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# Hello, how to solve this equation E*I*k^4-m*​v^2*k^2+2*​m*v*w*k+(m​+M)w^2=0 numerically where w is variable,

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shoaib Shoaib on 10 Dec 2019
Commented: shoaib Shoaib on 12 Dec 2019
can we solve this for k numerically, sorry this is fourth order equation not two order
Thanks

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### Accepted Answer

Star Strider on 10 Dec 2019
Edited: Star Strider on 10 Dec 2019
Supply all the scalar parameters, then:
Eqn = @(w) E*I*k^2-m*v^2*k^2+2*m*v*w*k+(m+M)w^2;
w0 = 42;
[w,fval] = fsolve(Eqn, w0)
Experiment with the correct value of ‘w0’ to get the correct result.
EDIT (Dec 10 2019 at 13:18)
The Symbolic Math Toolbox produces:
w = (k*(- E*I*k^2*m - E*I*M*k^2 + 2*m^2*v^2 + M*m*v^2)^(1/2) - k*m*v)/(M + m)
or to calculate both roots:
w = [(k*(- E*I*k^2*m - E*I*M*k^2 + 2*m^2*v^2 + M*m*v^2)^(1/2) - k*m*v)/(M + m)
-(k*(- E*I*k^2*m - E*I*M*k^2 + 2*m^2*v^2 + M*m*v^2)^(1/2) - k*m*v)/(M + m)]

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Star Strider on 12 Dec 2019
As always, my pleasure!
Walter Roberson on 12 Dec 2019
syms E I k m v w M
Eqn = E*I*k^4-m*v^2*k^2+2*m*v*w*k+(m+M)*w^2
sol_exact = solve(Eqn, k, 'MaxDegree', 4); %valid for symbolic variables, gives LONG exact solutions
sol_numeric = vpasolve(Eqn, k); %valid only if numeric values are known for everything except k, gives numeric solutions
shoaib Shoaib on 12 Dec 2019
Thanks Walter Roberson

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