] ---
**Author:** ![barking\_frog](https://avatars.discourse-cdn.com/v4/letter/b/96bed5/32.png) [@barking\_frog](https://boards.straightdope.com/u/barking_frog)\ **Post date:** [September 27, 2003, 9:46pm UTC](https://boards.straightdope.com/t/second-order-linear-homogeneous-odes/204203/4 "2003-09-27T21:46:10Z")
> [@](#): > > \*Originally posted by Cabbage \* > \*\*Have you used matrices to solve systems of linear equations before? That will do it, but it’s not absolutely necessary, since this only involves two equations with two unknowns, which is pretty straightforward. > > Say you have: > > y’’ - 5y’ + 6y = 0 > > with initial conditions > > y(0) = 1, y’(0) = 3 > > You know how to get the general solution: > > y(t) = Ae[sup]2t[/sup] + Be[sup]3t[/sup] > > From here, first plug in your IC information: > > y(0) = **A + B = 1** , > > and (since y’(t) = 2Ae[sup]2t[/sup] + 3Be[sup]3t[/sup]): > > y’(0) = **2A + 3B = 3**. > > Solve the two bolded equations for A and B: > > A = 1 - B > > 2(1 - B) + 3B = 3 > > 2 + B = 3 > > B = 1 > > A = 1 - 1 = 0 > > And so our final solution is: > > y(t) = e[sup]3t[/sup] \*\* Got it. Thanks.