Trial Solutions of Particular Integrals?

In summary, Davio was having trouble understanding an integral and was looking for help. He found that using his notes and equating coefficients with A0=1 and A1=0, he was able to solve the equation for cos x. However, he is still unsure of why he is getting incorrect results. He then tried using D operators and got the same answer as his original equation. He also found the general solution to be e^x (C cos x + D cos x) + 1/5 cos x + 2/5 sin x.
  • #1
Davio
65
0
Hiya, I'm having trouble understanding particular integrals at the moment. From what I understand, you have to try trial solutions, unfortunately I've got no idea how to start? Google doesn't turn up anything .. I think I need to examine the equation, and somehow determine its form?

For example this is what I'm currently trying to solve :

Y" + 2y` +2y = cos x

y=?

From my notes, if f(x) = A cos wx + A1 sin wx then y = a0 cos wx + a1 sin wx
From this, I'm guessing A1=0? so y=a1 sin wx
and y' is a1 w cos wx
and y'' is -a1 w^2 wx

And then I'm stuck, how do I now find the PI?

ps. Does anyone have any good notes on the PI and relavent methods to determine what form its in?

Sorry for the messy post, the working came to me as I was typing out the post!
Thanks in advance!
 
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  • #2
Davio said:
I'm guessing A1=0?

Hi Davio! :smile:

Don't guess!

Don't take short-cuts ("more haste, less speed")! :frown:

Learn the method in your notes, and follow it!

Use a0 and a1, and you should get a solution! :smile:
 
  • #3
After a bit of thought ... and more deciphering of my notes (missing just one lecture makes everything so much more hard!), I've come up with:
PI= 1/5 cos x + 2/5 sin x

from me putting in the equations, and equating coefficients with A0=1 and A1=0 (on the premise that 0 sin x = 0).
Putting that back into the original equation.. doesn't get me the right answer, I'll muse on it, thanks for the reply :D
 
  • #4
Hey Davio
Have you guys learned about D operators?
I found that method to help immensely when dealing with linear ODE's.
Good luck!

Edit: They're not going to help us here with the particular integral since [tex]\cos x[/tex] is infinitely differentiable. D operators are still worth a look though :)
 
Last edited:
  • #5
D operators? Nope never heard of them :-D, I'll learn it after my exams, since I think they would be quite strict if I used something they didn't teach me to solve their questions :-p

I now have PI= 1/10 cos x + 1/5 sin x, which equals to 1/2 cos x = 1 cos x ..
Still not sure why? I've checked my working a few times, but I don't see any obvious errors.

a0 (c-aw^2)+a1 . b . w =A0

-a0 . b .w + a1 (c-aw^2) =A1

With A1 =0 and A0 = 1
and a=1 b=2 c=2

a0=? a1=?
 
  • #6
Oh God, I was right the first time, thanks guys! :-p
 
  • #7
Hi Davio! :smile:

Your original solution, "PI= 1/5 cos x + 2/5 sin x", was correct.

Just put it back into the equation, and it gives cosx. Try it again! :smile:

Then I used your equations (with w = 1), and I got the same result, 1/5 cos x + 2/5 sin x.

I don't understand how you keep making these mistakes. :confused:

If you type out your whole working, I'll try to spot where you went wrong.

ok, now you've got a PI, what about the general solution?

(btw, I always use the D method … what method have you been taught?)
 
  • #8
Hey, yeah, I made a mistake the second time round, probabily due to me staring at the question too long, thanks for help though!
I worked out the gen solution to be e^x (C cos x + D cos x) + 1/5 cos x + 2/5 sin x. Which in comparson to my friends working, is correct! :-D
 

1. What is a trial solution of a particular integral?

A trial solution of a particular integral is a proposed solution to a differential equation that includes a general constant or constants. It is used to find the particular solution of the differential equation by substituting it into the equation and solving for the constants.

2. How do you determine the trial solution for a particular integral?

The trial solution for a particular integral is typically determined by examining the non-homogeneous term of the differential equation. The form of the trial solution will depend on the type of non-homogeneous term, such as a polynomial, exponential, or trigonometric function.

3. Can more than one trial solution be used for a particular integral?

Yes, multiple trial solutions can be used for a particular integral. This is especially useful when the non-homogeneous term is a combination of different types of functions, as each type may require a different trial solution.

4. What is the purpose of using a trial solution for a particular integral?

The purpose of using a trial solution is to simplify the process of finding the particular solution of a differential equation. By substituting the trial solution into the equation, the constants can be solved for and a specific solution can be obtained without having to solve the entire differential equation.

5. Are there any limitations to using trial solutions for particular integrals?

Yes, there are limitations to using trial solutions for particular integrals. They may not work for all types of non-homogeneous terms, and in some cases, the trial solution may need to be modified or a different approach may need to be used to find the particular solution.

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