Second-Order Linear ODE - General Form
e(x)d²y/dx² + f(x)dy/dx+ g(x)y
=h(x)
Homogeneous and Inhomogeneous Second-Order Linear ODEs
-a second order linear ODE is homogeneous if h(x)=0
-it is inhomogeneous if
h(x) ≠ 0
Constant Coefficient and Non-Constant Coefficient Second-Order Linear ODEs
- 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
Homogeneous Second Order Linear ODEs
Initial Value Problem
-given values of y and y’ at a single value of x
Homogeneous Second Order Linear ODEs
Boundary Value Problem
-given the values of y at two different values of x
Second Order Linear ODEs
Theorem for Initial Value Problems
- 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
Superposition Principle for Linear ODEs
Linear Superposition
- 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
Superposition Principe for Linear ODEs
Particular Integral and Complementary Functions
- 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
Reduction of Order
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
First Order Constant Coefficient Linear ODEs
General Form
dy/dx + ay = h(x)
First Order Constant Coefficient Linear ODEs
Complimentary Function
ycf = C e^(-ax)
General Solution, Complimentary Function, Particular Integral
-the general solution is the sum of the complimentary function and particular integral
y(x) = ycf + ypi
Complimentary Function
-the solution to the homogeneous form of the equation
Particular Integral
h(x) = constant
ypi = constant
Particular Integral
h(x) = polynomial
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)
Particular Integral
h(x) = exponential
- 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
Particular Integral
h(x) = sine or cosine
ypi is a combination of sines and cosines
Second Order Constant Coefficient Linear ODEs
General Equation - Homogeneous Case
y’’ + ay’ + by = 0
where a and b are constants
Second Order Constant Coefficient Linear ODEs
Solving the Homogeneous Case
-substitute y = e^λx -divide by e^λx to get the characteristic equation λ² + aλ + b = 0 -solve for lambda and find y
Second Order Constant Coefficient Linear ODEs
Solving the Homogeneous Case - Real Distinct Roots
a² - 4b > 0
-two real distinct roots λ1 and λ2
ycf = C1e^λ1x + C2e^λ2x
Second Order Constant Coefficient Linear ODEs
Solving the Homogeneous Case - Real Repeated Roots
a² - 4b = 0
-one real root that is repeated, λ
ycf = (A+Bx)e^λx
Second Order Constant Coefficient Linear ODEs
Solving the Homogeneous Case - Complex Roots
a² - 4b < 0
-two distinct complex roots
λ1 = p + iq
λ2 = p - iq
ycf = e^px(C1cos(qx)+C2sin(qx))
Second Order Constant Coefficient Linear ODEs
Homogeneous Case - Particular Integral
-as this is the homogeneous case, the particular integral is 0
y = ycf + ypi
y = ycf + 0
y = ycf
Second Order Constant Coefficient Linear ODEs
Inhomogeneous Case - General Equation
y’’ + ay’ +by = h(x)
where a and b are both constants
Second Order Constant Coefficient Linear ODEs
Inhomogeneous Case - Complimentary Function
- let y = Ce^λx
- substitute into the homogenous form of the equation to find the characteristic equation
- solve for λ
Second Order Constant Coefficient Linear ODEs
Inhomogeneous Case - Particular Integral
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
Second Order Constant Coefficient Linear ODEs
Inhomogeneous Case - General Solution
y = ycf + ypi
-in the inhomogeneous case ypi is not usually zero