Differential operators find widespread use in many applications and in solving differential equations. In my October 12 post I introduced some basic differential operators and gave the various forms in which they are written. There are also more advanced operators which can be used to solve differential equations and those are the topic of this post. The first thing to note is that there can be multiple variations on the same basic operator form but which can be used in different settings, problems.

Let P(D) represent a polynomial in D with constant coefficients then:

e ^{rx} P(D)y = P (D -r) [ e ^{rx} y]

Or we can write:

P(D)[ e ^{rx} y] = e ^{rx} P(D + r)y

Also; e ^{- rx} P(D )[ e ^{rx} y] = P(D + r) y

This can also be written:

[ e ^{rx} y] / P(D ) = e ^{rx }[y] /P(D + r)

Further: P(D) e ^{rx} = e ^{rx} P(r)

Or: e ^{rx} / P(D) = e ^{rx} / P(r) If P(r) =/ 0

And more generally:

[e ^{rx} y] / P(D) = e ^{rx} [y] / P(D + r)

Each of the above basically states the same property which we refer to as an "** exponential shift**". It's instructive here to see the application in solving differential equations. We look first at obtaining the general solution of the differential equation:

y"'- 4 y" + 4y' = 0

Which we will first convert to operator form, i.e.

(D^{ }^{3} -
4D ^{2} + 4D)
y =
D (D- 2)^{ }^{2} y
= 0

So that we get two initial results:

a) D y = 0 and b): (D- 2)^{ }^{2} y = 0

For which (a) has solution: y_{1} (x) = C

And for (b) we apply the exponential shift in the form:

e ^{- rx} P(D )[ e ^{rx} y] = P(D + r) y

Using **r = -2** to get:

(D- 2)^{ }^{2} y = e ^{2x} D ^{2} ( e ^{-2x} y) = 0

Alternatively:

D ^{2} ( e ^{-2x} y) = D ^{2} z = 0

Integrating twice yields:

D z = A

z = e ^{-2x} y = Ax + B

Or: y_{ 2 } = (Ax + B) e ^{2x}

Then the general solution becomes:

y_{ 1 } + y_{ 2 } = (Ax + B) e ^{2x} + C

__Example Problem:__

Solve the equation: y" + 4y = e ^{3x}

We apply the shifting form:

e ^{rx} / P(D) = e ^{rx} / P(r) With r = -1

Then write: (D ^{2} + 4) y = e ^{x}

y = (D ^{2} + 4) ^{-1 } e ^{x }= e ^{x }/5

**Suggested Problems**:

1) Solve the differential equation below using an appropriate differential operator:

y" - 4y + 4 = xe ^{2x}

2) Solve the operator form of the differential equation below:

(D- 1)^{ }^{3} y = xe ^{x }

^{}

^{3) Consider the differential equation:}

dy^{2}/dx^{2} + ^{2}

a) Show the equation in operator form, with operator specified.

b) Obtain the general solution from factorization of the full operator differential equation.

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