A mass weighing 32 lb. stretches a spring 32 ft. There is a damping constant of 2 `(lb*s)/(ft)`. There is a force of `4e^(t)` lb. Find the general solution to the differential equation: `m u''+gamma u'+ku=F(t)`; `u(0)=0`, `u'(0)=0`.







Now, from equation `W=mg` we find that `m=W/(g)=32/32=1(lb*s^2)/(ft)`.

Next, since `W=kL` then `k=W/L=32/32=1(lb)/(ft)`.

Differential equation has form `m u''+gammau'+ku=F(t)`.

Plugging found values we obtain that `u''+2u'+u=4e^t`, `u(0)=0`, `u'(0)=0`.

First step is to find solution to corresponding homogeneous differential equation: `u''+2u'+u=0`.

Characteristic equation is `r^2+2r+1=0` or `(r+1)^2=0`, so `r_1=-1` and `r_2=-1`.

Thus, `u_h=c_1e^-t+c_2te^-t`.

To find particular solution use method of undetermined coefficients. Assume that particular solution has form `u_p=Ae^t` then `u'_p=Ae^t` and `u''_p=Ae^t`.

Plugging these values into initial equation gives: `Ae^t+2Ae^t+Ae^t=4e^t` or `4Ae^t=4e^t`, so `A=1`.

Thus, particular solution is `u_p=e^t`.

General solution is, then, `u=u_h+u_p=c_1e^-t+c_2te^-t+e^t`.

Now, `u'=-c_1e^-t+c_2e^-t-c_2te^-t+e^t`.

Use initial conditions to eliminate constants:




From first equation `c_1=-1`, from second equation `c_2=c_1-1=-1-1=-2`.

Therefore, `u(t)=-e^-t-2te^-t+e^t`.