Questions: Find all values of (m) so that the function (y=e^mx) is a solution of the given differential equation. (Enter (2 y''+15 y'-8 y=0) (m=)

Solution Steps

Let \( y = e^{mx} \). The first derivative is given by \( y' = m e^{mx} \) and the second derivative is \( y'' = m^2 e^{mx} \).

Substituting \( y \), \( y' \), and \( y'' \) into the differential equation \( 2y'' + 15y' - 8y = 0 \) results in: \[ 2(m^2 e^{mx}) + 15(m e^{mx}) - 8(e^{mx}) = 0 \]

Factoring out \( e^{mx} \) (which is never zero), we simplify the equation to: \[ 2m^2 + 15m - 8 = 0 \]

We solve the quadratic equation \( 2m^2 + 15m - 8 = 0 \) using the quadratic formula: \[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 2 \), \( b = 15 \), and \( c = -8 \).

Calculating the discriminant and applying the quadratic formula yields the solutions: \[ m = -8 \quad \text{and} \quad m = \frac{1}{2} \]

Final Answer