What is the general solution of the differential equation? : (d^4y)/(dx^4) - 4 y = 0
2 Answers
Explanation:
With linear operator
Each of these factors has a simple solution, eg:
-
(D - sqrt2)y=0 implies y' = sqrt 2y implies y = alpha e^(sqrt 2 x) -
(D -i sqrt2 )y = 0 implies y' = i sqrt 2 y implies y = beta e^(i sqrt 2 x)
The complete solution is the superposition of individual solutions:
y =Ae^(sqrt(2)x) + Be^(-sqrt(2)x) + Ccos(sqrt(2)x)+Dsin(sqrt(2)x)
Explanation:
We have:
(d^4y)/(dx^4) - 4 y = 0 ..... [A]
This is a fourth order linear Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution,
Complementary Function
The Auxiliary equation associated with [A] is:
m^4-4 = 0
The challenge with higher order Differential Equation is solving the associated higher order Auxiliary equation. However, in this case we have a difference of two squares, thus we can immediately factorise the equation using
(m^2-2)(m^2+2) = 0
So we have:
m^2-2 = 0 = > m=+-sqrt(2) (real and distinct)
m^2+2 = 0 = > m=+-sqrt(2)i (pure imaginary)
The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.
- Real distinct roots
m=alpha,beta, ... will yield linearly independent solutions of the formy_1=Ae^(alphax) ,y_2=Be^(betax) , ... - Real repeated roots
m=alpha , will yield a solution of the formy=(Ax+B)e^(alphax) where the polynomial has the same degree as the repeat. - Complex roots (which must occur as conjugate pairs)
m=p+-qi will yield a pairs linearly independent solutions of the formy=e^(px)(Acos(qx)+Bsin(qx))
Thus the solution of the homogeneous equation [A] is:
y = Ae^(sqrt(2)x) + Be^(-sqrt(2)x) + e^(0x)(Ccos(sqrt(2)x)+Dsin(sqrt(2)x))
\ \ = Ae^(sqrt(2)x) + Be^(-sqrt(2)x) + Ccos(sqrt(2)x)+Dsin(sqrt(2)x)
Note this solution has