What is the general solution of the differential equation? : #(d^4y)/(dx^4) - 4 y = 0#

With linear operator #D = d/(dx)#, this is:

#(D^4 - 4)y=0#

#(D^2 - 2)(D^2 + 2)y=0#

#(D - sqrt2)(D+sqrt2)(D -i sqrt2 )(D + i sqrt2 )y=0#

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:

#implies y(x) = A e^(sqrt2 x) + B e^(- sqrt 2 x) + C e^(i sqrt2 x) + D e^(- i sqrt 2 x)#

#C e^(i sqrt2 x) + D e^(- i sqrt 2 x)# can also be re-written using the Euler formula , arriving at:

#y = A e^(sqrt2 x) + B e^(- sqrt 2 x) + C' cos ( sqrt2 x) + D' sin (sqrt 2 x)#

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, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the polynomial equation with the coefficients of the derivatives.

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 #A^2-B^2 -=(A-b)(A+B)#, so we get::

# (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 form #y_1=Ae^(alphax)#, #y_2=Be^(betax)#, ...
• Real repeated roots #m=alpha#, will yield a solution of the form #y=(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 form #y=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 #4# constants of integration and #4# linearly independent solutions, hence by the Existence and Uniqueness Theorem their superposition is the General Solution