y^3 = ( 4 * ( x ^ 2 ) ) + 28

Question

salma omarii 2014-10-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/157921-y-3-4-x-2-28

评论： salma omarii 2014-10-8

I have solved the following by hand but am having difficulties implementing the code. If anyone is able to assist me I would great appreciate it.

I would like to use Newton-Raphson to solve:

y^3 = ( 4 * ( x ^ 2 ) ) + 28

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Answer 1

Sean de Wolski 2014-10-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/157921-y-3-4-x-2-28#answer_154464

在 MATLAB Online 中打开

If you have symbolic math toolbox:

solve('y^3 = ( 4 * ( x ^ 2 ) ) + 28','x')
ans =
(y^3 - 28)^(1/2)/2
-(y^3 - 28)^(1/2)/2
solve('y^3 = ( 4 * ( x ^ 2 ) ) + 28','y')
ans =
                      4^(1/3)*(x^2 + 7)^(1/3)
4^(1/3)*((3^(1/2)*i)/2 - 1/2)*(x^2 + 7)^(1/3)
-4^(1/3)*((3^(1/2)*i)/2 + 1/2)*(x^2 + 7)^(1/3)

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salma omarii 2014-10-8

i don't need to solve it like this i wanna find the roots of this nonlinear equation by newton raphson method so i have to write a code on matlab if you can help me please :)

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