O.D.E. Undetermined coefficients solution

[Mechdude]

Homework Statement

i have this problem

$$y'' - y' - 2y= -6xe^{-x}$$

Homework Equations

this is the homogenous solution : $$y_{h}= c_{1} e^{-x} + c_{2} e^{2x}$$

The Attempt at a Solution

first i can do the problem but what is perplexing me is why of these two tentative particular solutions the latter is chosen, $$y_{p1}= Axe^{-x}$$
$$y_{p2}= x(A + Bx)e^{-x}$$
the former leads to an impossible situation but none the less its not appearing in the homogenous solution , the latter looks like the choice made when the roots of the auxilliary equation of the differential equation are real and the same , but in this case they are not real and the same , but this is what works , why would one choose the latter?

 

[Mark44]

This is the short answer, which might not be satisfactory to you. If the diff. equation happened to be y'' - y' -2y = -6e^-x, then the first choice for a particular solution would be the one that would work. But because the right side is -6xe^-x, you're going to need the next higher power of x to take that into account. IOW, y_p = (Ax + Bx²)e^-x. Hope that makes sense.

 

[Mechdude]

This is the short answer, which might not be satisfactory to you. If the diff. equation happened to be y'' - y' -2y = -6e^-x, then the first choice for a particular solution would be the one that would work. But because the right side is -6xe^-x, you're going to need the next higher power of x to take that into account. IOW, y_p = (Ax + Bx²)e^-x. Hope that makes sense.

thanks for the reply, makes perfect sense for the most part, but why do you have to take the next higher power of x into account?

 

[Mark44]

OK, here's a bit more detail that involves the concept of annihilators. If we look at the homogeneous version of your DE, the characteristic equation is r² - r -2 = 0, which can be factored into (r - 2)(r + 1) = 0.

Looking at the homogeneous equation in terms of derivative operators, it's not too hard to see that it can be written as (D² - D -2)y = 0. This equation can be factored into (D - 2)(D + 1)y = 0. The D + 1 operator annihilates e^-x or any constant multiple of it. In case you aren't following, (D + 1)e^-x means d/dx(e^-x + e^-x), which is clearly equal to zero. Similarly, the operator D - 2 annihilates e^2x.

When you have repeated roots in the characteristic equation, you have repeated annihilator factors. For example, if the homogeneous DE were y'' + 2y' + y = 0, the char. equation would be r² + 2r + 1 = 0, or (r + 1)² = 0.

The operator form of this DE would be (D + 1)²y = 0. One solution to this DE is e^-x, and as it turns out, xe^-x is another linearly independent solution. All solutions of this DE are of the form y = c₁e^-x + c₂xe^-x. The D + 1 operator annihilates e^-x, and the (D + 1)² operator annihilates e^-x and xe^-x.

If the DE were y''' + 3y'' + 3y' + y = 0, the characteristic equation would be (r + 1)³ = 0. Following the logic as before, three linearly independent solutions are e^-x, xe^-x, and x²e^-x. The (D + 1)³ operator annihilates e^-x, xe^-x, and x²e^-x.

OK, now back to your original nonhomogeneous problem: y'' - y' - 2y = -6xe^-x. This can be rewritten as (D² - D - 2)y = -6xe^-x, or equivalently as (D - 2)(D + 1)y = -6xe^-x.

My previous work shows that the (D + 1)² operator annihilates xe^-x, so we'll tack another couple of factors of (D + 1) to make the nonhomogeneous equation a homogeneous one.

(D - 2)(D + 1)³)y = (D + 1)²(-6xe^-x) = 0.

For this homogeneous equation, the set of linearly independent solutions is {e^2x, e^-x, xe^-x, x²e^-x}.

The first two are the solutions to the original homogeneous equation; the last two are solutions to the new homogeneous equation, and should therefore be selected as particular solutions to your nonhomogeneous problem.

 

[Mechdude]

Cool thanks, i had not met this perspective, thanks, il look at it further.