Example 3.4.11

Check the solution of the initial value problem:

sol=DSolve[{9y''[t]-12y'[t]+4y[t]==0,y[0]==2,y'[0]==-1},y[t],t]

y[t_]=y[t]/.Last[sol]

y'[t]

Simplify[%]

Plot[y[t],{t,0,3},AxesLabel->{"t","y(t)"}]

Example 3.4.16

Clear[y]

Check the solution of the initial value problem:

sol=DSolve[{y''[t]-y'[t]+y[t]/4==0,y[0]==2,y'[0]==b},y[t],t]

y[t_]=y[t]/.Last[sol]

Simplify[y'[t]]

sollist=Table[y[t]/.b->Range[-2,2,1/2]]

plot1=Plot[Evaluate[sollist],{t,0,80},AxesLabel->{"t","y(t)"}]

plot2=Plot[y[t]/.b->0.97,{t,0,80},AxesLabel->{"t","y(t)"},PlotStyle->{Red}]

plot3=Plot[y[t]/.b->1,{t,0,80},AxesLabel->{"t","y(t)"},PlotStyle->{Blue}]

Show[plot1,plot2,plot3,PlotRange->{{0,80},{-10^(13),10^(13)}},ImageSize->{500,Automatic}]

Notice how the red curve, which is for b=0.97, eventually becomes negative. The blue curve is the solution that separates the two behaviours (eventually negative versus eventually positive) and has b=1. It took a bit of experimenting to come up with curves that illustrates this fact to my liking.

Manipulate[Plot[y[t]/.b->b0,{t,0,80},AxesLabel->{"t","y(t)"},PlotRange->{{0,80},{-(10^13),(10^13)}}],{b0,0.95,1.05}]