Find the appropriate particular solution.

[Mukuro]

Homework Statement

Find the appropiate particular solution form for

y''' - y'' - 12y' = x - 2x*e^-(3x)

DO NOT SOLVE FOR THE COEFFICIENTS

Homework Equations

Just guessing proper solution forms for nonhomogeneous second order D.Es with undetermined coefficients.

The Attempt at a Solution

Ok, as far as I know the solution in my mind was...

Ax + B + (Cx^2 + Dx) e ^(-3x)

The homogeneous solution was yc = c1 + c2e^(4x) + c3e^(-3x)
BUT

the solution is Ax + Bx^2 + (Cx + Dx^2)e^(-3x)

I don;t know why the B has x^2, and I don't know why D and C are switched around. I've been going at this for a good hour.

 

[STEMucator]

So you want a particular solution to $L[y] = x - 2xe^{-3x}$

So by solving the homogeneous system $L[y] = 0$, you get a solution y_c(x).

Now, your particular solution, which I'll denote Y(x), cannot be the same as any of the solutions of the homogeneous system.

So since you've found your homogeneous solution to L[y] = 0 to be $y_c(x) = c_1 + c_2e^{-3x} + c_3e^{4x}$ your particular solution Y(x) cannot be equal to any of your terms y₁, y₂ or y₃.

 

[Mukuro]

So you want a particular solution to $L[y] = x - 2xe^{-3x}$

So by solving the homogeneous system $L[y] = 0$, you get a solution y_c(x).

Now, your particular solution, which I'll denote Y(x), cannot be the same as any of the solutions of the homogeneous system.

So since you've found your homogeneous solution to L[y] = 0 to be $y_c(x) = c_1 + c_2e^{-3x} + c_3e^{4x}$ your particular solution Y(x) cannot be equal to any of your terms y₁, y₂ or y₃.

I understand that e ^-3x is in the homogeneous solution, that's why it by x, however this still does not explain why C and D happen to be reversed or why b has an x^2 attached.

 

[STEMucator]

You stated your idea for your particular solution was : Ax + B + (Cx^2 + Dx) e ^(-3x)

Notice you have e^(-3x) in your particular solution when I said that if a solution appeared in your homogeneous solution, then it could not be a particular solution.

 

[Mukuro]

Oh I typed the actual solution wrong, which I'll have to fix.

The actual solution is ax + bx^2 (cx + dx^2) e^(-3x),

and I know that e^(-3x) is in the solution that is why I multiply it by and distribute it out.

x * (cx + b)*e^(-3x) is not equal to (cx +b)*e^(-3x).

 

[STEMucator]

No no, since e^(-3x) appeared in your solution to the homogeneous system, it cannot appear in your formulation of a particular solution. So e^(-3x) CAN'T be part of the particular solution.

There's also a very simple fix to this and it has to do with the order of your equation.

 

[Mukuro]

But the problem with that is, the book CLEARLY SAYS that e^(-3x) is apart of the particular solution, I have double checked two books, my own and my fellow classmates.

It also states that when this happens we can multiply the function by x to achieve Axe^(x) (standard example). This is what I have done.

 

[STEMucator]

Okay so you know what you're doing, you're very close actually.

Why is multiplying by just x insufficient? Look at the order of your equation.

 

[Mukuro]

Could you give me a hint as to what equation your referring to, my solution or the actual equation stated in the problem? I'm so very determined atm <.>

OK I recognize that it is a 3rd order equation... I do not however know what that does LOL. I usually deal with just 2nd orders.

 

[STEMucator]

Since you have a third order equation, it may be the case that you must multiply by at most x³ to find your particular solution.

Also note, you can break your equation $L[y] = x - 2xe^{-3x}$ into two different problems :

$L[y] = x$ and $L[y] = -2xe^{-3x}$ and solve both of those instead of solving it all at once ( If it helps ).

 

[Mukuro]

Well we don't have to solve it which I am happy about. However, I am still confused, even if I may have to multiply by at most x^3, it doesn't seem to explain the bx^2.

For instance the Ax + B is my guess to Ly = x, which I figure should be enough but seeing as it is a 3rd order equation how would I go about getting Ax + bx^2, I honestly have no idea how anyone could possibly guess that.

 

[STEMucator]

Well we don't have to solve it which I am happy about. However, I am still confused, even if I may have to multiply by at most x^3, it doesn't seem to explain the bx^2.

For instance the Ax + B is my guess to Ly = x, which I figure should be enough but seeing as it is a 3rd order equation how would I go about getting Ax + bx^2, I honestly have no idea how anyone could possibly guess that.

You want a particular solution to : $L[y] = x - 2xe^{-3x}$

Split it into two cases : So if Y is the particular solution, it must be the sum of the solutions Y₁ and Y₂ of the equations :

Case one : $L[Y_1] = x$

Case two : $L[Y_2] = -2xe^{-3x}$

Find solutions to those ones because it's easier, then add them together. The case with x is not too bad since after you take a few derivatives... it's easy. Then just add the two solutions of those equations together to get Y.

EDIT : I'm almost positive you should have a theorem like this in your book?