Final answer:
To solve the given initial-value problem, we find the characteristic equation by substituting y = e^(rx) into the differential equation. The roots of the characteristic equation are found to be r = 1/2 and r = -3. Using the general solution equation, we substitute the roots and apply the initial conditions to find the specific solution.
Step-by-step explanation:
To solve the initial-value problem 2y" + 5y - 3y = 0 with the initial conditions y(0) = -1 and y'(0) = 31, we can use the method of finding the characteristic equation and its roots. The characteristic equation is found by substituting y = e^(rx) into the differential equation, which gives 2r^2 + 5r - 3 = 0. Factoring this equation, we get (2r - 1)(r + 3) = 0. So, the roots are r = 1/2 and r = -3.
Using the general solution y(x) = c1e^(r1x) + c2e^(r2x), where r1 and r2 are the roots of the characteristic equation, we have y(x) = c1e^(1/2x) + c2e^(-3x).
Applying the initial conditions, we have y(0) = c1e^0 + c2e^0 = c1 + c2 = -1 and y'(0) = (c1/2)e^(1/2*0) + (-3c2)e^(-3*0) = (c1/2) - 3c2 = 31.
Solving these equations simultaneously, we find c1 = 14 and c2 = -15. Therefore, the solution to the initial-value problem is y(x) = 14e^(1/2x) - 15e^(-3x).