What should be the particular solution?

  • Thread starter tony_engin
  • Start date
  • #1
45
0
Hi all!
In this differentail equation, what particular should I try?
Is it correct to try
"[Asinx + Bcosx + C] exp(-2x)"?
 

Attachments

  • de.JPG
    de.JPG
    7.7 KB · Views: 377

Answers and Replies

  • #2
Pyrrhus
Homework Helper
2,179
1
Yes it's ok.
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
13,112
663
Or you can do it by the method of variation of constants.I've always thought of it as being elegant...

Daniel.
 
  • #4
45
0
what about this?
p.s. please open the word file as i don't know how to type mathematical symbols here..
 

Attachments

  • math problem2.doc
    22.5 KB · Views: 152
  • #5
Pyrrhus
Homework Helper
2,179
1
Yes you should try

[tex] x^{2}e^{-2x}(Ax \cos 2x + Bx \sin 2x + C) [/tex]

There are other methods for solving nonhomogenous constant coefficients ODEs such as Anulator Method by using the differential operator, and also Variation of parameters.

Here's a preview of the Anulator Method.

Imagine a ODE

[tex] y'' + 3y' + 2y = 4x^2 [/tex]

if we rewrite the above ODE with the differential operator D

[tex] (D^2 + 3D +2)y = 4x^2 [/tex]

Now if we use a differential operator [itex] D^3 [/itex] we could eliminate [itex] 4x^{2} [/itex]

[tex] D^3(D^2 + 3D +2)y = 4 D^3 x^2 [/tex]

[tex] D^3(D^2 + 3D +2)y = 0 [/tex]

Thus making a fifth grade auxiliary equation. There are more steps it was just a preview.
 
Last edited:
  • #6
45
0
For the first differential equation that I firtsly raised, the answer of the particular solution should be something like "[1+xcosx] exp(-2x)" as given by the book. So I think the trial form should not be "[Asinx + Bcosx + C] exp(-2x)", right? So, what should be the correct form?
 
  • #7
dextercioby
Science Advisor
Homework Helper
Insights Author
13,112
663
I've already given you the hint to do it.It's up to you if u prefer other method,just as long as it leads to the result.The correct one.

Daniel.
 
  • #8
45
0
um...you mean do it by variation of constants?
Since this is an exercise in trying the particular solution, I would like to know the form of the trial particular solution.
 
  • #9
dextercioby
Science Advisor
Homework Helper
Insights Author
13,112
663
One of the solution of the homogenous one contains that e^{-2x},so that should give a hint on how to pick the trial function.

Daniel.
 
  • #10
45
0
I previously thought that I should try "[Asinx + Bcosx + C] exp(-2x)"
but the answer given by the book shows that this is wrong..
 
  • #11
dextercioby
Science Advisor
Homework Helper
Insights Author
13,112
663
Of course.You need an "x" to multiply the exponential,because the exponential is contained both in the solution of the hom.eq. and in the nonhomogeneity term.

Daniel.
 
  • #12
45
0
So, it should be "[(Asinx + Bcosx)x + C] exp(-2x)" or "x[Asinx + Bcosx + C] exp(-2x)"?
 
  • #13
dextercioby
Science Advisor
Homework Helper
Insights Author
13,112
663
The first without the C,it's useless there.

Daniel.
 
  • #14
Pyrrhus
Homework Helper
2,179
1
If you're still refering to your ODE.

[tex] y_{p_{1}} = Ce^{-2x} [/tex]

[tex] y_{p_{2}} = xe^{-2x}(A \cos x + B \sin x) [/tex]

[tex] y_{p} = Ce^{2x} + xe^{-2x}(A \cos x + B \sin x) [/tex]
 
  • #15
saltydog
Science Advisor
Homework Helper
1,590
3
Tony, where are you with this? Getting it?

This is what I'd do (other than just plug it into Mathematica):

I tell you what, since the RHS is a particular solution to the homogeneous ODE:

[tex](D+2)(D^2+4D+5)y=0\quad\text{(1)}\quad[/tex]

you can apply this operator to both sides of the original equation to aniliate the RHS, that is:

If you have:

[tex]y^{''}+4y^{'}+5y=e^{-2x}(1+Cos(x))\quad\text{(2)}\quad[/tex]

Applying the differential operator in (1) to both sides yields:

[tex](D+2)(D^2+4D+5)(D^2+4D+5)y=0\quad\text{(3)}\quad[/tex]

Right?

The solution of (3) will contain the solution of (2), just gets messy that's all.

So the solution of (3) is:

[tex]y(x)=c_1e^{-2x}+c_2e^{-2x}Cos(x)+c_3e^{-2x}Sin(x)+c_4xe^{-2x}Cos(x)+c_5xe^{-2x}Sin(x)[/tex]

And thus a particular solution of (2) will be:

[tex]y_p(x)=Ae^{-2x}+Bxe^{-2x}Cos(x)+Cxe^{-2x}Sin(x)[/tex]

Now what? How about a plot of the solution (just make up some initial conditions)?
 
Last edited:
  • #16
saltydog
Science Advisor
Homework Helper
1,590
3
Just some follow up:

Well, if you substitute [itex]y_p(x)[/itex] into (2) above, it gets a little messy, do this and that, equate coefficients, some things cancel, and we're left with:

[tex]y_p(x)=e^{-2x}+\frac{1}{2}xe^{-2x}Sin(x)[/tex]

So that the general solution is:

[tex]y(x)=c_1e^{-2x}Cos(x)+c_2e^{-2x}Sin(x)+e^{-2x}+\frac{1}{2}xe^{-2x}Sin(x)[/tex]

Letting:

y(0)=1

y'(0)=1

we get:

[tex]y(x)=3e^{-2x}Sin(x)+e^{-2x}+\frac{1}{2}xe^{-2x}Sin(x)[/tex]

I've attached a plot of this solution. Tony, you can do all this right?
 

Attachments

  • ode6.JPG
    ode6.JPG
    3.6 KB · Views: 277

Related Threads on What should be the particular solution?

  • Last Post
Replies
2
Views
659
  • Last Post
Replies
2
Views
2K
Replies
1
Views
3K
  • Last Post
Replies
2
Views
896
  • Last Post
Replies
7
Views
8K
  • Last Post
Replies
2
Views
4K
Replies
1
Views
3K
  • Last Post
Replies
4
Views
796
Replies
1
Views
1K
Top