Diffe Q, a forced harmonic oscillator

1. May 13, 2006

2cansam

okay this is kind of a quick one when youre solving a forced harmonic oscillator for an initial condition by the method of undetermined coefficients, you do the part for the homogenoues equation first and come up with somethign like ke^at + ce^bt. Now when you goto solve for those constants k and c using your initial conditions do you use the general soltion form the nonhomogenous part as well? so it would either be ke^at + ce^bt = d -or- ke^at + ce^bt+re^dt = f

Thanks in advance and sorry about my rapant use of variables and poor math notation

2. May 14, 2006

HallsofIvy

Staff Emeritus
Yes, of course. The initial conditions apply to the function that satisfies the entire equation. You must use that function, both "homogeneous" and "non-homogeneous" parts to satisfy the initial conditions.

For example, to solve y"+ 4y= x, y(0)= 0, y'(0)= 0.
The associated homogeneous equation is y"+ 4y= 0 which has characteristic equation r2+ 4= 0 and so characteristic values 2i and -2i. The general solution to that homogeneous equation is
y(x)= kcos(2x)+ c sin(2x). Now, using the method of "undetermined coefficients, we seek a particular solution to the entire equation of the form y(x)= Ax+ B. Then y'= A and y"= 0 so the equation becomes
0+ 4(Ax+ B)= 4Ax+ 4B= x. Clearly, A= 1/4 and B= 0. The general solution to the entire equation is y(x)= kcos(2x)+ csin(2x)+ x/4. Of course, then, y'(x)= -2ksin(2x)+ 2ccos(2x)+ 1/4.

Now, we have y(0)= k= 0 and y'(0)= 2c+ 1/4= 0 so c= -1/8. The solution to the problem is y(x)= -(1/8)sin(2x)+ x/4.

Since y(x)= 0 (y identically equal to 0) satisfies any homogeneous differential equation, if we just used the homogeneous equation to satisfy these initial conditions we would have just got y(x)= 0 and so, adding the particular solution, y(x)= x/4 which does NOT satisfy
y'(0)= 0.

3. May 15, 2006

2cansam

thanks man , haha yeah i guess it is pretty obvious looking back at it now. Thanks a bunch man

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