## Question

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.

## Answer

F(t)=4e^tlb.

g=32(ft)/s^2.

W=32lb.

L=32ft.

gamma=2(lb*s)/(ft).

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:

{(u(0)=0=c_1e^-0+c_2*0*e^-0+e^0),(u'(0)=0=-c_1e^-0+c_2e^-0-c_2*0*e^-0+e^0):}

Or

{(c_1=-1),(-c_1+c_2=-1):}.

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.