2nd order nonhomogenous DEs

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  • #1
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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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Answers and Replies

  • #2
donpacino
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here is a hint that may help.
Euler's formula

e^(it)=cos(t)+i*sin(t)
 
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  • #3
SteamKing
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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
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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
BiGyElLoWhAt
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##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
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##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
BiGyElLoWhAt
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but you have no dy/dt expression in your diff eq?
 
  • #8
BiGyElLoWhAt
Gold Member
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oh I didn't see that in your initial conditions
 
  • #9
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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
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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.
 

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