Question

the solution for the partial differential equation (d^4-d'^4)z = 0 is​

Answer 1

Step-by-step 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)#