operator methods

MATH F211:MATHEMATICS–III

Presented by

Dr. M.S. Radhakrishnan

Email: [email protected]

BITS-PILANI HYDERABAD CAMPUS

9

Apr 22, 2023

2

Operator Methods for finding Particular Solutions of LDE’s with constant coefficients

Ch. 3 Section 23George F. Simmons, Differential Equations with Applications and Historical notes, Tata McGraw-Hill, 2nd Ed, 2003 (Twelfth reprint, 2008)

Lecture 9

Presented by Dr. M.S. Radhakrishnan BITS, Pilani

Apr 22, 2023

Presented by Dr. M.S. Radhakrishnan BITS, Pilani 3

You don’t have to be famous. You just have to make your mother and father proud of you.

Operator Methods for finding Particular Solutions of Linear Differential Equations with constant coefficients

Apr 22, 2023

4Presented by Dr. M.S. Radhakrishnan BITS, Pilani

Apr 22, 2023


In this section we give a very brief sketch of the use of differential operators for finding particular solutions of linear differential equations with constant coefficients.

These methods are more efficient than the previous methods and were mainly due to the English applied mathematician Oliver Heaviside.

Apr 22, 2023


Heaviside’s methods seemed so strange to the scientists of his time that he was widely regarded as a crackpot, which unfortunately is a common fate for thinkers of unusual originality.

Apr 22, 2023


Consider the th order linear differential equation with constant cofficients:

𝑎0𝑑𝑛 𝑦𝑑𝑥𝑛 +𝑎1

𝑑𝑛−1 𝑦𝑑𝑥𝑛− 1 +…+𝑎𝑛−1

𝑑𝑦𝑑𝑥

+𝑎𝑛 𝑦

Using the symbols etc.

we can write the above equation as

¿h (𝑥)

Apr 22, 2023


(𝑎¿¿ 0𝐷𝑛+𝑎1 𝐷𝑛− 1+…+𝑎𝑛− 1𝐷+𝑎𝑛)𝑦 ¿

Or more compactly as 𝑝 (𝐷 ) 𝑦=h(𝑥)where is the th order linear differential operator

𝑝 (𝐷 )≡𝑎0 𝐷𝑛+𝑎1 𝐷

𝑛−1+…+𝑎𝑛−1 𝐷+𝑎𝑛

¿h (𝑥)

Apr 22, 2023


A particular solution of the given d.e. is got by “solving” for

𝑦=1

𝑝 (𝐷 )h(𝑥 )

Thus 𝑦=1

𝑝 (𝐷 )h(𝑥 ) if and only if

𝑝 (𝐷 ) 𝑦=h(𝑥)

Apr 22, 2023


For example, a particular solution of

𝐷𝑦=cos 𝑥

is 𝑦=1𝐷

cos 𝑥¿ sin 𝑥

A particular solution of (𝐷2+4 )𝑦=𝑒2𝑥

is 𝑦=1

𝐷2+4𝑒2𝑥¿

18𝑒2𝑥

Apr 22, 2023


Commutativity of the differential operators

Suppose we can factorize as

𝑝 (𝐷 )= 𝑓 (𝐷 )𝑔 (𝐷)

Then we easily verify that

𝑝 (𝐷 ) 𝑦=¿ 𝑓 (𝐷 ) [𝑔 (𝐷 ) ] 𝑦¿𝑔 (𝐷 ) [ 𝑓 (𝐷 ) ] 𝑦

Apr 22, 2023


Hence a particular solution of

𝑝 (𝐷 ) 𝑦=h(𝑥)

is 𝑦=1

𝑝 (𝐷 )h(𝑥 )

¿1

𝑓 (𝐷 )1

𝑔 (𝐷 )h(𝑥)

Also ¿1

𝑔 (𝐷 )1

𝑓 (𝐷 )h(𝑥)

Apr 22, 2023


We also note that from the principle of superposition, we get

1𝑝 (𝐷 )

[𝑅 (𝑥 )+𝑆 (𝑥 ) ]

¿1

𝑝 (𝐷 )𝑅 (𝑥 )+ 1

𝑝 (𝐷 )𝑆 (𝑥 )

Also1

𝑝 (𝐷 ) [𝑐 {h (𝑥 ) }]=¿𝑐1

𝑝 (𝐷 )h (𝑥)

a constant.

Apr 22, 2023


We also note that from definition, we get

𝑝 (𝐷 ) 1𝑝 (𝐷 )

h (𝑥)=¿

i.e. and 1

𝑝 (𝐷 )are inverses to each other.

h (𝑥)

Apr 22, 2023


𝐷𝑒𝑎𝑥=𝑎𝑒𝑎𝑥

𝐷2𝑒𝑎𝑥=𝑎2𝑒𝑎𝑥

…𝐷𝑛𝑒𝑎𝑥=𝑎𝑛𝑒𝑎𝑥

Hence 𝑝 (𝐷 )𝑒𝑎𝑥=𝑝(𝑎)𝑒𝑎𝑥

Apr 22, 2023


Or

And so 1

𝑝 (𝐷 )𝑒𝑎𝑥=¿

1𝑝 (𝑎 )

𝑒𝑎𝑥if

1𝑝 (𝐷 )

𝑝 (𝑎)𝑒𝑎𝑥=¿𝑒𝑎𝑥

i.e. 𝑝 (𝑎)1

𝑝 (𝐷 )𝑒𝑎𝑥=¿𝑒𝑎𝑥

Apr 22, 2023


We easily verify that

1𝐷−𝑎

𝑒𝑎𝑥=¿

𝑥2

2!𝑒𝑎𝑥1

(𝐷−𝑎 )2𝑒𝑎𝑥=¿

𝑥𝑒𝑎𝑥

…𝑥𝑚

𝑚!𝑒𝑎𝑥1

(𝐷−𝑎 )𝑚𝑒𝑎𝑥=¿

Apr 22, 2023


𝐷2sin𝑎𝑥=¿ −𝑎2 sin𝑎𝑥

𝐷4 sin𝑎𝑥=¿ (−𝑎2)2 sin𝑎𝑥

…Thus 𝑝 (𝐷¿¿2)sin 𝑎𝑥=¿¿𝑝 (−𝑎2 ) sin𝑎𝑥

Hence1

𝑝 (𝐷¿¿ 2)sin𝑎𝑥=¿¿1

𝑝 (−𝑎¿¿2)sin𝑎𝑥 ¿

if

Apr 22, 2023


Similarly1

𝑝 (𝐷¿¿ 2)cos𝑎𝑥=¿¿1

𝑝 (−𝑎¿¿2)cos𝑎𝑥 ¿

if

We easily see that1

𝐷2+𝑎2 sin 𝑎𝑥=¿− 𝑥2𝑎

cos𝑎𝑥

1

𝐷2+𝑎2 cos𝑎𝑥=¿ 𝑥2𝑎

sin𝑎𝑥

Apr 22, 2023


If is a polynomial in we can find a particular solution of

𝑝 (𝐷 ) 𝑦=h(𝑥)

as 𝑦=1

𝑝 (𝐷 )h(𝑥 )

¿ [1+𝑏1𝐷+𝑏2 𝐷2+…]h(𝑥)

where is the power1

𝑝 (𝐷 ).series expansion of

Series Expansion of Operators

Apr 22, 2023


The Exponential Shift Rule

𝐷 [𝑒¿¿ 𝑎𝑥𝑔(𝑥 )]=¿¿

𝐷2[𝑒𝑎𝑥𝑔 (𝑥 )]=𝑒𝑎𝑥 (𝐷+𝑎)2𝑔 (𝑥)

…𝐷𝑛 [𝑒𝑎𝑥𝑔 (𝑥 )]=𝑒𝑎𝑥 (𝐷+𝑎 )𝑛𝑔(𝑥 )

Hence

¿𝑒𝑎𝑥 (𝐷+𝑎)𝑔(𝑥)

𝑝 (𝐷 )[𝑒¿¿𝑎𝑥𝑔 (𝑥)]=𝑒𝑎𝑥𝑝 (𝐷+𝑎)𝑔(𝑥 )¿

𝑒𝑎𝑥𝐷𝑔(𝑥 )+𝑎𝑒𝑎𝑥𝑔(𝑥)

Apr 22, 2023


Hence

𝑝 (𝐷){𝑒𝑎𝑥 1𝑝 (𝐷+𝑎)

𝑔(𝑥)}¿𝑒𝑎𝑥𝑝 (𝐷+𝑎)

1𝑝 (𝐷+𝑎)

𝑔 (𝑥)

¿𝑒𝑎𝑥𝑔(𝑥)

Apr 22, 2023


Thus we get the

1𝑝 (𝐷 )

[𝑒¿¿𝑎𝑥𝑔 (𝑥)]=¿¿𝑒𝑎𝑥 1𝑝 (𝐷+𝑎 )

𝑔 (𝑥 )

The Exponential Shift Rule

Apr 22, 2023


An Application

1𝐷−𝑎

𝑒𝑎𝑥=¿

¿ 𝑥2

2 !𝑒𝑎𝑥1

(𝐷−𝑎 )2𝑒𝑎𝑥=¿

¿ 𝑥𝑒𝑎𝑥

…

𝑒𝑎𝑥 1

𝐷𝑚 (1)1

(𝐷−𝑎 )𝑚𝑒𝑎𝑥=¿

𝑒𝑎𝑥 1𝐷

(1)

𝑒𝑎𝑥 1

𝐷2 (1)

¿ 𝑥𝑚

𝑚 !𝑒𝑎𝑥

Apr 22, 2023


PROBLEMS

Apr 22, 2023


1. Find a particular solution of

𝑦 ′ ′+4 𝑦=𝑒2𝑥

i.e. (𝐷¿¿2+4)𝑦=𝑒2𝑥¿

A particular solution is

𝑦=1

𝐷2+4𝑒2𝑥

¿18𝑒2𝑥

¿1

22+4𝑒2 𝑥

Apr 22, 2023



𝑦 ′ ′−5 𝑦 ′+6 𝑦=𝑒2𝑥

i.e. (𝐷¿¿2−5𝐷+6)𝑦=𝑒2 𝑥¿

A particular solution is𝑦=1

𝐷2−5𝐷+6𝑒2𝑥

¿−𝑥 𝑒2 𝑥

¿1

(𝐷−2)(𝐷−3)𝑒2𝑥

¿ 1(𝐷−2)

1(𝐷−3)

𝑒2 𝑥

¿1

(𝐷−2)(−1 )𝑒2𝑥¿−

1(𝐷−2)

𝑒2𝑥

Apr 22, 2023



𝐷2−5𝐷+6𝑒2𝑥

−𝑥𝑒2𝑥

¿1

(𝐷−2)(𝐷−3)𝑒2𝑥

¿ [ 1(𝐷−3)

−1

(𝐷−2) ]𝑒2𝑥

¿ (−1 )𝑒2𝑥

We can also do the above problem by “Partial Fractions”:

¿− (𝑥+1 )𝑒2𝑥

Apr 22, 2023



𝑦 ′ ′+𝑦 ′+𝑦=sin 2𝑥

i.e. (𝐷¿¿2+𝐷+1)𝑦=sin 2 𝑥¿


𝐷2+𝐷+1sin 2 𝑥

¿1

−22+𝐷+1sin 2 𝑥¿

1𝐷−3

sin 2𝑥

¿ (𝐷+3) [ 1

𝐷2−9sin 2 𝑥]

Apr 22, 2023


¿ (𝐷+3) [ 1

−22−9sin 2 𝑥 ]

¿ (𝐷+3) [− 11 3

sin 2𝑥 ]¿−

113

(𝐷+3 )sin 2 𝑥

¿−113

(2cos2𝑥+3sin 2𝑥 )

Apr 22, 2023



4 𝑦 ′ ′+𝑦=𝑥4

i.e. (4𝐷¿¿2+1)𝑦=𝑥4 ¿

A particular solution is 𝑦=1

4 𝐷2+1𝑥4

¿ 𝑥4−4 8 𝑥2+384

¿ [1−4𝐷2+16𝐷4−64𝐷6+… ]𝑥4

Apr 22, 2023



i.e. (𝐷¿¿2−2𝐷+2)𝑦=𝑒𝑥 sin𝑥 ¿


𝑦=1

𝐷2−2𝐷+2𝑒𝑥sin 𝑥

𝑦 ′ ′−2 𝑦 ′+2 𝑦=𝑒𝑥 sin 𝑥

¿𝑒𝑥 1

(𝐷+1)2−2(𝐷+1)+2sin 𝑥

Apr 22, 2023


¿𝑒𝑥 1

𝐷2+1sin 𝑥

¿𝑒𝑥 ¿¿−

𝑥2𝑒𝑥 cos 𝑥

Apr 22, 2023



i.e. (𝐷¿¿2−4𝐷+3)𝑦=𝑥3𝑒2𝑥 ¿


𝑦=1

𝐷2−4𝐷+3𝑥3𝑒2𝑥

𝑦 ′ ′−4 𝑦 ′+3 𝑦=𝑥3𝑒2 𝑥

¿𝑒2𝑥 1

(𝐷+2)2−4(𝐷+2)+3𝑥3

Apr 22, 2023


¿𝑒2𝑥 1

𝐷2−1𝑥3

¿−𝑒2 𝑥 [1+𝐷2+𝐷4+… ]𝑥3

¿−𝑒2 𝑥 [𝑥3+6 𝑥 ]

¿−𝑒2 𝑥 1

1−𝐷2 𝑥3

Apr 22, 2023



(𝐷−2)3 𝑦=𝑒2𝑥


𝑦=1

(𝐷−2)3 𝑒2𝑥

¿𝑒2𝑥 1

𝐷3 (1)¿𝑒2𝑥 1

𝐷2

1𝐷

(1)¿𝑒2𝑥 1

𝐷2 𝑥

¿𝑒2𝑥 1𝐷𝑥2

2¿𝑒2𝑥 𝑥3

3 !¿ 𝑥

3

3 !𝑒2 𝑥

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