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

2 Answers
May 12, 2018

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

Explanation:

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)

May 12, 2018

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, 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