- #1
mitch_1211
- 99
- 1
hi everyone,
im currently trying to teach myself the D operator technique, as opposed to the 'guess method' which i don't really like.
I stumbled upon this on yahoo answers:
(D^2 + 1)y = 4 cos x - sin x
Find the complementary function by solving the auxiliary equation:
m² + 1 = 0
m² = -1
m = ±i
yᶜ = Asinx + Bcosx
Find the particular integral by comparing coefficients:
yᵖ = Cxsinx + Dxcosx
yᵖ' = (C - Dx)sinx + (Cx + D)cosx
yᵖ'' = (2C - Dx)cosx - (Cx + 2D)sinx
yᵖ'' + yᵖ = 4cosx - sinx
(2C - Dx)cosx - (Cx + 2D)sinx + Cxsinx + Dxcosx = 4cosx - sinx
2Ccosx - 2Dsinx = 4cosx - sinx
2C = 4
C = 2
2D = 1
D = ½
yᵖ = 2xsinx + xcosx / 2
Find the general solution by combining these two parts:
y = yᶜ + yᵖ
y = Asinx + Bcosx + 2xsinx + xcosx / 2
y = (2x + A)sinx + (x + B)cosx / 2
That's all well and good, but from my understanding to solve a DE of the form
y'' -2y' -3y = cos3x
using the D operator technique I would say ok
ypi=1/ (D2-2D-3) cos3x
and then replace D2's with -(α)2 where α is 3 in this case
and then proceed to solve by using D2 terms
In the yahoo answers example are they using a different method because there are 2 trig functions on the RHS? I'm also unsure if they have used D as a constant and as the operator?
Sorry for the long post, bottom line is I want to able to use the D operator technique to solve a DE with multiple trig terms on the RHS, and is there a way similar to how you would solve for one trig term, as I explained?
Mitch
im currently trying to teach myself the D operator technique, as opposed to the 'guess method' which i don't really like.
I stumbled upon this on yahoo answers:
(D^2 + 1)y = 4 cos x - sin x
Find the complementary function by solving the auxiliary equation:
m² + 1 = 0
m² = -1
m = ±i
yᶜ = Asinx + Bcosx
Find the particular integral by comparing coefficients:
yᵖ = Cxsinx + Dxcosx
yᵖ' = (C - Dx)sinx + (Cx + D)cosx
yᵖ'' = (2C - Dx)cosx - (Cx + 2D)sinx
yᵖ'' + yᵖ = 4cosx - sinx
(2C - Dx)cosx - (Cx + 2D)sinx + Cxsinx + Dxcosx = 4cosx - sinx
2Ccosx - 2Dsinx = 4cosx - sinx
2C = 4
C = 2
2D = 1
D = ½
yᵖ = 2xsinx + xcosx / 2
Find the general solution by combining these two parts:
y = yᶜ + yᵖ
y = Asinx + Bcosx + 2xsinx + xcosx / 2
y = (2x + A)sinx + (x + B)cosx / 2
That's all well and good, but from my understanding to solve a DE of the form
y'' -2y' -3y = cos3x
using the D operator technique I would say ok
ypi=1/ (D2-2D-3) cos3x
and then replace D2's with -(α)2 where α is 3 in this case
and then proceed to solve by using D2 terms
In the yahoo answers example are they using a different method because there are 2 trig functions on the RHS? I'm also unsure if they have used D as a constant and as the operator?
Sorry for the long post, bottom line is I want to able to use the D operator technique to solve a DE with multiple trig terms on the RHS, and is there a way similar to how you would solve for one trig term, as I explained?
Mitch