Question

Find all values of m so that the function y = emx is a solution of the given differential equation. (Enter your answers as a comma-separated list.)
2y'' + 5y' − 3y = 0

Answer 1

Answer:

$m =\dfrac{1}{2}, -3$

Step-by-step explanation:

We are given the following in the question:

$y = e^{mx}$ solutio to the differential equation:

$2y'' + 5y' - 3y = 0$

$y' = \dfrac{dy}{dx} = \dfrac{d(e^{mx})}{dx} = me^{mx}\\\\y'' = \dfrac{d(y')}{dx} = \dfrac{d(me^{mx})}{dx} = m^2e^{mx}$

Putting these values in the differential equation, we have,

$2y'' + 5y' - 3y = 0\\2(m^2e^{mx}) + 5(me^{mx}) - 3e^{mx} = 0\\(2m^2 + 5m-3)(e^{mx} = 0\\e^{mx}\neq 0\\2m^2 + 5m-3 = 0\\2m^2 +6m-m-3=0\\2m(m+3)-1(m+3)=0\\(2m-1)(m+3) = 0\\(2m-1) = 0, (m+3) = 0\\\\m =\dfrac{1}{2}, m = -3$

are the required values of m.