Final answer:
The general solution of the given differential equation 4y'' - 9y = 0 is y = C₁e^(3x) + C₂e^(-3x).
Step-by-step explanation:
The given differential equation is 4y'' - 9y = 0. To find the general solution, we can assume the solution to be of the form y = e^(rx), where r is a constant.
Substitute this into the differential equation: 4(e^(rx))'' - 9(e^(rx)) = 0. Simplifying this equation, we get (r² - 9)e^(rx) = 0. Since e^(rx) is never zero, we have r² - 9 = 0. Solving for r, we get r = ±3.
Therefore, the general solution of the given differential equation is y = C₁e^(3x) + C₂e^(-3x), where C₁ and C₂ are arbitrary constants.