# Differential equations

1. May 4, 2006

### UrbanXrisis

y is a function of x if
$$y'''-6y''+8y'=9e^x$$
$$y(0)=13, y'(0)=17, y''(0)=16$$
solve for y(x)=

I first solved for $$y'''-6y''+8y'=0$$
I got the general solution:
$$y=A0.5e^{2x}+B0.25e^{4x}+C$$

I was just wondering if I should plug in the initial conditions now or should I solve the 9e^x and then solve for the initial conditions? Also, if I solved for 9e^x now, would the costant C just be zero?

2. May 4, 2006

### siddharth

You don't solve for the initial conditions now, because the equation you got, $y=Ae^{2x}+Be^{4x}+C$ is the solution to the corresponding homogeneous equation only, and not the solution of the original nonhomogeneous equation given in the initial problem.

You can now find the particular solution of the nonhomogeneous equation by the method of undetermined coefficients.

The general solution of the nonhomogeneous equation will be the sum of the solution of the homogeneous equation and the particular solution of the nonhomogeneous equation. You should substitute your initial conditions to this general solution