Question

check whether each of the following functions is a solution of the differential equation (4y + 7y = 3e^{-x}).
(a) (y = e^{-x})
(b) (y = e^{-x}+e^{-(7/4)x})
(c) (y = e^{-x}+ce^{-(7/4)x})
(a) find (4y), (7y), and (4y + 7y) for (y = e^{-x}).
(4y=)
(7y=)
(4y + 7y=)

Explanation:

Step1: Find the derivative of \(y = e^{-x}\)

The derivative of \(y = e^{-x}\) using the rule \((e^{ax})^\prime=ae^{ax}\) is \(y^\prime=-e^{-x}\). Then \(4y^\prime = 4\times(-e^{-x})=-4e^{-x}\)

Step2: Calculate \(7y\)

Since \(y = e^{-x}\), then \(7y = 7e^{-x}\)

Step3: Calculate \(4y^\prime + 7y\)

Substitute the values of \(4y^\prime\) and \(7y\) into \(4y^\prime + 7y\):
\(4y^\prime+7y=-4e^{-x}+7e^{-x}=( - 4 + 7)e^{-x}=3e^{-x}\)

Answer:

\(4y^\prime=-4e^{-x}\), \(7y = 7e^{-x}\), \(4y^\prime + 7y=3e^{-x}\)