HIGHER ORDER LINEAR
DIFFERENTIAL EQUATIONS
WITH CONSTANT COEFFICIENTS (COMPLEMENTARY FUNCTION)
BY PROF. TILOTTAMA BARHATE
MOTIVATION :
𝑑𝑦
Consider + 𝑎𝑦 = 0
𝑑𝑥
𝑑𝑦
= −𝑎𝑑𝑥
𝑦
𝑦 = 𝐶𝑒 −𝑎𝑥
Observation: The constant coefficient of 𝑦 in the given equation goes in
power of exponential in the solution with opposite sign.
LINEAR DIFFERENTIAL EQUATION:
A differential equation in which 𝑦 and its differentials have degree 1 is called Linear
Differential equation.
Linear differential equation (LDE) of order 𝑛 given by
𝑑𝑛 𝑦 𝑑𝑛−1 𝑦 𝑑𝑛−2 𝑦 𝑑𝑦
𝑛
+ 𝑎1 𝑛−1 + 𝑎2 𝑛−2 +. . . +𝑎𝑛−1 + 𝑎𝑛 𝑦 = ϕ(𝑥)
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
where the coefficients 𝑎1 , 𝑎2 , … are constants.
is called higher order linear Differential equation with constant coefficients
LINEAR DIFFERENTIAL EQUATION:
different ways to write it
The higher order linear Differential equation with constant coefficients
𝑑𝑛 𝑦 𝑑𝑛−1 𝑦 𝑑𝑛−2 𝑦 𝑑𝑦
𝑛
+ 𝑎1 𝑛−1 + 𝑎2 𝑛−2 +. . . +𝑎𝑛−1 + 𝑎𝑛 𝑦 = ϕ(𝑥)
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
Also written as,
𝑑
(𝐷𝑛 + 𝑎1 𝐷𝑛−1 + 𝑎2 𝐷𝑛−2 +. . . +𝑎𝑛−1 𝐷 + 𝑎𝑛 )𝑦 = ϕ(𝑥)where 𝐷 =
𝑑𝑥
Let 𝑓(𝐷) = 𝐷𝑛 + 𝑎1 𝐷𝑛−1 + 𝑎2 𝐷𝑛−2 +. . . +𝑎𝑛−1 𝐷 + 𝑎𝑛 = 0
So, 𝑓 𝐷 𝑦 = 𝜙(𝑥)
HOMOGENEOUS AND NON-HOMOGENEOUS LDE
• Defn: When 𝜙 𝑥 = 0 then LDE 𝑓 𝐷 𝑦 = 0 is called homogeneous LDE
• Defn: When 𝜙 𝑥 ≠ 0 then LDE 𝑓 𝐷 𝑦 = 𝜙(𝑥) is called non- homogeneous
Note: 1. The solution of 𝑓 𝐷 𝑦 = 0 is called Complementary function (C.F.)
1
2. 𝑓 𝐷 𝑦 = 𝜙(𝑥) 𝑦= 𝜙(𝑥) called particular integral (P.I.)
𝑓 𝐷
3. Solution of 𝑓 𝐷 𝑦 = 𝜙(𝑥) is given by, 𝑦 = 𝐶. 𝐹. +𝑃. 𝐼.
AUXILIARY EQUATION
Consider a homogeneous differential equation
(𝐷𝑛 + 𝑎1 𝐷𝑛−1 + 𝑎2 𝐷𝑛−2 +. . . +𝑎𝑛−1 𝐷 + 𝑎𝑛 )𝑦 = 0
Let 𝑓(𝐷) = 𝐷𝑛 + 𝑎1 𝐷𝑛−1 + 𝑎2 𝐷𝑛−2 +. . . +𝑎𝑛−1 𝐷 + 𝑎𝑛
Replace 𝐷 by 𝑚 in 𝑓(𝐷),
The equation 𝑓 𝑚 = 0 is called as auxiliary equation (A.E.).
Here, A.E. is 𝑚𝑛 + 𝑎1 𝑚𝑛−1 + 𝑎2 𝑚𝑛−2 +. . . +𝑎𝑛−1 𝑚 + 𝑎𝑛 = 0
COMPLEMENTARY FUNCTION (C.F.)
solution of 𝑓 𝐷 𝑦 = 0
Suppose 𝑓 𝐷 𝑦 = 0 is second order DE, then A.E. is quadratic equation with
two roots say 𝑚1 and 𝑚2
Sr. No. Roots of A.E. C. F.
1 𝑚1 And 𝑚2 real and distinct 𝑦 = 𝐶1 𝑒 𝑚1 𝑥 + 𝐶2 𝑒 𝑚2 𝑥
2 𝑚1 And 𝑚2 real and equal 𝑦 = (𝐶1 + 𝐶2 𝑥)𝑒 𝑚1 𝑥
3 𝑚1 And 𝑚2 complex
𝑚1 = 𝑎 + 𝑖𝑏 and 𝑚2 = 𝑎 − 𝑖𝑏 𝑦 = 𝑒 𝑎𝑥 (𝐶1 cos𝑏𝑥 + 𝐶2 sin𝑏𝑥)
Note: Number of constants in complementary function is equal to order of the
differential equation.
SOLVED EXAMPLES: (COMPLEMENTARY FUNCTION)
𝑑2 𝑦 𝑑𝑦
1. Solve +4 − 5𝑦 = 0
𝑑𝑥 2 𝑑𝑥
A.E. is 𝑚2 + 4𝑚 − 5 = 0 which gives 𝑚 = −5, 1
Let 𝑚1 = −5 and 𝑚2 = 1 real and distinct.
𝐶. 𝐹. = 𝐶1 𝑒 −5𝑥 + 𝐶2 𝑒 𝑥
𝑑2 𝑦 𝑑𝑦
2. Solve −4 + 4𝑦 = 0
𝑑𝑥 2 𝑑𝑥
A.E. is 𝑚2 − 4𝑚 + 4 = 0 which gives 𝑚 = 2
Only one solution for quadratic equation real and equal
Let 𝑚1 = 𝑚2 = 2 real and equal.
𝐶. 𝐹. = (𝐶1 + 𝐶2 𝑥)𝑒 2𝑥
SOLVED EXAMPLES: (COMPLEMENTARY FUNCTION)
𝑑2 𝑦 𝑑𝑦
3. Solve +2 + 5𝑦 = 0
𝑑𝑥 2 𝑑𝑥
A.E. is 𝑚2 + 2𝑚 + 5 = 0 which gives 𝑚 = − 1 ± 2𝑖
Let 𝑚1 = −1 + 2i and 𝑚2 = −1 − 2𝑖 complex roots.
𝐶. 𝐹. = 𝑒 −𝑥 (𝐶1 cos 2𝑥 + 𝐶2 sin 2𝑥)
4. Solve 𝐷2 − 4 2 𝑦 = 0
A.E. is (𝑚2 − 4)2 = 0 → 𝑚 − 2 2 𝑚+2 2 = 0 which gives 𝑚 = ±2
Degree 4 polynomial i.e. 4 roots 2 and -2 both are repeated
Let 𝑚1 = 𝑚2 = 2 real and equal and 𝑚3 = 𝑚4 = −2 real and equal.
𝐶. 𝐹. = 𝐶1 + 𝐶2 𝑥 𝑒 2𝑥 + (𝐶3 + 𝐶4 𝑥)𝑒 −2𝑥
SOLVED EXAMPLES: (COMPLEMENTARY FUNCTION)
5. Solve 𝐷 − 3 4 𝑦 = 0
A.E. is 𝑚 − 3 4 = 0 which gives 𝑚 = 3
Degree of A.E. is 4, so it should have 4 roots. All roots are equal to 3.
A differential equation of order 4 has four independent solutions.
Here those independent solutions are, 𝑒 3𝑥 , 𝑥𝑒 3𝑥 , 𝑥 2 𝑒 3𝑥 , 𝑎𝑛𝑑 𝑥 3 𝑒 3𝑥
𝐶. 𝐹 = 𝐶1 𝑒 3𝑥 + 𝐶2 𝑥𝑒 3𝑥 + 𝐶3 𝑥 2 𝑒 3𝑥 + 𝐶4 𝑥 3 𝑒 3𝑥
= 𝐶1 + 𝐶2 𝑥 + 𝐶3 𝑥 2 + 𝐶4 𝑥 3 𝑒 3𝑥
EXERCISE 6.1: (COMPLEMENTARY FUNCTION)
Solve the following differential equations:
1. (𝐷2 − 2D + 1)𝑦=0 (Ans: 𝑦 = 𝐶1 𝑒 𝑥 + 𝐶2 𝑥𝑒 𝑥 )
2. (𝐷2 − 5D + 6)𝑦=0 (Ans: 𝑦 = 𝐶1 𝑒 3𝑥 + 𝐶2 𝑒 2𝑥 )
3. (𝐷2 − 5D − 6)𝑦=0 (Ans: 𝑦 = 𝐶1 𝑒 6𝑥 + 𝐶2 𝑒 −𝑥 )
4. 𝐷3 − 2D + 4 𝑦 = 0 (Ans: 𝐶1 𝑒 −2x + 𝑒 𝑥 (𝐶2 cos𝑥 + 𝐶3 sin𝑥))
5. 𝐷3 − 3D2 + 4D − 2 𝑦 = 0 (Ans: 𝑦 = 𝐶1 𝑒 𝑥 + 𝑒 𝑥 (𝐶2 cos𝑥 + 𝐶3 sin𝑥))
𝑑3 𝑦 𝑑𝑦
6. −7 − 6𝑦 = 0 (Ans: 𝑦 = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −3𝑥 + 𝐶3 𝑒 −2𝑥 )
𝑑𝑥 3 𝑑𝑥
7. (𝐷 − 1)2 (𝐷2 + 1)𝑦 = 0 Ans: 𝑦 = (𝐶1 + 𝐶2 𝑥)𝑒 𝑥 + 𝐶3 cos𝑥 + 𝐶4 sin𝑥)
𝑥
2 − 5 5
8. 2𝐷 + 2𝐷 + 3 𝑦 = 0 (Ans: 𝑦 = 𝑒 2 𝐶1 cos 𝑥 + 𝐶2 sin 𝑥 )
2 2
9. (𝐷2 + 4)2 𝑦 = 0 Ans: 𝑦 = 𝐶1 + 𝐶2 𝑥 cos 2𝑥 + 𝐶3 + 𝐶4 𝑥 sin 2𝑥)
10. 𝐷 − 3 3 𝑦 = 0 (Ans:𝑦 = (𝐶1 + 𝐶2 𝑥 + 𝐶3 𝑥 2 )𝑒 3𝑥 )
THANK YOU
