# What should be the particular solution?

1. May 13, 2005

### tony_engin

Hi all!
In this differentail equation, what particular should I try?
Is it correct to try
"[Asinx + Bcosx + C] exp(-2x)"?

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2. May 13, 2005

### Pyrrhus

Yes it's ok.

3. May 13, 2005

### dextercioby

Or you can do it by the method of variation of constants.I've always thought of it as being elegant...

Daniel.

4. May 14, 2005

### tony_engin

p.s. please open the word file as i don't know how to type mathematical symbols here..

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5. May 14, 2005

### Pyrrhus

Yes you should try

$$x^{2}e^{-2x}(Ax \cos 2x + Bx \sin 2x + C)$$

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

$$y'' + 3y' + 2y = 4x^2$$

if we rewrite the above ODE with the differential operator D

$$(D^2 + 3D +2)y = 4x^2$$

Now if we use a differential operator $D^3$ we could eliminate $4x^{2}$

$$D^3(D^2 + 3D +2)y = 4 D^3 x^2$$

$$D^3(D^2 + 3D +2)y = 0$$

Thus making a fifth grade auxiliary equation. There are more steps it was just a preview.

Last edited: May 14, 2005
6. May 14, 2005

### tony_engin

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. May 14, 2005

### dextercioby

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. May 14, 2005

### tony_engin

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. May 14, 2005

### dextercioby

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. May 14, 2005

### tony_engin

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. May 14, 2005

### dextercioby

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. May 14, 2005

### tony_engin

So, it should be "[(Asinx + Bcosx)x + C] exp(-2x)" or "x[Asinx + Bcosx + C] exp(-2x)"?

13. May 14, 2005

### dextercioby

The first without the C,it's useless there.

Daniel.

14. May 14, 2005

### Pyrrhus

If you're still refering to your ODE.

$$y_{p_{1}} = Ce^{-2x}$$

$$y_{p_{2}} = xe^{-2x}(A \cos x + B \sin x)$$

$$y_{p} = Ce^{2x} + xe^{-2x}(A \cos x + B \sin x)$$

15. May 15, 2005

### saltydog

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:

$$(D+2)(D^2+4D+5)y=0\quad\text{(1)}\quad$$

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

If you have:

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

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

$$(D+2)(D^2+4D+5)(D^2+4D+5)y=0\quad\text{(3)}\quad$$

Right?

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

So the solution of (3) is:

$$y(x)=c_1e^{-2x}+c_2e^{-2x}Cos(x)+c_3e^{-2x}Sin(x)+c_4xe^{-2x}Cos(x)+c_5xe^{-2x}Sin(x)$$

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

$$y_p(x)=Ae^{-2x}+Bxe^{-2x}Cos(x)+Cxe^{-2x}Sin(x)$$

Now what? How about a plot of the solution (just make up some initial conditions)?

Last edited: May 15, 2005
16. May 15, 2005

### saltydog

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

$$y_p(x)=e^{-2x}+\frac{1}{2}xe^{-2x}Sin(x)$$

So that the general solution is:

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

Letting:

y(0)=1

y'(0)=1

we get:

$$y(x)=3e^{-2x}Sin(x)+e^{-2x}+\frac{1}{2}xe^{-2x}Sin(x)$$

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

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