Question

Need some help please..
Obtain the general solution:
y'(y''-y)=x
Need help on the Yp solution..

Answer 1

The problem is shown as:
\[(D^3-D)y=x\]
Its undetermined coefficients and I need help finding the Yp solution..

Answer 2

Would Yp=A+Bx ?

Answer 3

this is confusing me:
(D^3 - D)y = x
Is this notation such that:
\( Dy = \dfrac{d}{dx} y \) \(D^3 y = \dfrac{d^3}{dx^3} y \)

idk why this is throwing me off so much lol

Answer 4

Lol i believe it's an old form of notation..
I think it means the same as y'(y''-y)=x

Answer 5

I believe you distribute it through, like this:
\( (D^3 - D) y = D^3 y - Dy = \dfrac{d^3}{dx^3} y - \dfrac{dy}{dx} \)

Answer 6

if you factored it, you'd have to "create" a y to go with both factors really.

Answer 7

Ok, gotcha.. now im getting my Yc as:
\[Yc=C1+C2e^{x}+C3e^{-x}\]

Answer 8

hmm, where'd you get that from?

Answer 9

For the left hand side only..
\[(D^3-D)y=0\]
\[e^{mx}(m^3-m)=0\]
\[m(m^2-1)=0\]
\[m(m-1)(m+1)=0\]
\[m=0,1,-1\]
\[Solution: e^{0x},e^x,e^{-x}\]
\[Yc=C1+C2e^x+C3e^{-x}\]

Answer 10

ohh you got it, was thinking m^3 - 1 but its Dy not y. :p
so Y_c looks good

Answer 11

for Y_p, I'm thinking about how the equation plays out:
third derivative of polynomail - first derivative of polynomial = x
The smallest order derivative is that first order, and so we should test something whose first derivative is linear i am thinking.

Answer 12

the third degree isn't quite important because our only interest is matching the right-hand side's linear term, x.
as long as our first derivative contains an x term, we should be fine at getting a match. anything residual will be 0.

Answer 13

I see... thats why im thinking that Yp would = A+Bx, but it doesnt look right when i try to plug it in...

Answer 14

well, Y_p = A + Bx contains a linear term, but its first derivative is a constant, right?

Answer 15

Exactly... So that would lead me to think that:
\[Yp=Ax^2+Bx+C\]

Answer 16

i think that will work out better for you. :)

Answer 17

\[Yp=\frac{ 1 }{ 2 }x^2\]?

Answer 18

i think you have one small sign to recheck there

Answer 19

(third derivative = 0) minus (first derivative = 2A x + B) = x

Answer 20

You guys are on the right path, are you familiar with reduction of order, ie for example just let v(x) = y'(x)?

(I believe guys are basically doing the same thing, but don't take my word for it)

Answer 21

oh, i see what you're getting at @Crixpack
from y' ' ' - y ' = x <-- just reducing the order here with v = y'
to v ' ' - v = x

and then finding v there / referring back to v = y' to find y?

Answer 22

Your gold @AccessDenied !!! Got it!

Answer 23

Yup, exactly, and you get the same answer you guys do. Cheers :)

Answer 24

good work :)
and i do believe you could reduce order as well, in fact i hadn't even noticed it until it was pointed out! but either way i think you get the same answer. :)