代码搜索:initial

找到约 10,000 项符合「initial」的源代码

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www.eeworm.com/read/475765/6777694

m ex5_1.m

% EX5_1.M Solve Dy=1/(t-1) using Symbolic Math Toolbox % for Example 5.1 and 5.2 y=dsolve('Dy=1/(t-1)') y1=dsolve('Dy=1/(t-1)','y(2)=-10') % Initial values t=2 y2=dsolve('Dy=1/(t-1)','y(2)=0')
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m ex5_9.m

% EX5_9.M Solve the system of equations % Dx1(t)= x1 + x2 % Dx2(t)=4x1 + x2 % Use dsolve with x=x1 and y=x2; Initial [1 1] xvec='Dx=x+y,Dy=4*x+y' % Print the results initc='x(0)=1,y(0)=1' [x
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m ex5_14.m

% EX5_14.M Verify Example 5.14 using dsolve % Dx1=x1+exp(t); Dx2=-x1 +3*x2; In result x=x1, y=x2. % (This requires Symbolic Math Toolbox) fprintf('Solution of system of Example 5.14 x=x1,y=x2 \n'
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m ex5_12.m

% EX5_12.M Verify Example 5.12 using dsolve % Dx1=x1+x2; Dx2=x2; In result x=x1, y=x2. % (This requires Symbolic Math Toolbox) fprintf('Solution of system of Example 5.12 x=x1,y=x2 \n') [x,y]=ds
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m ex6_7.m

% EX6_7.M Test Euler method on Dy(t)=y(t) t=[0 1]. % Compare exact value with Euler solution % y(n+1)=y(n)+hy(n)=(1+h)y(n); h=0.1 clear n=10; % Number of intervals y(1)=1;
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m main.m

initial; global G; G=1; adapting; % 适应度值 for G=2:200 selection; % 选择 crossover; % 交叉 mutation; % 变异 adapting; % 适应度值 end result;
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#include void main() { struct childrec/*定义结构体*/ { char initial; /* 姓名首字母 */ int age; /* 年龄 */ int grade; /* 考试成绩 */ }boy,girl;
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txt ho2_1.txt

ho2_1.txt >> ho2 Enter the title name TTL:Falling with simple air resistance Enter the number calculation steps desired NPTS: 200 Enter the run time TMAX: 20 Enter value of gravity G: 9.8
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m ho1.m

%ho1.m %Calculation of position, velocity, and acceleration for a harmonic %oscillator versus time. The equations of motion are used for small time intervals clear; %NPTS=100;TMAX=1.0;%example Max
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m least_action.m

%least_action.m - simulates Hamilton's Least Action principle for a %particle under the action of gravity. The trajectory is compared %with what is expected analytically clear; tol=1.e-7; %clear,