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- Thread starter tony_engin
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Pyrrhus

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Yes it's ok.

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Daniel.

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Pyrrhus

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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.

[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.

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Daniel.

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Since this is an exercise in trying the particular solution, I would like to know the form of the trial particular solution.

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Daniel.

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but the answer given by the book shows that this is wrong..

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Daniel.

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

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

Daniel.

Daniel.

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Pyrrhus

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[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]

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saltydog

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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)?

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)?

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saltydog

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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?

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?

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