please solve this differential equation problem and select t

please solve this differential equation problem and select the right choice

Solution

y\" + y\' = 3e^3t

Characteristic equation is given by:r² + r =0 => r(r+1) = 0 => r=0,-1

The solution to homogeneous equation y\" + y\' =0 is of the form:

y_c = Ae^ot + Be^-t = A + Be^-t

Now guess of particular solution is y_p = Ce^3t

y_p\' = 3Ce^3t, y_p\" = 9Ce^3t

Plug in these values in the given differential equation, we get the value of C

3Ce^3t+9Ce^3t = 3e^3t

=> 12Ce^3t = 3e^3t

Comparing coefficients of e^3t on both sides:

12C = 3 => C = 1/4

Therefore the solution is given by: y(t) = y_c + y_p = A + Be^-t + 1/4e^3t

Now y(0) =1 => 1 = A + B + 1/4 => A + B = 3/4

Also y\'(0) = 2 => 2 = 0 - B + 3/4 => -B = 5/4 => B = -5/4 and so A = 3/4+5/4 = 8/4 = 2

The solution to the given initial value problem is given by:

y(t) = 2-5/4e^-t+1/4e^3t

Option (a) is correct