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

• mitch_1211
In summary, the conversation discusses the use of the D operator technique in solving differential equations with multiple trigonometric terms on the right hand side (RHS). The person asking for help is trying to understand the method and asks for clarification on the process. The response explains that the idea is to apply 1/P(D) (where P(D) is the operator definition on the LHS) to each term on the RHS and use the property of linearity to solve each term individually. This simplifies the process and makes it easier to apply the D operator technique.
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

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)

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

## 1. What is the D operator technique?

The D operator technique is a method used to solve differential equations (DEs) that involve trigonometric functions. It involves replacing the trigonometric functions with their corresponding D operator expressions and solving the resulting polynomial equation.

## 2. How does the D operator technique work?

The D operator technique works by converting a trigonometric DE into a polynomial equation in terms of the D operator. This allows for the use of algebraic methods to solve the equation, which is typically easier than solving the original DE.

## 3. When is the D operator technique most useful?

The D operator technique is most useful when solving DEs that involve trigonometric functions, as it simplifies the process and often results in a more manageable equation. It is also useful when the DE has a high degree or involves complex trigonometric expressions.

## 4. Are there any limitations to the D operator technique?

While the D operator technique is a useful tool, it does have limitations. It can only be used for DEs with constant coefficients and cannot be applied to DEs that involve non-trigonometric functions. It also may not provide a complete solution in some cases.

## 5. Can the D operator technique be used for higher-order DEs?

Yes, the D operator technique can be applied to higher-order DEs by using it repeatedly or by converting the higher-order DE into a system of first-order DEs. However, the complexity of the resulting equations may make it more challenging to find a solution.

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