Find the solution of the given initial value problem y 6y

Find the solution of the given initial value problem. y\" + 6y\' + 10y = 0, y (pi/2) = 0, y\'(pi/2) = 8 y(t) =

y\" +6y\' +10y =0

characteristic equation r^2 +6r +10 =0

r = (-6 +/- sqrt(36 -40)/2 = (-3 +/- i)

Since roots are complex , general solution : y = c1e^(-3x)*[cosx] + c2e^(-3x)*[sinx]

solve for constants c1 and c2 : :y (pi/2) =0

0 = c1(0) +c2e^(-3pi/2) c2 =0

So, y = c1e^(-3x)*[cosx]

y\' = - c1e^(-3x)sinx -3c1e^(-3x)cosx

y\'(pi/2) =8 ; 8 = -c1e^-3pi/2*1 -3*0

8 = -c1e^(-3pi/2)

c1 = -8e^(3pi/2)

y = -8e^(-3x+3pi/2)*[cosx]

$Find the solution of the given initial value problem. y\$