Solución
El método Runge-Kutta establece que yn+1=yn+6h(k1+2k2+2k3+k4), donde tn+1=tn+h, k1=f(tn,yn), k2=f(tn+2h,yn+2hk1), k3=f(tn+2h,yn+2hk2), y k4=f(tn+h,yn+hk3).
Tenemos que h=52, t0=0, y0=1, y f(t,y)=sin(ty).
Paso 1
t1=t0+h=0+52=52
k1=f(t0,y0)=f(0,1)=0
k2=f(t0+2h,y0+2hk1)=f(0+252,1+2(52)⋅(0))=f(51,1)=0.198669330795061
k3=f(t0+2h,y0+2hk2)=f(0+252,1+2(52)⋅(0.198669330795061))=f(51,1.03973386615901)=0.206451342596583
k4=f(t0+h,y0+hk3)=f(0+52,1+(52)⋅(0.206451342596583))=f(52,1.08258053703863)=0.419625061196877
y1=y(t1)=y(52)=y0+6h(k1+2k2+2k3+k4)=1+652(0+2⋅0.198669330795061+2⋅0.206451342596583+0.419625061196877)=1.08199109386534
Paso 2
t2=t1+h=52+52=54
k1=f(t1,y1)=f(52,1.08199109386534)=0.419411035089935
k2=f(t1+2h,y1+2hk1)=f(52+252,1.08199109386534+2(52)⋅(0.419411035089935))=f(53,1.16587330088333)=0.643853534490712
k3=f(t1+2h,y1+2hk2)=f(52+252,1.08199109386534+2(52)⋅(0.643853534490712))=f(53,1.21076180076349)=0.664225362212255
k4=f(t1+h,y1+hk3)=f(52+52,1.08199109386534+(52)⋅(0.664225362212255))=f(54,1.34768123875025)=0.881081971595253
y2=y(t2)=y(54)=y1+6h(k1+2k2+2k3+k4)=1.08199109386534+652(0.419411035089935+2⋅0.643853534490712+2⋅0.664225362212255+0.881081971595253)=1.34310114720475
Paso 3
t3=t2+h=54+52=56
k1=f(t2,y2)=f(54,1.34310114720475)=0.879343087787042
k2=f(t2+2h,y2+2hk1)=f(54+252,1.34310114720475+2(52)⋅(0.879343087787042))=f(1,1.51896976476216)=0.998657304313516
k3=f(t2+2h,y2+2hk2)=f(54+252,1.34310114720475+2(52)⋅(0.998657304313516))=f(1,1.54283260806746)=0.999609040694986
k4=f(t2+h,y2+hk3)=f(54+52,1.34310114720475+(52)⋅(0.999609040694986))=f(56,1.74294476348275)=0.867452549636552
y3=y(t3)=y(56)=y2+6h(k1+2k2+2k3+k4)=1.34310114720475+652(0.879343087787042+2⋅0.998657304313516+2⋅0.999609040694986+0.867452549636552)=1.72598970236746
Paso 4
t4=t3+h=56+52=58
k1=f(t3,y3)=f(56,1.72598970236746)=0.877394887797677
k2=f(t3+2h,y3+2hk1)=f(56+252,1.72598970236746+2(52)⋅(0.877394887797677))=f(57,1.90146867992699)=0.461368005308125
k3=f(t3+2h,y3+2hk2)=f(56+252,1.72598970236746+2(52)⋅(0.461368005308125))=f(57,1.81826330342908)=0.561356508370458
k4=f(t3+h,y3+hk3)=f(56+52,1.72598970236746+(52)⋅(0.561356508370458))=f(58,1.95053230571564)=0.020739477392444
y4=y(t4)=y(58)=y3+6h(k1+2k2+2k3+k4)=1.72598970236746+652(0.877394887797677+2⋅0.461368005308125+2⋅0.561356508370458+0.020739477392444)=1.92222859520394
Paso 5
t5=t4+h=58+52=2
k1=f(t4,y4)=f(58,1.92222859520394)=0.06597893710495
k2=f(t4+2h,y4+2hk1)=f(58+252,1.92222859520394+2(52)⋅(0.06597893710495))=f(59,1.93542438262493)=−0.33553324651362
k3=f(t4+2h,y4+2hk2)=f(58+252,1.92222859520394+2(52)⋅(−0.33553324651362))=f(59,1.85512194590122)=−0.196342927593455
k4=f(t4+h,y4+hk3)=f(58+52,1.92222859520394+(52)⋅(−0.196342927593455))=f(2,1.84369142416656)=−0.519093645672128
y5=y(t5)=y(2)=y4+6h(k1+2k2+2k3+k4)=1.92222859520394+652(0.06597893710495+2(−0.33553324651362)+2(−0.196342927593455)−0.519093645672128)=1.82110412475186