Solving Second Order Nonhomogeneous DEs with Initial Conditions

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In summary: Thanks for pointing out my mistake!In summary, the conversation discusses finding the particular solution to a differential equation and applying initial conditions. The hint of using Euler's formula is mentioned, and the correct approach of setting the derivative equal to zero at t=0 is suggested. Eventually, the individual realizes their mistake and is able to solve the problem by equating real and complex terms.
  • #1
Plaetean
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Homework Statement


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Homework Equations



y=yPI+yCF

The Attempt at a Solution


First issue is I was under the impression that a particular solution is the final solution to a DE; a solved DE with initial conditions applied, but it would be weird for that to be the first part of the question so I'm interpreting a) as asking for the particular integral (maybe they're the same thing :S). Anyway for the first part of the question I've done:

yPI=Ceit

so dyPI/dt = iCeit and d2yPI/dt2= -Ceit

-C+4C=1

which gives: C=1/3

so yPI=(1/3)eit

Using a trial solution for the complementary function gives yCF=Ae2it+Be-2it

and a general solution of:

y=Ae2it+Be-2it+(1/3)eit

At this point I have no idea how to start applying initial conditions as I have an expression that's full of 'i's, so maybe I don't understand the first part of the question and have gone about the whole thing the wrong way. Some help would be really appreciated!
 
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  • #2
here is a hint that may help.
Euler's formula

e^(it)=cos(t)+i*sin(t)
 
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  • #3
Sorry. You need to brush up on your ODEs.

The 'final solution' to a linear ODE, to use that unfortunate term, is composed of the solution to the homogeneous ODE plus the solution to the particular ODE. Because there are various constants associated with these solutions, application of the initial values is required to determine the exact solution to the specified ODE and initial conditions.

This problem is a little inverted, in that it asks you to find a particular solution first, then find the homogeneous solution.

To apply your two initial conditions, take your general solution y(t) = MESS and set t = 0, since y(0) = 0.
Then, for y'(0) = 0, calculate the derivative y'(t) and set t = 0 there. You'll wind up with two equations for the two unknown constants A and B. Solve for same.
 
  • #4
donpacino said:
here is a hint that may help.
Euler's formula

e^(it)=cos(t)+i*sin(t)

I tried using this, but when they say y(0)=0, shouldn't they give the value as a complex number? Is it just the real part that =0, or is the complex part 0 as well?
 
  • #5
##0 + 0i = 0##
if: ##y(t) = e^{it} + C##,
then ##y(0) = e^{0i} + C = 1+C##
or: ##y(t) = e^{it} + C = \text{cos}(t) + i \text{sin}(t) + C##,
and ##y(0) = \text{cos}(0) + i \text{sin}(0) + C = 1 + 0i + C##
 
  • #6
BiGyElLoWhAt said:
##0 + 0i = 0##
if: ##y(t) = e^{it} + C##,
then ##y(0) = e^{0i} + C = 1+C##
or: ##y(t) = e^{it} + C = \text{cos}(t) + i \text{sin}(t) + C##,
and ##y(0) = \text{cos}(0) + i \text{sin}(0) + C = 1 + 0i + C##

In this problem its not y(t)=eit+c but ceit. I'm stuck because I can get an expression for the coefficients with y(0)=0 as below:

y(0)=1/3+A+B

But if I use Euler's method and differentiate the expression, all the real terms become sine terms, so saying that dy/dt(0)=0 doesn't help.
 
  • #7
but you have no dy/dt expression in your diff eq?
 
  • #8
oh I didn't see that in your initial conditions
 
  • #9
Plaetean said:
In this problem its not y(t)=eit+c but ceit. I'm stuck because I can get an expression for the coefficients with y(0)=0 as below:

y(0)=1/3+A+B

But if I use Euler's method and differentiate the expression, all the real terms become sine terms, so saying that dy/dt(0)=0 doesn't help.

That's where you mistake is! Don't use Euler method Find the complete complex form for dy/dt and set it to zero at t=0.
 
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  • #10
Ah I got it, didn't need to use Euler's at all, had a massive brain fart and forgot that I could just equate real and complex terms anyway to solve for constants.
 

1. What is a 2nd order nonhomogeneous differential equation?

A 2nd order nonhomogeneous differential equation (DE) is a type of mathematical equation that involves a function, its derivatives, and a non-zero function on one side of the equation. It is called nonhomogeneous because the function on one side of the equation is not equal to zero, unlike in a homogeneous DE.

2. How do you solve a 2nd order nonhomogeneous DE?

To solve a 2nd order nonhomogeneous DE, you can use the method of undetermined coefficients or the method of variation of parameters. These methods involve finding a particular solution and then combining it with the general solution of the corresponding homogeneous DE to get the complete solution.

3. What is the difference between a 2nd order homogeneous and nonhomogeneous DE?

The main difference between a 2nd order homogeneous and nonhomogeneous DE is the presence of a non-zero function on one side of the equation in the nonhomogeneous DE. This makes the solution more complex, as it requires finding a particular solution in addition to the general solution of the corresponding homogeneous DE.

4. What are some real-world applications of 2nd order nonhomogeneous DEs?

2nd order nonhomogeneous DEs have many real-world applications in fields such as physics, engineering, and economics. Some examples include modeling the motion of a spring-mass system, predicting the growth of a population, and analyzing the behavior of electrical circuits.

5. Can a 2nd order nonhomogeneous DE have multiple solutions?

Yes, a 2nd order nonhomogeneous DE can have multiple solutions. This is because the particular solution and the general solution of the corresponding homogeneous DE can be combined in different ways to form different solutions. These solutions may also differ depending on the initial conditions given in the problem.

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