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
Steve M
Apr 24, 2018

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, #y#, using the chain rule and the product rule:

# 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 #sin2x# and #cos2x#

# (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 # y = cos2x \ e^(-x) # is a solution, QED

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