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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

y_{pi}=1/ (D^{2}-2D-3) cos3x

and then replace D^{2}'s with -(α)^{2}where α is 3 in this case

and then proceed to solve by using D^{2}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

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# Using the D operator technique to solve a trig based DE

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