Respuesta :
Answer: y'' + 2y' + 5y is 8e^(-x)sin(2x) - 4x^2e^(-x)sin(2x) + 4e^(-x)cos(2x).
Step-by-step explanation:
Using the product rule, we have:
y' = e^(-x)sin(2x) - 2xe^(-x)sin(2x)
Using the product rule again, we have:
y'' = e^(-x)sin(2x) - 4xe^(-x)sin(2x) - 4x^2e^(-x)sin(2x) + 4e^(-x)cos(2x)
Now we can simplify the expression:
y'' + 2y' + 5y = (e^(-x)sin(2x) - 4xe^(-x)sin(2x) - 4x^2e^(-x)sin(2x) + 4e^(-x)cos(2x)) + 2(e^(-x)sin(2x) - 2xe^(-x)sin(2x)) + 5(e^(-x)sin(2x))
y'' + 2y' + 5y = e^(-x)sin(2x) - 4xe^(-x)sin(2x) - 4x^2e^(-x)sin(2x) + 4e^(-x)cos(2x) + 2e^(-x)sin(2x) - 4xe^(-x)sin(2x) + 5e^(-x)sin(2x)
y'' + 2y' + 5y = 8e^(-x)sin(2x) - 4x^2e^(-x)sin(2x) + 4e^(-x)cos(2x)
Therefore, y'' + 2y' + 5y is 8e^(-x)sin(2x) - 4x^2e^(-x)sin(2x) + 4e^(-x)cos(2x).
