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

1. Aug 4, 2012

### mitch_1211

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

2. Aug 5, 2012

### chiro

Hey mitch_1211.

The idea will be to apply 1/P(D) (where P(D) is the operator definition on the LHS) to each term on the RHS. You have the property of linearity which means you can separate the terms and solve each one individually.

Just to clarify, this means 1/P(D)[f(x) + g(x)] = [1/P(D)]f(x) + [1/P(D)]g(x)

3. Aug 5, 2012

### mitch_1211

Oh I didn't even think of that! Thank you so much that makes things much easier