explain methods of solving high order linear differntial equation
Answers
Answer:
Higher Order Linear Nonhomogeneous Differential Equations with Constant Coefficients. ... The general solution y(x) of the nonhomogeneous equation is the sum of the general solution y0(x) of the corresponding homogeneous equation and a particular solution y1(x) of the nonhomogeneous equation: y(x)=y0(x)+y1(x).
Step-by-step explanation:
Objectives: Solve n-th order homogeneous linear equations
any
(n) + an−1y
(n−1) + · · · + a1y
′ + a0y = 0,
where an, · · · , a1, a0 are constants with an 6= 0.
Solution Method:
• Find the roots of the characteristic polynomial:
anλ
n + an−1λ
n−1 + · · · + a1λ + a0 = 0.
• Each root λ produces a particular exponential solution e
λt of the differential equation.
• A repeated root λ of multiplicity k produces k linearly independent solutions e
λt, teλt
, · · · , tk−1
e
λt
.
Warning: The above method of characteristic roots does not work for linear equations with
variable coefficients. As matter of fact, the explicit solution method does not exist for the
general class of linear equations with variable coefficients.
Example 1: (a) Find general solutions of y
′′′ + 4y
′′ − 7y
′ − 10y = 0.
(b) Solve y
′′′ + 4y
′′ − 7y
′ − 10y = 0, y(0) = −3, y′
(0) = 12, y′′(0) = −36.
Solution: (a) Solve the characteristic polynomial:
λ
3 + 4λ
2 − 7λ − 10 = 0.
The roots are λ1 = −1, λ2 = 2, λ3 = −5. Each root gives a particular exponential solution of
the differential equation. Combined, the general solutions are
y = C1e
−t + C2e
2t + C3e
−5t
,
where C1, C2, C3 are free parameters (arbitrary constants).
(b) We use the initial conditions to determine the values of the constants C1, C2, C3 in the
general solution formula. The initial conditions y(0) = −3, y′
(0) = 12, y′′(0) = −36 yield
C1 + C2 + C3 = −3,
−C1 + 2C2 − 5C3 = 12,
C1 + 4C2 + 25C3 = −36.
Solve this linear system for C1, C2, C3:
C1 = −5/2,
C2 = 1,
C3 = −3/2.
Thus, the solution to the initial value problem is
y = −
5
2
e
−t + e
2t −
3
2
e
−5t
.