Show all work for full credit Use the Laplace transform and

Show all work for full credit !

Use the Laplace transform and particle fractions (show all steps) to solve the differential equation with initial conditions: y\" + 2y\' + 5y = 15 t(0) = 2, y\'(0) = 1

Solution

s^2Y(s)-sy(0)-y\'(0)+2sY(s)-2y(0)+5Y(s)=15/s [s^2+2s+5]Y(s)-2s-1-4=15/s [s^2+2s+5]Y(s)=15/s + 2s+5 So Y(s)=15/s(s^2+2s+5) + 2s+5/(s^2+2s+5) Take inverse laplace