南邮Matlab实验报告二

实验二MATLAB的符号运算符号运算的特点：一、运算以推理解析的方式进行，因此不受计算误差积累问题困扰；二、符号计算，或给出完全正确的封闭解，或给出任意精度的数值解（当封闭解不存在时）；三、符号计算指令的调用比较简单；四、计算所需时间较长。在MATLAB中，符号计算虽以数值计算的补充身份出现，但实际符号计算的指令使用、计算符操作、计算结果可视化、程序编制以及在线帮助系统都是十分完整、便捷的。实验目的熟悉符号对象和符号表达式；掌握符号表达式和符号函数的操作；学习利用符号代数方程的求解。实验基本知识基本运算指令；掌握符号计算的应用。上级练习第一题：（1）>>symst>>y=abs(sin(t))y=abs(sin(t))>>d=diff(y)d=sign(sin(t))*cos(t)>>do_=limit(d,t,0,'left')do_=-1>>dpi_2=limit(d,t,pi/2)dpi_2=0>>y=abs(sin(t))y=abs(sin(t))第二题：>>symsxy>>f=x^2+y^2;>>r=int(int(f,y,1,x^2),x,1,2)r=1006/105第三题：>>symsxy>>f=x^2+y^2;>>dfdvn=diff(f,y,2)dfdvn=2>>intf=int(f,x,-1,1)intf=2*y^2+2/3第四题：>>symsx>>f=3*x^2+4*x-6;>>S=solve(f,x)S=-22^(1/2)/3-2/322^(1/2)/3-2/3第五题：>>clearall>>y=dsolve('D3y+3*Dy+2*y-6=0','x')y=C2/exp(x*(1/(2^(1/2)-1)^(1/3)-(2^(1/2)-1)^(1/3)))+C3*cos((3^(1/2)*x*(1/(2^(1/2)-1)^(1/3)+(2^(1/2)-1)^(1/3)))/2)*exp(x*(1/(2*(2^(1/2)-1)^(1/3))-(2^(1/2)-1)^(1/3)/2))+C4*sin((3^(1/2)*x*(1/(2^(1/2)-1)^(1/3)+(2^(1/2)-1)^(1/3)))/2)*exp(x*(1/(2*(2^(1/2)-1)^(1/3))-(2^(1/2)-1)^(1/3)/2))+3>>y=dsolve('D3y+3*Dy+2*y-6=0','D2y(0)=1,Dy(0)=2,y(0)=3','x')y=(2*(2^(1/2)-1)^(1/3)-2/(2^(1/2)-1)^(1/3)+1)/(exp(x*(1/(2^(1/2)-1)^(1/3)-(2^(1/2)-1)^(1/3)))*(3/(2^(1/2)-1)^(2/3)+3*(2^(1/2)-1)^(2/3)-3))-(cos((3^(1/2)*x*(1/(2^(1/2)-1)^(1/3)+(2^(1/2)-1)^(1/3)))/2)*exp(x*(1/(2*(2^(1/2)-1)^(1/3))-(2^(1/2)-1)^(1/3)/2))*(2*(2^(1/2)-1)^(1/3)-2/(2^(1/2)-1)^(1/3)+1))/(3/(2^(1/2)-1)^(2/3)+3*(2^(1/2)-1)^(2/3)-3)+(sin((3^(1/2)*x*(1/(2^(1/2)-1)^(1/3)+(2^(1/2)-1)^(1/3)))/2)*exp(x*(1/(2*(2^(1/2)-1)^(1/3))-(2^(1/2)-1)^(1/3)/2))*(1/(2^(1/2)-1)^(1/3)-(2^(1/2)-1)^(1/3)+2/(2^(1/2)-1)^(2/3)+2*(2^(1/2)-1)^(2/3)))/(3^(1/2)*(2^(1/2)-1)+3^(1/2)/(2^(1/2)-1))+3第六题：>>symsxy>>S=solve('x^2+y^2=1','x*y=2','y','x')S=x:[4x1sym]y:[4x1sym]>>disp('S.x'),disp(S.x),disp('S.y'),disp(S.y)S.x(1/2+(15^(1/2)*i)/2)^(1/2)/2-(1/2+(15^(1/2)*i)/2)^(3/2)/2-(1/2+(15^(1/2)*i)/2)^(1/2)/2+(1/2+(15^(1/2)*i)/2)^(3/2)/2(1/2-(15^(1/2)*i)/2)^(1/2)/2-(1/2-(15^(1/2)*i)/2)^(3/2)/2-(1/2-(15^(1/2)*i)/2)^(1/2)/2+(1/2-(15^(1/2)*i)/2)^(3/2)/2S.y(