Answer:
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}
