Differential Equation Question

Question

Determine the specific solution of the following initial-value problems. y^''+ 4y^' + 20y = 0, y(pi/2)=0 and y^'(pi/2)=2

Answer 1

Characteristic equation is

k2+4k+20=0.

k1=-2-4i, k2=-2+4i

 Therefore the general  solution is

y(x)=A e^-2xcos(4x) + B e^-2xsin(4x)

Take derivative, find A and B as in your previous question and get a specific solution (A=0, B=(1/2)e^Pi)

y(x)=(1/2)e^Pi-2xsin(4x)