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

Question

Charles Anctil · Accepted Answer

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

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#

There is an easy fix for each of these factors, such as:

#(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 sum of the parts that make up the whole solution is:

#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)#

Emily Ammerman · Answer

# y =Ae^(sqrt(2)x) + Be^(-sqrt(2)x) + Ccos(sqrt(2)x)+Dsin(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 Role The following is the auxiliary equation for [A]: # 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# Thus, we have: # m^2-2 = 0 = > m=+-sqrt(2) # (real and distinct) # m^2+2 = 0 = > m=+-sqrt(2)i # (pure imaginary) Parts of the solution are determined by the auxiliary equation's roots; if these parts are linearly independent, the solutions' superposition forms the complete general solution. Thus, the homogeneous equation [A] has the following solution: # 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

Luke Alvarez · Answer

undefined The general solution of the differential equation $\frac{{d^4y}}{{dx^4}} - 4y = 0$ is:

$$y = C_1 e^{2x} + C_2 e^{-2x} + C_3 \cos(2x) + C_4 \sin(2x)$$

Where $C_1$, $C_2$, $C_3$, and $C_4$ are arbitrary constants.