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page 70

2.6.3.1.3 - Homogeneous Equations and Substitution

• These techniques depend upon finding some combination of the variables in the equation that can be replaced with another variable to simplify the equation. This technique requires a bit of guessing about what to substitute for, and when it is to be applied.

e.g.,

dy

y

 

the equation given

 

----- =

- – 1

 

 

dx

x

 

 

 

 

y

 

the substitution chosen

 

u =

x-

 

 

Put the substitution in and solve the differential equation,

 

dy

u – 1

 

 

----- =

 

 

dx

 

 

 

 

 

du

= u – 1

 

 

u + x-----

 

 

 

dx

 

 

 

du

–1

 

 

----- = -----

 

 

dx

x

 

 

du

=

1

 

 

–-----

--

 

 

dx

 

x

 

u = ln (x )+ C

Substitute the results back into the original substitution equation to get rid of ’u’,

y

= ln (x )+ C

x--

y = – x ln (x )– Cx

2.6.3.2 - Second Order Differential Equations

These equations have at least one second order derivative.

In engineering we will encounter a number of forms,

-homogeneous

-nonhomogeneous

page 71

2.6.3.2.1 - Linear Homogeneous

• These equations will have a standard form,

 

d

2

 

 

d

 

 

 

 

 

0

 

 

----

 

y + A ----

y + B =

 

 

dt

 

 

dt

 

 

 

 

 

 

 

• An example of a solution is,

 

 

 

 

 

 

 

e.g.,

d

2

 

 

d

 

 

 

 

 

0

 

----

 

y + 6 ----

y + 3 =

 

 

dt

 

 

dt

 

 

 

 

 

 

 

 

Guess,

 

 

 

 

 

 

 

 

 

 

 

 

y = eBt

 

 

 

 

 

 

 

 

 

 

d

 

 

Be

Bt

 

 

 

 

 

 

----

y =

 

 

 

 

 

 

 

 

dt

 

 

 

 

 

 

 

 

 

 

 

 

d

2

 

2

e

Bt

 

 

 

 

 

----

 

y = B

 

 

 

 

 

 

 

 

dt

 

 

 

 

 

 

 

 

 

substitute and solve for B,

 

 

 

B2eBt + 6BeBt + 3eBt = 0

 

 

 

B2 + 6B + 3 = 0

 

 

 

 

 

B = – 3 + 2.449j,– 3 – 2.449j

 

substitute and solve for B,

 

 

 

y = e(– 3 + 2.449j )t

 

 

 

 

 

y = e–3te2.449jt

 

 

 

 

 

y = e–3t(cos (2.449t )+ j sin (2.449t ))

 

 

 

 

 

 

 

 

Note: if both the roots are the same,

 

 

 

 

 

 

y = C1eBt + C2teBt

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2.6.3.2.2 - Nonhomogeneous Linear Equations

page 72

• These equations have the general form,

d

2

d

 

----

 

y + A ----

y + By = Cx

dt

dt

 

• to solve these equations we need to find the homogeneous and particular solutions and then add

the two solutions.

 

 

 

 

 

y = yh + yp

 

 

 

 

to find yh solve,

 

 

 

 

 

d

2

 

d

 

0

 

----

 

y + A ----

y + B =

 

 

dt

 

dt

 

 

 

to find yp guess at a value of y and then test for validity, A good table of guesses is,

 

Cx form

 

 

 

Guess

 

 

 

 

 

 

 

 

 

 

 

 

 

A

 

 

 

 

 

C

 

Ax + B

 

 

 

 

Cx + D

 

eAx

 

 

 

 

 

CeAx CxeAx

 

B sin (Ax )

or B cos (Ax )

 

C sin (Ax )+ D cos (Ax )

 

 

 

 

 

 

 

or Cx sin (Ax )+ xD cos (Ax )

• Consider the example below,

 

 

 

 

 

 

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