General Solution for ODE: y'' + 6y' + 9y = x*exp(-3x)3x

[itsjared]

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

Homework Statement

y'' + 6y' + 9y = x*exp(-3x)3x

Homework Equations

Find the general solution.

 

[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.

 

[itsjared]

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.

thanks for the reply.

im working through this question. I am not to sure I am 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?

 

[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.