Question

Use the principal of superposition to find a particular solution of y"+2y'+10y=4+26x+6x^2+10x^3+e^(x/2).

Answer 1

@freckles @mathstudent55 @phi @Compassionate @shifuyanli @paki @campbell_st

Answer 2

ok
so we want to look at solving both of the following for the particulars:
\[y''+2y'+10y=e^\frac{x}{2} \\ y''+2y'+10y=4+26x+6x^2+10x^3\]

Answer 3

and then take those particulars and add them together to find the particular of our differential equation above

Answer 4

So that's the method for using the principal of superposition?

Answer 5

yep
if y1 and y2 are solutions to the dq, then y1+y2 is also a solution to the dq

Answer 6

So what do I do with those ^ above?

Answer 7

find the particular solutions for both

Answer 8

I would probably use y=Aexp(x/2) for the first one
and probably use y=Kx^3+Bx^2+Cx+D for the second one
as my choice for particular solutions
of course you need to find A,K,B,C, and D

Answer 9

Let me do them real quick, please wait.

Answer 10

I think that for second one we have to choose a polynomial of fifth degree, namely:
\(P(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5\)

Answer 11

I still think we only need to do a 3rd degree polynomial since we have that that right hand side of our differential equation is a 3rd degree polynomial
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 12

@Michele_Laino , what's your method and why 5th degree?

Answer 13

you can do a 5th degree polynomial but I feel like you would be wasting your time just a bit @Idealist10 since you will get the coefficients of x^4 and x^5 are zero

Answer 14

I choose a fifth degree polynomial, since after two derivatives, as indicated by your ODE, I get a polynomial of third degree

Answer 15

and the right side is a third degree polynomial

Answer 16

If we compute the second derivative we get:
\[y'' = 2{a_2} + 6{a_3}x + 12{a_4}{x^2} + 20{a_5}{x^3}\]

Answer 17

but you have no x^4 and x^5 's on the right hand side

Answer 18

the coefficients would have to be 0 of x^4 and x^5

Answer 19

and if \(a_4,a_5\) are both zero, then we can not have a third degree polynomial

Answer 20

Okay, so I've got the right answer. But what if we just use \[y _{p}=A+Bx+Cx^2+Dx^3+Ae ^{\frac{ x }{ 2 }}\] as the substitution instead of doing it 2 times like @freckles suggested at the beginning?

Answer 21

we could have but I just wanted to use the principle they mentioned

Answer 22

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&uact=8&ved=0CDcQFjAEahUKEwiZg6GCp9TIAhXLzoAKHQB_BD0&url=http%3A%2F%2Fwww.utdallas.edu%2Fdept%2Fabp%2FPDF_Files%2FDE_Folder%2FNonhomogeneousEquations.pdf&usg=AFQjCNH30L6D4YqOtPEKp0GYIlfU_XDbcQ&sig2=Ef-LYuEJCJJfEhf8V4e9EA
this explains that principle pretty well

Answer 23

Lets use shortcut methods in evaluating the particular solutions

Answer 24

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=6&cad=rja&uact=8&ved=0CD4QFjAFahUKEwiZg6GCp9TIAhXLzoAKHQB_BD0&url=http%3A%2F%2Fwww.math.uah.edu%2Fhowell%2FDEtext%2FPart3%2FGuess_Method.pdf&usg=AFQjCNHWZPScO9PvUGzEQknnAp1jiwZR1Q&sig2=ygZwowdnc6RSZeFWBT25iQ
page 415
actually this source might explain it better

Answer 25

the principle of superposition

Answer 26

The best way is to write the solution with both methods

Answer 27

\[y''+p(x)y'+q(x)y=g(x)\]

basically it says if g(x)=g1+g2
then
\[y''+py'+qy=g_1 \text{ will give solution } y_1 \\ \text{ and } y''+py'+qy=g_2 \text{ will give solution } y_2 \\ \text{ then the solution to } \\ y''+py'+qy=g \text{ is } y_1+y_2\]

Answer 28

Now I don't like to use the principal of superposition. It's kinda irritating. :)

Answer 29

I got this:
\(P(x)=2x+x^3\)

Answer 30

check if you can compute the remaining part. or else I will compute

Answer 31

Thanks a lot for the help, guys!