Results 1 to 8 of 8

- Dec 26th 2010, 12:25 AM #1

- Dec 26th 2010, 12:37 AM #2

- Dec 26th 2010, 12:41 AM #3
You don't need to, if you make the substitution this transforms the DE into

which is second order linear constant coefficient nonhomogeneous.

Homogeneous solution:

The characteristic equation is

.

So the homogeneous solution is .

Nonhomogeneous solution:

Assume a solution of the form .

Then and .

Substituting into the DE gives

and

and .

So the nonhomogeneous solution is .

The general solution is the sum of the homogeneous and nonhomogeneous solutions.

So .

From this we know .

From the initial conditions:

and

and

and

.

So .

Therefore

.

From the final initial condition

.

So finally our final solution is

.

- Dec 26th 2010, 01:42 AM #4

- Dec 26th 2010, 07:25 AM #5

- Dec 26th 2010, 07:31 AM #6

- Dec 26th 2010, 08:30 AM #7

- Dec 26th 2010, 08:37 AM #8