facit11.mws

Lektion 11

>    with(DEtools);

Opg 1

>    ode1:=diff(y(x),x)-y(x)=11/8*exp(-x/3);

ode1 := diff(y(x),x)-y(x) = 11/8*exp(-1/3*x)

>    odeadvisor(ode1);

[[_linear, `class A`]]

>    ans1:=dsolve({ode1,y(0)=0});

ans1 := y(x) = (-33/32*exp(-4/3*x)+33/32)*exp(x)

Opg 2

>    ode2:=diff(y(x),x) + y(x)=2;

ode2 := diff(y(x),x)+y(x) = 2

>    odeadvisor(ode2);

[_quadrature]

>    ans2:=dsolve({ode2,y(0)=0});

ans2 := y(x) = 2-2*exp(-x)

Opg 3

>    ode3:=x^2*diff(y(x),x)+x*y(x)=sin(x);

ode3 := x^2*diff(y(x),x)+x*y(x) = sin(x)

>    odeadvisor(ode3);

[_linear]

>    ans3:=dsolve({ode3,y(1)=0});

ans3 := y(x) = (Si(x)-Si(1))/x

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);

ode4 := diff(Q(t),t) = -1/200*Q(t)

>    dsolve({ode4,Q(0)=100}, Q(t));

Q(t) = 100*exp(-1/200*t)

>    solve(100*exp(-t/200)=10,t);

200*ln(10)

>    evalf(%);

460.5170186

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);

ode5 := diff(Q(t),t) = 2-3*Q(t)/(60-t)

>    odeadvisor(ode5);

[_linear]

>    dsolve({ode5, Q(0)=0});

Q(t) = 1/3600*(-60+t)^3+60-t

>    plot((-60+t)^3/3600 + 60 -t,t=0..60);

[Maple Plot]

>    solve(diff((-60+t)^3/3600 + 60 -t,t)=0,t);

60+20*3^(1/2), 60-20*3^(1/2)

>    evalf(60-20*3^(1/2));

25.35898384

Opg 6

>    assume(k>0);

>    ode6:=diff(x(t),t,t)=-k*diff(x(t),t);

ode6 := diff(x(t),`$`(t,2)) = -k*diff(x(t),t)

>    ans6a:=dsolve({ode6,x(0)=x_0,D(x)(0)=v_0});

ans6a := x(t) = 1/k*(k*x_0+v_0)-1/k*v_0*exp(-k*t)

>    simplify(%);

x(t) = -(-k*x_0-v_0+v_0*exp(-k*t))/k

>    limit(%, t=infinity);

limit(x(t),t = infinity) = 1/k*(k*x_0+v_0)

Opg 7

>    ode7:=diff(Q(t),t)=-ln(2)/12*Q(t);

ode7 := diff(Q(t),t) = -1/12*ln(2)*Q(t)

a) Halveringstiden er 12 timer.