Show #y=cos2xe^-x# is a solution to the differential # (d^2y)/(dx^2) + 2dy/dx+5y = 0 #?
#y=cos2xe^-x#
# (d^2y)/(dx^2) + 2dy/dx+5y = 0 #
1 Answer
We wish to show that:
# y = cos2x \ e^(-x) #
is a solution of the ODE:
# (d^2y)/(dx^2) + 2dy/dx+5y #
We can differentiate,
# dy/dx \ = (cos2x)(-e^(-x)) + (-2sin2x)(e^(-x)) #
# \ \ \ \ \ \ \= (-2sin2x-cos2x)e^(-x) #
And doing the same again:
# (d^2y)/(dx^2) = (-2sin2x-cos2x)(-e^(-x)) + (-4cos2x+2sin2x)(e^(-x)) #
# \ \ \ \ \ \ \= (2sin2x+cos2x-4cos2x+2sin2x)e^(-x) #
# \ \ \ \ \ \ \= (4sin2x-3cos2x)e^(-x) #
Then inserting the result into the given expression we have:
# (d^2y)/(dx^2) + 2dy/dx+5y = (4sin2x-3cos2x)e^(-x) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + 2(-2sin2x-cos2x)e^(-x) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + 5cos2x e^(-x) #
Collecting terms in
# (d^2y)/(dx^2) + 2dy/dx+5y = {(4-4)sin2x+(-3-2+5)cos2x}e^(-x) #
Hence, we have:
# (d^2y)/(dx^2) + 2dy/dx+5y = 0 #
Confirming that