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