
Differential Equations Ordinary and Partial Differential Equations Math Forum 
 LinkBack  Thread Tools  Display Modes 
July 7th, 2011, 09:32 AM  #1 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond  Differential equation problem
[attachment=0:3arcjy5l]untitled_1.png[/attachment:3arcjy5l] I have the following problem from a book I'm reading, it occurs several chapters before integration and DE's are covered! The current in the RCLcircuit above must satisfy the equation If where and are constants, show that a solution is where and are any constants whatsoever and 
July 7th, 2011, 12:53 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs  Re: Differential equation problem
Perhaps they intend for you to take the given solution, find its first and second derivatives with respect to t, then substitute into the ODE to show that it satisfies the equation. However, here is how I would derive the solution from the given information: Given then and the ODE is then: First, we find the solution to the corresponding homogeneous equation, whose associated auxiliary equation is: Application of the quadratic formula yields: With we have Since the auxiliary equation has complex conjugate roots, this means: where A and B are constants. Now we may use the method of undetermined coefficients to find the particular solution . We begin by assuming it will be of the form: where are the coefficients to be determined. Differentiating with respect to t, we find: Substituting into the original ODE we have: Equating coefficients across the equation yields the system: Solving the 2nd equation for we find: Substituting for into the 1st equation gives: Solving for gives: And hence: Thus, our particular solution is: Using the linear combination identity we may write this as: where Defining we may now write: Now, the general solution f(t) is found by the superposition: 
July 7th, 2011, 04:30 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs  Re: Differential equation problem
Here are a few words on some of the methods I used. First, consider the following homogeneous equation: Because of the nature of the equation, it is reasonable to assume a solution will be of the form: where Differentiating with respect to x, we find: Substituting into the ODE gives: Since we are left with the auxiliary or characteristic equation: Thus, by the quadratic formula, we find the roots of the auxiliary equation to be: If we define: Then we have two solutions: where are arbitrary constants. Thus, we know: Adding the two equations, we get: Now, if we define: then and we may now write: where Now, if the roots of the auxiliary equation are complex conjugates, then we may express them as: where: Thus, we may express the solution as: Now, using Euler's formula we may express the solution as: Since and we may write: Since are arbitrary constants, we may write: Now, as we saw above, we know: Are both linearly independent solutions to the ODE, thus if we define: then we have: Since is a solution, then so is and we may express the general solution as: Now, consider the inhomogeneous equation: and the corresponding homogeneous equation: Suppose is a particular solution to the inhomogeneous equation and is the solution to the homogeneous equation. Note that the two solutions must be linearly independent, we then have: Adding the two equations, we get: With we now have: The particular solution will depend on the form taken by g(x) naturally, and your book should have a table listing the form it will take. 
July 7th, 2011, 04:35 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond  Re: Differential equation problem
Thanks! 
July 7th, 2011, 05:26 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,064 Thanks: 1621 
What is represented by the little yellow blob on the lefthand side of the diagram?

July 7th, 2011, 05:35 PM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond  Re: Differential equation problem
I don't see it. I don't think it means anything, whatever it is, the diagram is just a resistor, inductor and capacitor with some other component E, connected by lines. There is some discoloration on the lines.

July 7th, 2011, 06:47 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,064 Thanks: 1621 
I assume the gaps in the wiring lines are similarly unintended. The "other component" you refer to is presumably a power supply.

July 7th, 2011, 08:10 PM  #9 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs  Re: Differential equation problem
Thanks! It's interesting that an underdamped springmass system subject to an external sinusoidal force is governed by exactly the same type of ODE. Note that as , we have thus this is called the transient solution, and the other term is called the steadystate solution. 
July 8th, 2011, 06:15 AM  #10  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond  Re: Quote:
Quote:
 

Tags 
differential, equation, problem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Differential Equation Problem  stealth4933  Differential Equations  3  February 13th, 2014 06:01 AM 
Differential equation problem  rob747uk  Differential Equations  2  March 26th, 2012 08:55 AM 
Help with differential equation problem  sivela  Differential Equations  1  January 21st, 2011 05:53 PM 
differential equation problem  noul  Differential Equations  4  December 19th, 2010 04:34 AM 
Differential equation problem!  David McLaurin  Differential Equations  3  July 8th, 2009 07:50 AM 