請問我該如何用RK法寫??
這是我已經降階好的非線性2階方程式，我已經降成一皆了y'(1)=y(2);
y'(2)=(-6.15*10.^6*(y(3)-y(1)))/13.2;
y'(3)=y(4);
y'(4)=((8.1*10.^7*(y(5)-y(3)))-(6.15*10.^6*(y(3)-y(1))))/(1.3);
y'(5)=y(6);
y'(6)=((6.03*10.^6*(y(5)-y(3)))-(1.545*10.^4*0.176*(0.176*y(7)-0.176*y(5))))/0.76;
y'(7)=y(8);
y'(8)=((1.545*10.^4*0.176*(0.176*y(7)-0.176*y(5)))+(1.545*10.^4*0.176*(0.176*y(9)-0.176*y(7))))/0.76;
y'(9)=y(10);
y'(10)=((1.545*10.^4*0.176*(0.176*y(9)-0.176*y(7)))-(8.1*10.^7*(y(11)-y(9))))/0.76;
y'(11)=y(12);
y'(12)=((8.1*10.^7*(y(11)-y(9)))+(1.618*10.^4*0.150*(0.150*y(13)-0.150*y(11))))/0.56;
y'(13)=y(14);
y'(14)=((1.618*10.^4*0.150*(0.150*y(12)-0.150*y(11)))-(6.03*10.^6*(y(15)-y(13))))/0.56;
y'(15)=y(16);
y'(16)=(6.03*10.^6*(y(15)-y(13)))/3.66;
y'(17)=y(18);
y'(18)=(-1.159*10.^4*0.156*(0.084*y(19)-0.156*y(17)))/0.25;
y'(19)=y(20);
y'(20)=(-2.01*10.^6*(y(21)-y(19)))/0.03;
y'(21)=y(22);
y'(22)=(2.01*10.^6*(y(21)-y(19)))/1.5;
小女子寫不出來 煩請各位大俠幫幫忙
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