What should be the particular solution?

[tony_engin]

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

[Pyrrhus]

Yes it's ok.

[dextercioby]

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

Daniel.

[tony_engin]

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

[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.

[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?

[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.

[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.

[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.

[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..

[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.

[tony_engin]

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

[dextercioby]

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

Daniel.

[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)

[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)?

[saltydog]

Just some follow up:

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?