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Flashcards in Chapter 4 (a&b) Deck (27)
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1
Q

Second-Order Linear ODE - General Form

A

e(x)d²y/dx² + f(x)dy/dx+ g(x)y

=h(x)

2
Q

Homogeneous and Inhomogeneous Second-Order Linear ODEs

A

-a second order linear ODE is homogeneous if h(x)=0
-it is inhomogeneous if
h(x) ≠ 0

3
Q

Constant Coefficient and Non-Constant Coefficient Second-Order Linear ODEs

A
  • a second order linear ODE is constant coefficient if e(x), f(x) and g(x) are all constants, rather than functions of x
  • otherwise the equation is non-constant coefficient
4
Q

Homogeneous Second Order Linear ODEs

Initial Value Problem

A

-given values of y and y’ at a single value of x

5
Q

Homogeneous Second Order Linear ODEs

Boundary Value Problem

A

-given the values of y at two different values of x

6
Q

Second Order Linear ODEs

Theorem for Initial Value Problems

A
  • let e(x), f(x), g(x) and h(x) be continuous on an interval I of the x-axis and let e(x) ≠ 0 for all x in I
  • if the initial condition is specified at a point in I then the solution exists and is unique
7
Q

Superposition Principle for Linear ODEs

Linear Superposition

A
  • suppose that y1(x) and y2(x) are two solutions of the homogenous second order linear ODE
  • then C1y1 + C2y2 is called a linear superposition of the two functions
8
Q

Superposition Principe for Linear ODEs

Particular Integral and Complementary Functions

A
  • suppose that y0(x) is a solution of the inhomogeneous second order linear ODE
  • and y1(x) is a solution of the homogeneous form of the same equation
  • then y0(x) + C1y1(x) is also a solution of the inhomogeneous form of the ODE
  • y0 is the particular integral and C1y1, C2y2, or C1y1+c2y2 are complimentary functions
9
Q

Reduction of Order

A

1) let y1(x) be a solution that we know to the homogeneous form of the equation, now let
y(x) = u(x) y1(x)
2) calculate the first and second derivatives of y
3) substitute y, y’ and y’’ into the inhomogeneous form of the equation and collect terms containing u, u’ and u’’
4) the coefficient of the u term should be the second order ODE solved by y1 which we know is homogeneous so equals 0
5) only terms in u’ and u’’ remain, use a substitution
v = u’ this turns the second order ODE into a first order ODE
6) solve for v by the integrating factor method
7) integrate v to find u remembering to add the constant of integration
8) substitute u into y=uy1 to find y
9) if given use any initial or boundary values to find the values of the constants

10
Q

First Order Constant Coefficient Linear ODEs

General Form

A

dy/dx + ay = h(x)

11
Q

First Order Constant Coefficient Linear ODEs

Complimentary Function

A

ycf = C e^(-ax)

12
Q

General Solution, Complimentary Function, Particular Integral

A

-the general solution is the sum of the complimentary function and particular integral
y(x) = ycf + ypi

13
Q

Complimentary Function

A

-the solution to the homogeneous form of the equation

14
Q

Particular Integral

h(x) = constant

A

ypi = constant

15
Q

Particular Integral

h(x) = polynomial

A

ypi = polynomial of the same degree as h(x)

-but if the h(x) is a polynomial times the complimentary function then ypi is a polynomial one degree higher than h(x)

16
Q

Particular Integral

h(x) = exponential

A
  • if h(x) is an exponential different from the complimentary function then ypi is the same exponential as h(x) but multiplied by a constant
  • if h(x) is the same exponential as the complimentary function then ypi is x times the complimentary function
17
Q

Particular Integral

h(x) = sine or cosine

A

ypi is a combination of sines and cosines

18
Q

Second Order Constant Coefficient Linear ODEs

General Equation - Homogeneous Case

A

y’’ + ay’ + by = 0

where a and b are constants

19
Q

Second Order Constant Coefficient Linear ODEs

Solving the Homogeneous Case

A
-substitute
y = e^λx
-divide by e^λx to get the characteristic equation
λ² + aλ + b = 0
-solve for lambda and find y
20
Q

Second Order Constant Coefficient Linear ODEs

Solving the Homogeneous Case - Real Distinct Roots

A

a² - 4b > 0
-two real distinct roots λ1 and λ2

ycf = C1e^λ1x + C2e^λ2x

21
Q

Second Order Constant Coefficient Linear ODEs

Solving the Homogeneous Case - Real Repeated Roots

A

a² - 4b = 0
-one real root that is repeated, λ

ycf = (A+Bx)e^λx

22
Q

Second Order Constant Coefficient Linear ODEs

Solving the Homogeneous Case - Complex Roots

A

a² - 4b < 0
-two distinct complex roots
λ1 = p + iq
λ2 = p - iq

ycf = e^px(C1cos(qx)+C2sin(qx))

23
Q

Second Order Constant Coefficient Linear ODEs

Homogeneous Case - Particular Integral

A

-as this is the homogeneous case, the particular integral is 0
y = ycf + ypi
y = ycf + 0
y = ycf

24
Q

Second Order Constant Coefficient Linear ODEs

Inhomogeneous Case - General Equation

A

y’’ + ay’ +by = h(x)

where a and b are both constants

25
Q

Second Order Constant Coefficient Linear ODEs

Inhomogeneous Case - Complimentary Function

A
  • let y = Ce^λx
  • substitute into the homogenous form of the equation to find the characteristic equation
  • solve for λ
26
Q

Second Order Constant Coefficient Linear ODEs

Inhomogeneous Case - Particular Integral

A

1) if h(x) does not include anything that looks like the CF try setting ypi to a function that looks like h(x) but with all coefficients undetermined
2) if h(x) does include functions that look like the CF try setting ypi to a function that looks like xh(x) but with all coefficients undetermined
3) if the CF contains xh(x) try setting ypi equal to x²h(x)
3) if h(x) is a sum or product of more than one function try adding or multiplying together the corresponding guesses
4) once calculated ypi should not contain any arbitrary constants

27
Q

Second Order Constant Coefficient Linear ODEs

Inhomogeneous Case - General Solution

A

y = ycf + ypi

-in the inhomogeneous case ypi is not usually zero