Substitute yequalse Superscript rx into the given differential equation to determine all values of the constant r for which yequalse Superscript rx is a solution of the equation. y double prime plus 4 y prime minus 12 y equals 0

y (x) = C_1·e^{2x} + C_2·e^{-6x}

Step-by-step explanation:

From Exercise we have the differential equation

y''+4y'-12=0.

This is a characteristic differential equation and we are solved as follows:

y''+4y'-12=0

m²+4m-12=0

m_{1,2}=/frac{-4±/sqrt{16+48}}{2}

m_{1,2}=/frac{-4±/sqrt{64}}{2}

m_{1,2}=/frac{-4±8}{2}

m_1=2

m_2=-6

The general solution of this differential equation is in the form

y (x) = C_1·e^{m_1 ·x} + C_2·e^{m_2 ·x}

Therefore, we get

y (x) = C_1·e^{2x} + C_2·e^{-6x}