MATLAB 求解y′=siny 3x2

z对x的偏导=cosx+cos(x+y)=0时,cosx=-cos(x+y)=cos(pi-x-y),所以x=pi-x-y.同理z对y的偏导=0时,有y=pi-x-y.所以x=y=pi/3.此时z=3

这是个微分方程,需要有初始条件才能求解.假设初始条件y(0)=0.5,则代码如下：dy=@(x,y)-y+y^(2/3);ode45(dy,0:0.1:1,0.5)再问：如果我想输出y值的矩阵呢？再答

dsolve('Dy=(k^(t-1)*y-d)*y','t') ans =      

symsxy0>>y=dsolve('Dy=y+1/y','y(0)=y0','x')y=(-1+exp(2*x)*(1+y0^2))^(1/2)-(-1+exp(2*x)*(1+y0^2))^(1/

1.Warning:Explicitsolutioncouldnotbefound;implicitsolutionreturned.这表示该微分方程无显式解2.symsaI=int(100/(200

两边求导：cos(x+y)*(1+y')=cosx+cosy*y'y'=(cosx-cos(x+y))/(cos(x+y)-cosy)e^x+1=e^y*y'+y'y'=(e^x+1)/(e^y+1)

sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2（sinx）^2-1+2(siny）^2)=（si

∫e^ysinydy=-∫e^yd(cosy)=-[e^y*cosy-∫cosyd(e^y)]=∫cosy*e^ydy-e^ycosy=∫e^yd(siny)-e^ycosy=e^ysiny-∫sin

1+y'=cosy*y'y'=1/(cosy-1)dy/dx=1/(cosy-1)

Mathematica语句如下：ContourPlot[4y+x*y/Sin[y]==Pi^2,{x,0,1},{y,0,6}]就能给出函数图象了,其中y的取值范围你可以自己调整,直接将上述语句复制到

解arcsiny=x中y是自变量,x是因变量∴(arcsiny)'=x'=1/√(1-y^2)≠1例如y=sinx,(sinx)'=y'≠1

前四行保存为m文件命令行输入 [t,y]=ode45(@rigid,[0 100],[0.785 0]);plot(t,y(:,1)) 结果

两边对x求导有1-y'+y'cosy=0所以y'=1/(cosy-1)

x*e^y+siny=0e^y+x*e^y*y'+cosy*y'=0=>y'=-e^y/[xe^y+cosy]再问：你好！我数学太烂。。能不能补充一下完整的答案。。。再答：x*e^y+siny=0两边

y=siny+(sinx)^2-1.(sinx^2+cosx^2=1)=1/3-sinx+(sinx)^2-1=(sinx)^2-sinx-2/3=(sinx-1/2)^2-2/3-1/4=(sinx

x=0则siny=2y所以y=0对x求导2-2y'+cosy*y'=0y'=2/(2-cosy)所以x=0y'=2/(2-1)=2再问：siny=2y所以y=0？？再答：嗯f(y)=siny-2yf'

%y1=y'%y2=y%----------函数文件fun.mfunctiondy=fun(t,y)dy=zeros(2,1);dy=[-0.147*sin(y(2));...y(1)];%-----

functiondy=myfunc(x,y)a=b=c=dy=zeros(2,1)dy(1)=y(2)dy(2)=(b*dy(2)^2-c*sin(y(1)))/a这里的dy(1)是一阶导数,所以你的

两边对x求导有1-y'+y'cosy=0所以y'=1/(cosy-1)