Uppgift 4.2: 9 

Typesetting:-mrow(Typesetting:-mi( 

Typesetting:-mrow(Typesetting:-mi( 

(D(x))(t) = `+`(`*`(2, `*`(x(t))), `-`(`*`(3, `*`(y(t)))), `*`(2, `*`(sin(`+`(`*`(2, `*`(t))))))), (D(y))(t) = `+`(x(t), `-`(`*`(2, `*`(y(t)))), `-`(cos(`+`(`*`(2, `*`(t)))))) (2.1)
 

Anges inga begynnelsedata fÃ¥r man lösningen med "Maplekonstanter" 

Typesetting:-mrow(Typesetting:-mi( 

{x(t) = `+`(`*`(3, `*`(exp(t), `*`(_C2))), `*`(exp(`+`(`-`(t))), `*`(_C1)), `-`(`*`(`/`(4, 5), `*`(sin(`+`(`*`(2, `*`(t))))))), `-`(`*`(`/`(7, 5), `*`(cos(`+`(`*`(2, `*`(t)))))))), y(t) = `+`(`*`(exp(... (2.2)
 

Dessa kan bestämmas ur begynnelsedata pÃ¥ följande sätt 

Typesetting:-mrow(Typesetting:-mi( 

{2 = `+`(_C2, _C1, `-`(`/`(2, 5))), 1 = `+`(`*`(3, `*`(_C2)), _C1, `-`(`/`(7, 5)))} (2.3)
 

Typesetting:-mrow(Typesetting:-mi( 

{_C2 = 0, _C1 = `/`(12, 5)} (2.4)
 

Med funktionen "assign" kan ges sedan koefficienterna de beräknade värdena. 

Typesetting:-mrow(Typesetting:-mi( 

DÃ¥ blir lösningen 

Typesetting:-mrow(Typesetting:-mi( 

{x(t) = `+`(`*`(`/`(12, 5), `*`(exp(`+`(`-`(t))))), `-`(`*`(`/`(4, 5), `*`(sin(`+`(`*`(2, `*`(t))))))), `-`(`*`(`/`(7, 5), `*`(cos(`+`(`*`(2, `*`(t)))))))), y(t) = `+`(`*`(`/`(12, 5), `*`(exp(`+`(`-`(... (2.5)
 

Funktioner och plottning sker sedan som vanligt. 

Typesetting:-mrow(Typesetting:-mi( 

proc (t) options operator, arrow; `+`(`*`(`/`(12, 5), `*`(exp(`+`(`-`(t))))), `-`(`*`(`/`(2, 5), `*`(cos(`+`(`*`(2, `*`(t))))))), `-`(`*`(`/`(4, 5), `*`(sin(`+`(`*`(2, `*`(t)))))))) end proc (2.6)
 

Typesetting:-mrow(Typesetting:-mi( 

proc (t) options operator, arrow; `+`(`*`(`/`(12, 5), `*`(exp(`+`(`-`(t))))), `-`(`*`(`/`(4, 5), `*`(sin(`+`(`*`(2, `*`(t))))))), `-`(`*`(`/`(7, 5), `*`(cos(`+`(`*`(2, `*`(t)))))))) end proc (2.7)
 

Typesetting:-mrow(Typesetting:-mi( 

Plot_2d
 

Det kan ocksÃ¥ vara intressant att rita lösningarna som parameterkurvor t↦[f(t),g(t)] i ett xy-plan. Det gör man i stället sÃ¥ här: 

Typesetting:-mrow(Typesetting:-mi( 

Plot_2d