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chaofan1231

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[求助] matlab解代数方程的问题

哪位大侠帮忙看下这是怎么回事啊？谢谢啊
y=2;
>> subs(solve('y=exp((1/0.0000005461-1/0.000000700)*0.014388/x+ln((x-993.74731)/2869.03078)-5*ln(700/546.1))','x'))
Warning: Explicit solution could not be found.
我把y=2直接代进去的话solve('2=exp((1/0.0000005461-1/0.000000700)*0.014388/x+ln((x-993.74731)/2869.03078)-5*ln(700/546.1))','x')
ans =matrix([[14197.808632879686391435779145547]])，但是我在origin中将这个方程画成曲线的话，y=2时x的两个值分别在1100和1900附近，这是怎么回事啊？

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1楼 2012-05-14 08:29:15

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【答案】应助回帖

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chaofan1231: 金币+5, ★★★★★最佳答案 2012-05-15 14:53:43
chaofan1231: 回帖置顶 2012-05-15 14:54:44

>> evalin(symengine, 'numeric::solve(2=exp((1/0.0000005461-1/0.000000700)*0.014388/x+ln((x-993.74731)/2869.03078)-5*ln(700/546.1)), x = 1000..2000)')

ans =

1092.7957902606343947864233870012

>> evalin(symengine, 'numeric::solve(2=exp((1/0.0000005461-1/0.000000700)*0.014388/x+ln((x-993.74731)/2869.03078)-5*ln(700/546.1)), x = 1500..2000)')

ans =

1826.2296709384982930177909403237

Solve the following equation:

syms x;
solve('sin(x) = x^2 - 1')

The symbolic solver cannot find an exact symbolic solution for this equation, and therefore, it calls the numeric solver. Because the equation is not polynomial, an attempt to find all possible solutions can take a long time. The numeric solver does not try to find all numeric solutions for this equation. Instead, it returns only the first solution that it finds:

ans =
-0.63673265080528201088799090383828

Plotting the left and the right sides of the equation in one graph shows that the equation also has a positive solution:

ezplot(sin(x), -2, 2);
hold on;
ezplot(x^2 - 1, -2, 2)
hold off

You can find this solution by calling the MuPAD numeric solver directly and specifying the interval where this solution can be found. To call MuPAD commands from the MATLAB Command Window, use the evalin or feval function:

evalin(symengine, 'numeric::solve(sin(x) = x^2 - 1, x = 0..2)')

ans =
1.4096240040025962492355939705895