Math help (Diff Eq.)

[Suspicious-Teach8788]

How do I solve y'' + 4y = cosx

y(0) = 0, y(pi) = 0

I understand how to do it for homogeneous eq, cuz u would get y = c1cos2x + c2sin2x.. but yea.. now with the cos x at the end i get confused

 

[Syringer]

Plug it back in to the original equation.

 

[cirthix]

hmm...

 

[MrBailey]

step 1) find homogeneous solution...done

step 2) solve for the coefficients (c1 and c2) in the homogeneous solution using the initial conditions

step 3) the function on the right side, f(x), is a trig function...you can use method of undetermined coefficients.

let y = A cos(x) + B sin(x)

find y' and y'' (using the above equation)

step 3) plug y'' and y into your original equation...and now solve for the A and B coefficients.

step 4) now, between the homogeneous solution from step 1) and what you find in step 3), you will have the solution.


Hope this helps some.

Bailey

 

[simms]

Originally posted by: MrBailey
step 1) find homogeneous solution...done

step 2) solve for the coefficients (c1 and c2) in the homogeneous solution using the initial conditions

step 3) the function on the right side, f(x), is a trig function...you can use method of undetermined coefficients.

let y = A cos(x) + B sin(x)

find y' and y'' (using the above equation)

step 3) plug y'' and y into your original equation...and now solve for the A and B coefficients.

step 4) now, between the homogeneous solution from step 1) and what you find in step 3), you will have the solution.


Hope this helps some.

Bailey

That sounds about right. You have to find the homogeneous soln first, then solve for yp, the particular solution..

 

[hollowman]

i took the class just last semester and i don't remember a damn thing! man.. that's really sad. (i even got an A in that class too. lol)

what's wrong with me. blah~

 

[speg]

Wow, I really need to learn how to do this before I do my DE course for a third time... (third time's the charm!)

 

[MrBailey]

it's not as bad as it looks...simple algebra and calc.

I blew this crap off while an undergrad (many moons ago)...but I've had to relearn it for my first Master's. Now it's essential for my second Master's (mathematics).

Bailey

 

[Suspicious-Teach8788]

Originally posted by: speg
Wow, I really need to learn how to do this before I do my DE course for a third time... (third time's the charm!)

seconod time here =P

k just woke up for some hw finishing.

 

[George P Burdell]

Originally posted by: hollowman
i took the class just last semester and i don't remember a damn thing! man.. that's really sad. (i even got an A in that class too. lol)

what's wrong with me. blah~

lol, i'm in the same boat... but my excuse is that i had that class two years ago

 

[vrbaba]

judiciously guess A*cos(x) for your particular solution

 

[Suspicious-Teach8788]

Originally posted by: vrbaba
judiciously guess A*cos(x) for your particular solution

mmm thats what the answer says, but i dont know where that comes from =(

 

[Hyperion042]

Laplace transform ftw.

 

[vrbaba]

Originally posted by: DLeRium

Originally posted by: vrbaba
judiciously guess A*cos(x) for your particular solution

mmm thats what the answer says, but i dont know where that comes from =(

yeah, i never got those either. But its one of the diferent "ways" of solving diff. eq. You use it when others dont work, and you know something funky is going on. and since there is a cos(x), you try to use a cos or sin as your guess.

 

[vrbaba]

Here is a better (or wordier ) explanation by chuckywang:
the nonhomogenous part is cos(x). ask yourself what type of functions has a second derivative added to the function itself gives you cos(x). The answer is just a cosine. Usually you need to guess A*cos(x)+B*sin(x), but note that in this case y''+4y is only in terms of cos(x). However, if you had a y' on the left hand side, you would need to guess A*cos(x)+B*sin(x). You don't, in this case, so just guessing A*cos(x) suffices.