Example 3.5.3

Check the solution of the differential equation:

sol=DSolve[{y''[t]-2y'[t]-3y[t]==-3t Exp[-t]},y[t],t]

Notice that this does not look like the solution worked out by hand. They are actually the same, as the following shows:

y[t]=(3/64) (e^(-t)) (1+4 t+8 (t^2))+(e^(-t)) C[1]+(e^(3 t)) C[2]
= (3/64) (e^(-t)) + (12/64) (e^(-t)) t+(24/64) (e^(-t)) (t^2)+(e^(-t)) C[1]+(e^(3 t)) C[2]
= (e^(3 t)) C[2]+((3/64)  + C[1])(e^(-t))+ (12/64) (e^(-t)) t+(24/64) (e^(-t)) (t^2)
= (e^(3 t)) C[2]+ k[1](e^(-t))+ (3/16) (e^(-t)) t+(3/8) (e^(-t)) (t^2)

where the constant k[1]=C[1]+3/64.

Example 3.5.7

Check the solution of the differential equation:

sol=DSolve[{2y''[t]+3y'[t]+y[t]==(t^2)+3Sin[t]},y[t],t]

Example 3.5.17

Check the solution of the initial value problem

sol=DSolve[{y''[t] +4y[t]==3Sin[2t],y[0]== 2,y'[0]== -1},y[t],t]

Simplify[%]

Plot[(2-((3 t)/4)) Cos[2 t]-(1/8) Sin[2 t],{t,0,20}]

For large value of t the -3/4*t cos 2t term dominates the solution.

Plot[{(2-((3 t)/4)) Cos[2 t]-(1/8) Sin[2 t],-((3 t)/4) Cos[2 t]},{t,0,30},PlotStyle->{Red,Blue},PlotPoints->100]