second order ODE

Question 1

general form of second order linear ODE

Answer 1

a2 y’’ + a1 y’ + a0y = ø(x)
where ø is an arbitrary function of x
we take a2 ≠ 0 and a1, a0 constant
if y1 and y2 are 2 solution then so is y = Ay1 + By2 for A,B constant

Question 2

homogeneous

Answer 2

if ø= 0 we call the ODE homogeneous
a2 y’’ + a1 y’ + a0y = 0

Question 3

characteristic / auxiliary equation

Answer 3

• distinct real roots: y = Ae^λ1x + Be^λ2x
• one repeated root: y = (Ax + B)e^λx
• 2 complex roots: e^⍺x(cosβx + sinβx) where solutions are ⍺ +- βi

Question 4

method of undetermined coefficients

Answer 4

general solution: y = yCF + yPI
yCF is the complementary function
yPI is the particular integral
know yCF from char
guess yPI based on form of ø
(only if its polynomial, exponential or trig or a sum/product of these)

Question 5

how to guess yPI based on ø

Answer 5

e^µx -> ae^µx
x^n -> a0 + a1x + … + anx^n
cosµx -> a1cosµx + a2sinµx
sin µx -> “ same
if ø is a sum/ product of the above terms try the corresponding sum/ product
- if the form to try is already in yCF then putting in gives 0 so try yPI timesed by x
- can apply this twice (times by x^2) if once still in yCF

Question 6

initial value problem

Answer 6

when we require y(x0) = y0 and y’(x0) = µ for given constants
to solve find GS and find unknown constants which satisfy these

Question 7

boundary value problem

Answer 7

is when we require y(x0) = y0 and y(x1) = y1 for given constants
to solve find GS and find unknown constants which satisfy these

Question 8

wronskian

Answer 8

W(y1,y2) =
| y1 y2|
| y1’ y2’| = y1y2’ - y2y1’

Question 9

method of variation of parameters

Answer 9

u1 = - ∫ ø y2 /a2 W
u2 = ∫ ø y1 /a2 W
yPI = u1y1 + u2 y2

Question 10

wronskian if LI

Answer 10

if y1 and y2 are linearly dependent then the wronskian = 0
if not 0 then linearly independent

Question 11

system of 1st order linear ODEs

Answer 11

a system of n coupled 1st order linear ODEs for n dependent variables can be rewritten as a single nth order linear ODE for a single dependent variable by eliminating the other dependent variables
eg y’ = y - 2z + 2x
z’ = 3y -4z + 2x
rearrange 1 to make z =
then differentiate to find z’
sub into eq 2 to get second order in terms of x and y
solve for y
then sub back using eq 1 to find z