Final answer:
The value of r in the general solution to the differential equation 5y'' = 4y' is 0.8 or 4/5. This is found by converting the differential equation to its characteristic equation and solving for r.
Step-by-step explanation:
The general solution to the second-order differential equation 5y'' = 4y' indicates a homogeneous linear differential equation with constant coefficients. To find the general solution, we first convert the differential equation into its characteristic equation.
The characteristic equation is based on the standard form ay'' + by' + cy = 0. Comparing this with our equation 5y'' = 4y', we can see that c = 0 and our equation becomes 5r^2 = 4r. To find r, we solve this quadratic equation.
Divide both sides by r (assuming r is not zero), which gives 5r = 4. Solving for r yields r = 4/5, which is the value we were asked to find. So, the value of r in the general solution y(x) = c1 e^(rx) + c2 is 0.8 or 4/5.