Question

solve the given differential equation by undetermined coefficients

y''' - 6y'' + 3y' - y = x - 4e^(x)
Please show work and thank you

Answer 1

Did you already solve the related homogenous DE?

\[\Large y'''-6''+y'-y=0 \]

Answer 2

I'm sorry, it's y''' - 3y'' + 3y' - y = x - 4e^(x)

Answer 3

Because your guess of the undetermined coefficients depends upon that solution. In this case, you have on the RHS of the DE

\[\Large x-4e^{x} \]
So a fair guess would be:
\[\Large Ax+Be^{x} \]

Answer 4

No, I couldn't solve it.

Answer 5

\[\Large r^3-3y^2+3r-1=0 \]
\[r=1 \]
leads to zero

Answer 6

the remaining roots might be complex. You will figure that out after you perform a long hand division, or use synthetic division.

Answer 7

What would I be dividing with the long division?

Answer 8

by r-1, but if you look a bit more carefully at the expression above, it seems to be just the cube of (r-1)^3, which means that r=1 is a root of order three. Do you know already how to deal with repeated roots?

Answer 9

\[\Large r^3-3r^2+3r-1=(r-1)^3=0 \]

Answer 10

Yes, I see now. I understand. You have already given me the particular solution, too. That's all I needed, thank you!

Answer 11

I need to add an x to that particular solution, right?

Answer 12

you're welcome. Just keep in mind that you still need to play with the solution a bit, and yes exactly, you need to add successive terms of x.

\[\Large e^x, xe^x,x^2e^x \]

Answer 13

I have Ax + B - Cxe^(x), is that right?

Answer 14

That is very good, but not good enough, you solve the homogenous DE first in order to compare your solutions with your guess of undetermined coefficients. so Lets do that together, solutions so far are:

\[\Large c_1e^x, c_2xe^x,c_3x^2e^x \]

So your guess:
\[\Large Ax+B+Cxe^x \]
(You don't need the minus, the minus can be absorbed by the constant, because minus just means minus one)

So your guess of the linear term is fine, but unfortunately \(xe^x\) is already solution of the given homogenous DE, so you don't want to include that in your guess, raise it by another exponent, gives you:
\[\Large Ax+B+Cx^2e^{x} \]
But unfortunately, \(x^2e^2\) is also already a solution of the homogenous differential equation, so we raise it by another power, you see where this is going?

Answer 15

I see it! I have it, thanks.

Answer 16

you're welcome