> # Teht.5
> with(DEtools): with(linalg):
Warning, new definition for norm
Warning, new definition for trace
> yhtalo5:=vector([x5(t) + 6*y5(t) + 11 , 5*x5(t) + 2*y5(t) - 1]);

       yhtalo5 := [x5(t) + 6 y5(t) + 11, 5 x5(t) + 2 y5(t) - 1]

> diffy5:={D(x5)(t)=yhtalo5[1],D(y5)(t)=yhtalo5[2]};

diffy5 := {

    D(x5)(t) = x5(t) + 6 y5(t) + 11, D(y5)(t) = 5 x5(t) + 2 y5(t) - 1

    }

> type(diffy5,set);

                                 true

> dsolve(diffy5,{x5(t),y5(t)});

{y5(t) = 5/11 _C1 exp(7 t) - 5/11 _C1 exp(-4 t) + 5/11 _C2 exp(-4 t)

     + 6/11 _C2 exp(7 t) - 2, x5(t) = 6/11 _C1 exp(-4 t)

     + 5/11 _C1 exp(7 t) + 6/11 _C2 exp(7 t) - 6/11 _C2 exp(-4 t) + 1

    }

> DEplot(diffy5,[x5(t),y5(t)],t=0..5,[[x5(1)=-1,y5(1)=1]],x5=-5..5,y5=-5..5,arrows=MEDIUM);

> # Kuvassa näkyy satulapiste
> 
> # Teht.6
> yhtalo6:=vector([x6(t)-(x6(t))^2+2*y6(t)+3,-x6(t)+2*y6(t)]);

                [             2                                ]
     yhtalo6 := [x6(t) - x6(t)  + 2 y6(t) + 3, -x6(t) + 2 y6(t)]

> diffy6:={D(x6)(t)=yhtalo6[1],D(y6)(t)=yhtalo6[2]};

                                   2
diffy6 := {D(x6)(t) = x6(t) - x6(t)  + 2 y6(t) + 3,

    D(y6)(t) = -x6(t) + 2 y6(t)}

> type(diffy6,set);

                                 true

> dsolve(diffy6,{x6(t),y6(t)});
> # Ei onnistu suoraan näin
> # Käytetään sen sijaan hyväksi alkuviikolla tehtyjä linearisointeja
> A6a:=matrix(2,2,[3,2,-1,2]);

                                  [ 3    2]
                           A6a := [       ]
                                  [-1    2]

> yhtalo6a:=evalm(A6a &* vector([v1(t),v2(t)]));

          yhtalo6a := [3 v1(t) + 2 v2(t), -v1(t) + 2 v2(t)]

> diffy6a:={D(v1)(t)=yhtalo6a[1],D(v2)(t)=yhtalo6a[2]};

diffy6a :=

    {D(v2)(t) = -v1(t) + 2 v2(t), D(v1)(t) = 3 v1(t) + 2 v2(t)}

> type(diffy6a,set);

                                 true

> dsolve(diffy6a,{v1(t),v2(t)});

              1/2                                               1/2
{v1(t) = 4/7 7    exp(5/2 t) %1 _C1 + _C2 exp(5/2 t) cos(1/2 t 7   )

            1/2
     + 1/7 7    exp(5/2 t) %1 _C2, v2(t) =

                              1/2         1/2
    _C1 exp(5/2 t) cos(1/2 t 7   ) - 1/7 7    exp(5/2 t) %1 _C1

            1/2
     - 2/7 7    exp(5/2 t) %1 _C2}

                 1/2
%1 := sin(1/2 t 7   )

> DEplot(diffy6a,[v1(t),v2(t)],t=0..3,[[v1(1)=1,v2(1)=-1]],v1=-5..5,v2=-5..5,arrows=MEDIUM);

> # Epästabiili fokus
> A6b:=matrix(2,2,[-5,2,-1,2]);

                                  [-5    2]
                           A6b := [       ]
                                  [-1    2]

> yhtalo6b:=evalm(A6b &* vector([w1(t),w2(t)]));

          yhtalo6b := [-5 w1(t) + 2 w2(t), -w1(t) + 2 w2(t)]

> diffy6b:={D(w1)(t)=yhtalo6b[1],D(w2)(t)=yhtalo6b[2]};

diffy6b :=

    {D(w1)(t) = -5 w1(t) + 2 w2(t), D(w2)(t) = -w1(t) + 2 w2(t)}

> type(diffy6b,set);

                                 true

> dsolve(diffy6b,{w1(t),w2(t)});

                                 1/2                 1/2
{w1(t) = 1/2 _C1 %2 + 7/82 _C1 41    %1 - 7/82 _C1 41    %2

                               1/2                 1/2
     + 1/2 _C1 %1 - 2/41 _C2 41    %1 + 2/41 _C2 41    %2, w2(t) =

               1/2                 1/2
    1/41 _C1 41    %1 - 1/41 _C1 41    %2 + 1/2 _C2 %2

                  1/2                 1/2
     - 7/82 _C2 41    %1 + 7/82 _C2 41    %2 + 1/2 _C2 %1}

                       1/2
%1 := exp(- 1/2 (3 + 41   ) t)

                      1/2
%2 := exp(1/2 (-3 + 41   ) t)

> DEplot(diffy6b,[w1(t),w2(t)],t=0..3,[[w1(1)=1,w2(1)=1]],w1=-5..5,w2=-5..5,arrows=MEDIUM);

> # Satula
>