Need help finding general solution of an initial value problem

[maxfails]

The equation is

y'' + 4y' + 4y = (3 + x)e^-x

and initial conditions y(0) = 2, y'(0)=5so from the associated homogenous equation
I think the fundamental set of solutions is {e^-2x, xe^-2x} and so yc would be

Y_c = c₁e^-2x + c₂xe^-2x

but now I don't know how to get Y_p, particular solution or what form they are.

Do I need to expand the (3 + x)e^-x, and then solve for 2 different particular solutions?

so solve for a Y_p when y'' + 4y' + 4y = 3e^-x as well as
y'' + 4y' + 4y = xe^-x, or do I only need to solve for 1 Yp?

 

[genericusrnme]

You need to have something other than just your $Y_c$

If you think about your vector spaces and your matrices and you remember the fact that the general solution contains linear combinations of the vectors of the nullspace?
Same thing here, your $Y_c$ is the nullspace part.

Now, just from looking at that equation, the solutions that aren't going to be in the nullspace are going to be of the form $x\ e^{-x}$ and $e^{-x}$ since the function of x on the right hand side contains only these terms, there aren't any e's with powers of +- 2x there.

If I was doing things, I'd plug in the terms I suggested and then add them to the nullspace terms and play about with the coefficients of the those terms to get the initial conditions.

 

[HallsofIvy]

Since you already have $e^{-x}$ and $xe^{-x}$ as solutions to the homeneous equation, look for a particular solution of the form $(Ax^2+ Bx^3)e^{-x}$.

 

[Char. Limit]

Since you already have $e^{-x}$ and $xe^{-x}$ as solutions to the homeneous equation, look for a particular solution of the form $(Ax^2+ Bx^3)e^{-x}$.

Hate to catch you on this, HoI, but not quite. His homogenous solution is $c_1 e^{-2x} + c_2 x e^{-2x}$. His particular solution needs to be of the form $(A+Bx)e^{-x}$.

 

[HallsofIvy]

Thanks, I completely mis-read!

 

[maxfails]

Hate to catch you on this, HoI, but not quite. His homogenous solution is $c_1 e^{-2x} + c_2 x e^{-2x}$. His particular solution needs to be of the form $(A+Bx)e^{-x}$.

So do I only need just one particular solution to find the general one?

 

[Dick]

So do I only need just one particular solution to find the general one?

Yes, you only need one particular solution. Then when you add it to your fundamental solutions, you have all solutions. Then use your initial conditions to find the one that fits.

 

[maxfails]

Is the form of the Y_p supposed to come from the fundamental set of solutions or from the right side of the equation, the (3 + x)e^-x ?

 

[Dick]

Is the form of the Y_p supposed to come from the fundamental set of solutions or from the right side of the equation, the (3 + x)e^-x ?

You find the particular solution by substituting $(A+Bx)e^{-x}$ into your differential equation and solving for A and B, as Char Limit pointed out.