Homogeneous Linear DE of 2
ndOrder
– Homogeneous DE
– Example of a non-homogeneous linear DE
– Example of a homogeneous linear DE
– Example of a non-linear DE
1 x
2 y 2 xy 6 y 0
y p x y q x r x
4
sin
xy y e x
2 2 0
x y y y y y
0
y p x y q x
Homogeneous Linear DE of 2
ndOrder
– Example of a non-linear DE
• Superposition or Linearity Principle
– DE
– Solution
– Linear combination of the known solutions y1, y2
,
,
x
xy e y e
1 1 2 2
y c y c y
0 y y
2
1
y y
3 2
5
x
xy e e
Superposition Principle
--- Any linear combination of two solutions is again a solution on I.
Sums and constant multiples of solutions are again solutions.
Does not hold for non-homogeneous and nonlinear DE – Example of non-homogeneous DE
– Solution
– But, not a solution
1 cos , 1 sin
y x y x
1 y y
2 1 cos x , 1 cos x 1 sin x
Superposition Principle
– Example of non-linear DE
– Solution
– But, not a solution
2
, 1
y x y
2 2
, 1
x x
0
y y xy
Initial Value Problem (1)
– Solution of 2nd order homogeneous linear DE
– Two IC’s
– Example of IVP
– Two known solutions ex, e-x – Particular solution
0, 0 4, 0 2
y y y y
1 1 2 2
y c y c y
0
0,
0
1y x K y x K
1 2
x
xy c e c e
3
x
xy e e
Initial Value Problem (2)
– Wrong selection of known solutions ex, e-x
Cannot solve the DE
1 2
x
xy c e c le
Basis (Fundamental System)
– General sol.of homogeneous linear DE
– Not proportional to the known solutions
– Particular solution can be obtained by assigning c1, c2
– Linearly independent y1, y2 on I
1 1
2 2 0
k y x k y x
1 1 2 2
y c y c y
1
,
2y y
Basis1 1 2 2
y c y c y
1
0,
2 0
k k
Basis (Fundamental System)
– Linearly dependent
2
1 2
1
k
y y
k
Basis (Fundamental System)
– General sol.of homogeneous linear DE
--- a pair y1, y2 on I of linearly independent sol.
– Example – General sol.
– Example
– General sol.
0 y y
1 2
x
xy c e c e
0 y y
1
cos
2sin
y c x c x
Method of Reduction of Order (1)
– How to obtain basis if one solution is known – set
– Differentiating
– Original DE
– Rearranging in u
2 1 1
2 1
2
1 1
y u y uy
y u y u y uy
2
1y uy
1
2
1 1 1 1 10
u y u y uy p u y uy quy
1
2
1 1 1 1 10
u y u y py u y py qy
0
Method of Reduction of Order (2)
– Divide by y1 and set
– Then
– Separation
1 1
2 0
U y p U
y
1 1
2
dU y
U y p dx
ln U 2 ln y
1 pdx
, u U u U
1 1
1
2 0
y py u u
y
Method of Reduction of Order (3)
– Taking exponentials
– Desired sol. y2
--- where
2
1
1
y uy y Udx
2 1
1
pdxU e
y
2 1
y u Udx
y const.
– 2nd order DE – Standard form
– One known sol.
– Formulation of the other sol.
ln 2
1 1
x
U e
x x
2
0 x y xy y
2
1 1
0
y y y
x x
1
y x
2
ln
y ux x Udx x x
Example of Reduction of Order
– 2nd order DE
– 1st order DE and its solution
– Substitution
ky e x
0 y ay by
,
2
y e x
y e x y e x
0 y ky
2 a b e
x 0
Characteristic equation2
ndOrder DE with Const. Coeff. (1)
– Roots of the characteristic eqn.
– Bases of the original DE
– Three cases
(Case I) two real roots if
(Case II) a real double root if
(Case I) complex conjugate roots if
• Two Distinct Real Roots
– Bases of the original DE
1 2
1
,
2
y e x y
e x
2
ndOrder DE with Const. Coeff. (2)
2
2
1 2
1 1
4 , 4
2 2
a a b a a b
2
4 0
a b
2
4 0
a b
2
4 0
a b
1 2
1
,
2
y e x y
e x
– General sol.
– Example
– Characteristic Eqn., its roots
– General sol.
0 y y
Case I. Two Distinct Real Roots (1)
1 2
1 2
x
xy c e
c e
2
1 2
1 0,
1, 1
1 2
x
xy c e c e
– Initial Value Problem
– Characteristic Eqn., its roots
– General sol.
– Particular sol.
2
2 0,
2
1 2
x
xy c e c e
Case I. Two Distinct Real Roots (2)
2 0, 0 4, 0 5
y y y y y
1 1
1 1
1 9 1, 1 9 2
2 2
3
2
x
xy e e
– Only one sol.
– To obtain a second independent solution y2, use reduction of order.
– Set
– Substituting and rearranging
Since
Case II. Real Double Root (1)
2
1
a xy e
2
1y uy
2 1 1
2 1
2
1 1
y u y uy
y u y u y uy
1
2
1 1 1 1 10
u y u y py u y py qy
0 0
2
1 1
2 y ae
ax ay
– Thus
– Double integrating
– Select an independent sol.
– Corresponding general solution
Case II. Real Double Root (2)
1
0, 0
u y u
1 2
u c x c
2
1,
y uy u x
1 2
2
axy c xc e
– DE
– Characteristic Eq.
– Double root
– Corresponding general solution
Example of Real Double Root (1)
8 16 0
y y y
2
8 16 0
4
1 2
4
xy c xc e
– Initial Value Problem – Characteristic Eq.
– Double root
– Corresponding general solution
– Particular solution
Example of Real Double Root (2)
4 4 0, 0 3, 0 1
y y y y y
2
4 4 0
2
1 2
2
xy c xc e
3 5
2
xy x e
Case III. Complex Conjugate Roots (1)
– 2nd order DE
– Characteristic Eq.
– Roots of the characteristic eqn.
– (Case III) complex conjugate roots if
0 y ay by
2
a b 0
2
2
1 2
1 1
4 , 4
2 2
a a b a a b
2
4 0
a b
Fundamental of Complex Roots (1)
– 2nd order DE
– Characteristic Eq.
– Roots of the characteristic eqn.
– Two complex bases
0 y y
2
1 0
1 i
ix
,
ixe e
Fundamental of Complex Roots (2)
– Complex exponential function
– Adding
– Subtracting
– The other bases:
cos sin e
ix x i x
cos ,sin x x cos sin e
ix x i x
cos 1
2
ix ix
x e e
Euler formula
sin 1
2
ix ix
x e e
i
Complex Exponential Function
– Real case:
– Maclaurin series
cos t i sin t
2 3 4 51 2! 3! 4! 5!
it
it it it it
e it
cos sin
z s it s it s
e e
e e e t i t z s
2 4 3 5
1 2! 4! 3! 5!
t t t t
i t
Case III. Complex Conjugate Roots (2)
– One root of the characteristic eqn.
– Roots of the characteristic eqn.
where
1 2
1 1
2 a i , 2 a i
2 2
1
1 1 1 1
2 a 2 a 4 b 2 a i b 4 a
1
2b 4 a
Case III. Complex Conjugate Roots (3)
– Using complex exponential function
– Adding and divide by 2 y1
– Subtracting and divide by 2i y2
– General solution
2
1 ax
cos
y e
x
1
2
2 2
2 2
cos sin
cos sin
a x i x a x
x
a x i x a x
x
e e e x i x
e e e x i x
2
2 ax
sin
y e
x
2
cos sin
y e
axA x B x
New bases
Example of Complex Conjugate Roots (1)
– Initial value problem
– Characteristic eqn.
– Roots of the characteristic eqn.
– General solution
0.1 2i
0.2 4.01 0, 0 0, 0 2
y y y y y
0.1x
cos 2 sin 2 y e
A x B x
2
0.2 4.01 0
Example of Complex Conjugate Roots (2)
– First IC
– Then, remaining and its derivative
– Second IC
– Particular solution
0 2 2, 1
y B B
0 0
y A
0.1x
sin 2 y e
x
0.1x
sin 2 y Be
x
0.1
0.1xsin 2 2
0.1xcos 2
y B e
x e
x
Example of Complex Conjugate Roots (3)
e
0.1x e
0.1xDamped vibrations
– DE
– General solution
2
0
y y
cos sin
y A x B x
Example of Complex Conjugate Roots (4)
Summary
Case
Roots of
Characteristic Eq.
Basis of DE General sol.
I
Distinct real
II
Real double root
III
Complex conjugate
1
,
2 e
1x, e
2xy c e
1 1x c e
2 2x1
2 a
e
ax 2, xe
ax 2
1 2
2
axy c c x e
1,2
1
2 a i
2 2
cos , sin
ax ax
e x
e x
2
cos sin
y e
axA x B x
Boundary Value Problem
– DE
– Example
– General sol.
– Particular sol.
1
1,
2
2y P k y P k
Boundaryconditons
0, 0 3, 3
y y y y
1cos
2sin
y x c x c x
3cos
2sin
y x x c x
Not determined yet