Lektion 11
> | with(DEtools); |
Opg 1
> | ode1:=diff(y(x),x)-y(x)=11/8*exp(-x/3); |
> | odeadvisor(ode1); |
> | ans1:=dsolve({ode1,y(0)=0}); |
Opg 2
> | ode2:=diff(y(x),x) + y(x)=2; |
> | odeadvisor(ode2); |
> | ans2:=dsolve({ode2,y(0)=0}); |
Opg 3
> | ode3:=x^2*diff(y(x),x)+x*y(x)=sin(x); |
> | odeadvisor(ode3); |
> | ans3:=dsolve({ode3,y(1)=0}); |
Si(x) staar for sinus intergralet int(sin(t)/t, t=0..x).
Opg 4
Lad Q(t) være indholdet af salt i beholderen. Så er
> | ode4:=diff(Q(t),t)=-5/1000*Q(t); |
> | dsolve({ode4,Q(0)=100}, Q(t)); |
> | solve(100*exp(-t/200)=10,t); |
> | evalf(%); |
Det tager ca 460 min før beholderen indeholder 10 g salt.
Opg 5
Antag at beholderen indeholder Q(t) gram salt til tiden t. Så er Q(0)=0 og Q(t) tilfredsstiller
> | ode5:=diff(Q(t),t) = 2 - 3*Q(t)/(60-t); |
> | odeadvisor(ode5); |
> | dsolve({ode5, Q(0)=0}); |
> | plot((-60+t)^3/3600 + 60 -t,t=0..60); |
> | solve(diff((-60+t)^3/3600 + 60 -t,t)=0,t); |
> | evalf(60-20*3^(1/2)); |
Opg 6
> | assume(k>0); |
> | ode6:=diff(x(t),t,t)=-k*diff(x(t),t); |
> | ans6a:=dsolve({ode6,x(0)=x_0,D(x)(0)=v_0}); |
> | simplify(%); |
> | limit(%, t=infinity); |
Opg 7
> | ode7:=diff(Q(t),t)=-ln(2)/12*Q(t); |
a) Halveringstiden er 12 timer.