Finding Particular Integral for Non-Homogenous Differential Equation

[Kawakaze]

Homework Statement

Find a particular integral of the non-homogenous differential equation

$$\frac{d^2y}{dx^2}+6\frac{dy}{dx}+10y=-6sin(x)+9cos(x)$$

The Attempt at a Solution

Solution is of the sort
$$qsin(x)+pcos(x)$$

Derivatives are
$$\frac{dy}{dx}=qcos(x)-psin(x)$$

$$\frac{d^2y}{dx^2}=-qsin(x)-pcos(x)$$

Substituting these in I eventually get that I require

$$(6p+9p)cos(x)+(10q-7p)sin(x)=-6sin(x)+9cos(x)$$

Comparing sin and cos terms I get

$$10q-7p=-6$$ (1)
$$6q+9p=9$$ (2)

Then I get the two simultaneous equations, rearranging 2 and putting it into 1 i get

$$p=\frac{21}{22}$$

This isn't the nice round number I have come expect from "simple" examples, so i stopped here and ask you all to put me back on track please. :)

 

[dextercioby]

I think you've done a mistake. Where did the 6q term come from ?

 

[Kawakaze]

Yes you are correct, I typed the question in wrong. :S

Edited

 

[HallsofIvy]

Homework Statement

Find a particular integral of the non-homogenous differential equation

$$\frac{d^2y}{dx^2}+6\frac{dy}{dx}+10y=-6sin(x)+9cos(x)$$

The Attempt at a Solution

Solution is of the sort
$$qsin(x)+pcos(x)$$

Derivatives are
$$\frac{dy}{dx}=qcos(x)-psin(x)$$

$$\frac{d^2y}{dx^2}=-qsin(x)-pcos(x)$$

so y''+ 6y'+ 10y= -q sin(x)- pcos(x)+ 6qcos(x)- 6psin(x)+ 10qsin(x)+ 10pcos(x)
= (-q- 6p+ 10q)sin(x)+ (-p+ 6q+ 10p)cos(x)
= (9q- 6p)sin(x)+ (6q+ 9p)cos(x)

Substituting these in I eventually get that I require

$$(6p+9p)cos(x)+(10q-7p)sin(x)=-6sin(x)+9cos(x)$$

Then I suggest you do that over again. You seem to have gotten a "p" and "q" confused in the coefficient of sin(x).

Comparing sin and cos terms I get

$$10q-7p=-6$$ (1)
$$6q+9p=9$$ (2)

Then I get the two simultaneous equations, rearranging 2 and putting it into 1 i get

$$p=\frac{21}{22}$$

This isn't the nice round number I have come expect from "simple" examples, so i stopped here and ask you all to put me back on track please. :)

 

[Kawakaze]

Thats great, thanks a lot. Its always the little things that are the hardest to track down.

I get p=1 and q=0

Does that tally at your end? :)

 

[Kawakaze]

Oops forgot to ask, how does this apply to the general solution? I get

$$C + De^{6x} + cos(x)$$

 

[dextercioby]

Oops forgot to ask, how does this apply to the general solution? I get

$$C + De^{6x} + cos(x)$$

Where did you get e^(6x) from ?

 

[Kawakaze]

The 6 comes from the second term of the original equation. The b so to speak.

 

[dextercioby]

The solution of the homogenous ODE must include $e^{\alpha x}$ as solutions, where

$$\alpha^2 + 6\alpha + 10 = 0$$

So I don't see any e^(6x) as 6 is not a solution of the quadratic.

 

[Kawakaze]

Sorry i don't understand, do i need to find the auxillary equation as well? Or is it simply

$$C + De^{x} + cos(x)$$

 

[dextercioby]

Perhaps we're not speaking the same language here. The full solution of the ODE is a sum between the general solution (which I showed you how to find it) and the particular solution which was the purpose of this thread to which both Halls and me gave you some support.

 

[Kawakaze]

I just reread thngs again, i see what you mean. I am sorry for being a little slow, this stuff is new to me and its overloading my fragile little braincell. :) Thanks for your help I do really appreciate it.

 

[Kawakaze]

Take 2 :)

$$\alpha^2 + 6\alpha + 10 = 0$$

This is the complimentary function right? I would solve this with the standard a, b, c formula. To save precious time I am using wolphram alpha :)

This gives x = -3 - i and x = -3 + i

How does this apply to my particular integral?

 

[tiny-tim]

Hi Kawakaze!

This gives x = -3 - i and x = -3 + i

Yes, so your general solution is Ae^{(-3 + i)x} + Be^{(-3 - i)x} …

you can simplify that to e^-3x(Ae^ix + Be^-ix),

which it's more convenient to write as e^-3x(Ccosx + Dsinx)

 

[Kawakaze]

Bingo, I get it. Thanks guys for your patience. If it frustrates me to spend so long on what turns out to be relatively quick if i had know what to do, i don't know where you get your patience from. :)

 

[Kawakaze]

This question turns to an initial value problem now, the general solution is

$$y=e^{-3x}(Ccos(x)+Dsin(x))+cos(x)$$

I need to find the initial values for when y(0)=3 and y'(0)=12

Differentiate general solution and plug in the values, get 2 simultaneous equations to solve? The reason i ask is that looks like a nightmare to differentiate.

 

[tiny-tim]

I need to find the initial values for when y(0)=3 and y'(0)=12

Differentiate general solution and plug in the values, get 2 simultaneous equations to solve? The reason i ask is that looks like a nightmare to differentiate.

Scaredy-cat!

 

[Kawakaze]

Scaredy-cat!

No arguments here! :P

$$\frac{dy}{dx}=-3e^{-3x}C-sin(x)+3e^{-3x}Dcos(x)-sin(x)$$

close?

 

[tiny-tim]

Hi Kawakaze!

(just got up :zzz: …)

hmm … I think I'd better do this one for you, and then you can try d²y/dx² …

to differentiate e^-3x(Ccosx + Dsinx),

you use the product rule … differentiate the left bit on its own (and multiply by the right), then add on the result of differentiating the right bit on its own (and multiply by the left) …

so that's -3e^-3x(Ccosx + Dsinx) + e^-3x(Dcosx - Csinx),

= e^-3x((D - 3C)cosx - (C + 3D)sinx) …

ok, now you try d²y/dx² ​

 

[Kawakaze]

Good morning, sleeping beauty, is this a new meaning to sleeps with the fishes? :)

Product rule, *facepalm* yes its glaring me in the face, i know that one. Working a full 8hr day and then going on to 5am on an assessment that is due really shuts your brain down!

$$\frac{d^2y}{dy^2}=-3e^{-3x}((D - 3C)cosx - (C + 3D)sinx) + e^{-3x}((D - 3C)-sinx + (C + 3D)cosx)-cos(x)$$

but where did the cos(x) go, from the end of the solution?

 

[tiny-tim]

Good morning, sleeping beauty, is this a new meaning to sleeps with the fishes? :)

oh yes, I've heard of that saying …

we have a similar saying, "he sleeps with the apes!"

but where did the cos(x) go, from the end of the solution?

I knew you could do that, so I left it off

ok, now rearrange all the cos together, and all the sin together, inside that big bracket,

then go back to the original equation, put cos = 1, sin = 0, and e^-3x = 1, and solve!

 

[Kawakaze]

oh yes, I've heard of that saying …

we have a similar saying, "he sleeps with the apes!"

Not heard that one, I am sure my missus will agree that she sleeps with an ape :D

$$e^{-3x)(-3Dcos(x)+9Ccos(x)+3Ccos(x)+9Dsin(x)-Dsin(x)+3Csin(x)-Ccos(x)-3Dcos(x)-cos(x))$$
$$e^{-3x}((-6D+11C)cos(x)+(8D+3C)sin(x)-cos(x))$$

this is not what wolphram alpha gives me, :S

 

[Kawakaze]

then go back to the original equation, put cos = 1, sin = 0, and e^-3x = 1, and solve!

where does this come from?

 

[tiny-tim]

where does this come from?

From putting x = 0 to get the initial values …

I need to find the initial values for when y(0)=3 and y'(0)=12

(oh, you don't need y''(0) after all … it was on page 1, and I didn't check it )