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
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