Characteristic equation of differential one: #r^3-3r^2+2r=0#
#r*(r-1)*(r-2)=0#
Its roots are #r_1=2#, #r_2=1# and #r_3=0#
So, homogenic part of differential equation is;
#z_h=c_1*e^(2t)+c_2*e^t+c_3#
According to variation of parameters method, particular solution of it;
#z_p=e^(2t)*u_1+e^t*u_2+u_3#
Also, according to this method, #u_1#, #u_2# and #u_3# must prove these equations,
#e^(2t)*(u_1)'+e^t*(u_2)'+(u_3)'=0#
#2e^(2t)*(u_1)'+e^t*(u_2)'=0# and
#4e^(2t)*(u_1)'+e^t*(u_2)'=e^(3t)/(e^t+1)#
After solving system of them,
#(u_1)'=1/2*e^t/(e^t+1)#, #(u_2)'=-e^(2t)/(e^t+1)# and #(u_3)'=1/2*e^(3t)/(e^t+1)#
Consequently,
#u_1=1/2int e^t/(e^t+1)*dt=1/2Ln(e^t+1)#
#u_2=-int e^(2t)/(e^t+1)*dt#
=#-int e^t/(e^t+1)*e^t*dt#
=#-int (e^t+1-1)/(e^t+1)*e^t*dt#
After using #s=e^t+1# and #ds=e^t*dt# transforms, this integral became
#u_2=-int (s-1)/s*ds#
=#int (ds)/s-int ds#
=#Lns-s#
=#Ln(e^t+1)-e^t-1#
#u_3=1/2int e^(3t)/(e^t+1)*dt#
=#1/2int e^(2t)/(e^t+1)*e^t*dt#
=#1/2int (e^t+1-1)^2/(e^t+1)*e^t*dt#
After using #p=e^t+1# and #dp=e^t*dt# transforms, this integral became
#u_3=1/2int (p-1)^2/p*dp#
=#1/2int p*dp-int dp+1/2int (dp)/p#
=#1/4*p^2-p+1/2Lnp#
=#1/4*(e^t+1)^2-(e^t+1)+1/2Ln(e^t+1)#
=#1/4*e^(2t)-1/2*e^t+1/2Ln(e^t+1)-3/4#
Hence,
#z_p=e^(2t)*1/2Ln(e^t+1)+e^t[Ln(e^t+1)-e^t-1]+1/4*e^(2t)-1/2*e^t+1/2Ln(e^t+1)-3/4#
=#1/2*(e^(2t)+2e^t+1)*Ln(e^t+1)#-#3/4e^(2t)-3/2e^t-3/4#
Thus,
#z=z_h+z_p#
=#c_1*e^(2t)+c_2*e^t+c_3#+#1/2*(e^(2t)+2e^t+1)*Ln(e^t+1)#-#3/4e^(2t)-3/2e^t-3/4#
=#Ae^(2t)+Be^t+C#+#1/2*(e^(2t)+2e^t+1)*Ln(e^t+1)#
Note: #A=c_1-3/4#,#B=c_2-3/2# and #C=c_3-3/4#
