Find the solution of the initial value problem y 6y 10y 0

Find the solution of the initial value problem y\'\' + 6y\' + 10y = 0, y(pi/2) = 0 and y\'(pi/2) = 5.

Solution

The characteristic equation:

r^2 + 6r + 10 = 0
r = -3 ± i <-- from the quadratic formula

The general solution for DEs with complex roots of the form a ± bi:

y(x) = Ce^(ax) sin(bx) + Ke^(ax) cos(bx)

Thus we have a = -3 and b = 1:

y(x) = Ce^(-3x) sin(x) + K e^(-3x) cos(x)