# Thread: Form of particular solution For Nonhomo. L.D.E. w/ Constant Coefficients

1. ## Form of particular solution For Nonhomo. L.D.E. w/ Constant Coefficients

Given that,

$m^{3}(m+4)^{2}(m^{2}+1) = 0$ is the auxiliary equation associated with the linear differential equation

$\phi(D)y = 1 + cos3x + x^{3}e^{-4x}$

what is the form of a particular solution $y_{p}(x)?$

DO NOT EVALUATE THE COEFFICIENTS IN $y_{p}(x).$

We have these 2 general rules that we follow to determine the form of the particular solution of a nonhomogenous L.D.E. with constant coefficients.

We look at each of the pieces not part $\phi (D)y$,

i.e. $1 + cos3x + x^{3}e^{-4x}$

The only part I was confused about is the "1" term in the RHS of the differential equation.

How do we apply our "rules" to this piece?

Also, if someone could quickly check my work that'd be great! (See figure attached)

Thanks again!

2. Try

$x^3/96$

CB

3. Originally Posted by CaptainBlack
Try

$x^3/96$

CB
Does that mean I did something wrong?

4. Originally Posted by jegues
Does that mean I did something wrong?
I don't see where you have tried anything for the component of the PI corresponding to the 1 on the right hand side.

CB

5. Originally Posted by CaptainBlack
I don't see where you have tried anything for the component of the PI corresponding to the 1 on the right hand side.

CB
I was confused about that 1 on the RHS of the equation, I wasn't really sure if I had to look at it or not.

If I assume I do,

Is 1 part of $y_{h}$? Yes, it is. $\text{( i.e. }C_{i} = 0, \quad i \neq 1, \quad C_{1} = 1 )$

So by rule 2,

$H \quad \text{(Just a constant)}$

is part of $y_{p}$, but we already have this piece in $y_{h}$ so I shouldn't be including it in $y_{p}$.

Can you see why I get confused when we work with that piece? Hopefully you can clarify my confusion.

EDIT: Here are the two rules as per my textbook. (See figure attached)

6. Originally Posted by jegues
I was confused about that 1 on the RHS of the equation, I wasn't really sure if I had to look at it or not.

If I assume I do,

Is 1 part of $y_{h}$? Yes, it is. $\text{( i.e. }C_{i} = 0, \quad i \neq 1, \quad C_{1} = 1 )$

So by rule 2,

$H \quad \text{(Just a constant)}$

is part of $y_{p}$, but we already have this piece in $y_{h}$ so I shouldn't be including it in $y_{p}$.

Can you see why I get confused when we work with that piece? Hopefully you can clarify my confusion.

EDIT: Here are the two rules as per my textbook. (See figure attached)
I can not make sense of your rules (I expect I could if I sent fifteen minutes on them, but the dog is waiting for his walk) , but you cannot ignore the 1 on the right, and inspection shows that what I posted above is a particular integral corresponding to a 1 on the right.

CB

7. Originally Posted by CaptainBlack
I can not make sense of your rules (I expect I could if I sent fifteen minutes on them, but the dog is waiting for his walk) , but you cannot ignore the 1 on the right, and inspection shows that what I posted above is a particular integral corresponding to a 1 on the right.

CB
Does anyone else care to share their input?

I'm not doubting what you've said CB, it seems you know what you're doing, but in the course I'm currently taking we haven't learnt about any of those things and these two "rules" are all I have to go off.

I have to try to make sense of things with the knowledge I've be taught throughout the course.

8. Still looking to figure out how to deal with the 1.

Any ideas?