Finding homogeneous soln to y'' + 3y' + 2y = sin(2x)

the_chemist

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The question is this:
Classify in terms of order, degree, and linearity, and find the homogeneous solutions to...

y'' + 3y' + 2y = sin(2x)

My problem is this: I don't know how to even begin this problem. I thought maybe I could multiply through by some term of y...like say (1/y) so that I can have y on the right hand side to bring it over to the left but to be honest I don't know how to even think about this problem. I am presently taking an intro course to differentials and I can't seem to understand this problem. I've tried another problem that I thought was similar to this one but this one is more difficult and slightly different. What I would really appreciate is someone telling me how I should be approaching this problem in order to solve it because I have others like these that I would love to know how to do, lol. Also, perhaps I don't understand what a homogeneous solution is but I thought it was when there is a y or at least its derivative everywhere in the equation. What this means specifically, I'm not sure but I know I don't know how to make this equation or any like it homogeneous. LOL In any case, any help/advice anyone can offer me would be greatly appreciated! I am absolutely stumped!
 
Re: Finding the homogeneous solution

If you are taking a course in differential equations then all of this information is in your textbook, in the first chapter or so.
 
Re: Finding the homogeneous solution

the_chemist said:
The question is this:
Classify in terms of order, degree, and linearity, and find the homogeneous solutions to...

y'' + 3y' + 2y = sin(2x)


My problem is this: I don't know how to even begin this problem. I thought maybe I could multiply through by some term of y...like say (1/y) so that I can have y on the right hand side to bring it over to the left but to be honest I don't know how to even think about this problem. I am presently taking an intro course to differentials and I can't seem to understand this problem. I've tried another problem that I thought was similar to this one but this one is more difficult and slightly different. What I would really appreciate is someone telling me how I should be approaching this problem in order to solve it because I have others like these that I would love to know how to do, lol. Also, perhaps I don't understand what a homogeneous solution is but I thought it was when there is a y or at least its derivative everywhere in the equation. What this means specifically, I'm not sure but I know I don't know how to make this equation or any like it homogeneous. LOL In any case, any help/advice anyone can offer me would be greatly appreciated! I am absolutely stumped!
These are good questions to ask your teacher -- face to face.

Homogeneous equation:

y'' + 3y' + 2y = 0

D^2 + 3D + 2 = 0

(D+2)(D+1) = 0

D = -2 or D = -1

Then the homogeneous solution is:

y = A*e^(-x) + B*e^(-2x)

For a "complete" solution - "you need to add "particular"solution to this and evaluate A & B from given boundary and/or initial conditions.
 
I appreciate the replies. I realize that I didn't word what my question correctly to indicate what was and where my confusion actually lies. To begin, I misread the problem. What was causing my confusion was the fact that the question asks for the homogeneous solution to the nonhomogeneous one given. So...I thought I had to solve for some particular solution (which I...am still unsure on how to). I had thought that I needed to take into account that y(x) = yhomogeneous(x) + yparticular(x)...and then do all sorts of crazy math. That is assuming all I had to do was only find yhomogeneous. :D
 
the_chemist said:
I appreciate the replies. I realize that I didn't word what my question correctly to indicate what was and where my confusion actually lies. To begin, I misread the problem. What was causing my confusion was the fact that the question asks for the homogeneous solution to the nonhomogeneous one given. So...I thought I had to solve for some particular solution (which I...am still unsure on how to). I had thought that I needed to take into account that y(x) = yhomogeneous(x) + yparticular(x)

for this type of equations

y(p) = M*sin(2x) + N*cos(2x)

Solve for M & N by putting this into

y(p)" + 3y(p)' + 2y(p) = sin(2x)...............................(1)

You have

y'(p) = -2N*sin(2x) + 2M*cos(2x)

y"(p) = -4M*sin(2x) - 4N*cos(2x)

Then from (1)

(-2M -6N) sin(2x) + (6M-2N) cos(2x) = sin(2x).............(2)

from (2)

2M + 6N = -1
6M - 2N = 0

Solve for M & N from above and you have your y(p).

...and then do all sorts of crazy math. That is assuming all I had to do was only find yhomogeneous. :D
 
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