General solution ODE

1. Aug 5, 2010

itsjared

Im having trouble with this question. can anyone explain please?

1. The problem statement, all variables and given/known data
y'' + 6y' + 9y = x*exp(-3x)3x

2. Relevant equations

Find the general solution.

2. Aug 5, 2010

Sir Beaver

Well, the general solution is the sum of the homogeneous solution and a particular solution.

The homogeneous solution (gotten from putting the right hand side = 0) is obtained using the characteristic equation.

To find a particular solution, one can make a clever guess how the solution must look like. Since the right hand side has a part $$e^{-3x}$$ one can safely assume that also $$y$$ must have this part. To deal with the $$x$$ one can guess that the solution must contain a polynomial of some degree. Thus, a clever guess would be $$y=p(x)e^{-3x}$$. All that is left is to see which $$p(x)$$ that will satisfy the solution.

3. Aug 5, 2010

itsjared

im working through this question. im not to sure im doing it right though.

K^2+6K+9 = 0
therefore K= -3
and y= A*exp(-3x) + Bx*exp(-3x)
y'=-3A*exp(-3x) - 3Bx*exp(-3x) + B*exp(-3x)

y(p)= Ax*exp(-3x)
y'(p)= A*exp(-3x) - 3Ax*exp(-3x)

am i on the right track?

4. Aug 5, 2010

Sir Beaver

The homogeneous solution looks correct, although you do not have to differentiate it. You merely have to add the solutions together in the end, $$y=y_H+y_{P}$$, so you are done with $$y_H$$.

In the particular solution you have assumed that the polynomial is $$p(x)=Ax$$, which is not the general case. In fact, a polynomial of larger order is needed. I think it is easier if you treat $$p(x)$$ as a general polynomial, and then you will get an expression for the polynomial in the end, when you have put in $$y=p(x)e^{-3x}$$ in the differential equation.