Question

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

Answer 1

`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)`

Answer 2

`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