Solving first order Ordinary differential equations

u' = v, v' = (1/m1+1/m2)[1/2*y2*∂L/∂u−μ1m1g sgn(v)−k0(u−G)H(u −G) −cv −k1(u−X0)+ (m1+ m1) g], w' = x, x' = 1/m2[1/2*y2*∂L/∂u+ μ1m1g sgn(v)+ k0(u−G)H(u −G) + cv + k1(u−X0) −μ2(m1+ m2)g sgn(x)+(m1+ m1) g] , y' = z, z' = 1/L[Ω Pctr Vs cos(Ωt)−(R +2∂L/∂uv)z−(1/c+ (∂^2 L)/(∂u^2 )v2+ ∂L/∂u v')y]. Only u,v,w,x,y,z and t are variables in this case.

Hi guys, I need to solve these 3 equations above and use u that we find in equation 1 then plot it( the plot will look like a cosine graph).

I have been working on this for weeks, but still no luck. Please help me. thanks a lot. Feel free to contact me for more info.

Answer 1

[Walter Roberson]

There are 6 equations there, it appears to me.

You indicate that only u,v,w,x,y,z and t are variables are variables, but you show u' and v' and so on, which implies that u, v,w, x, y, and z are functions rather than variables.

Is y2 a constant or is it y^2 ?

Is ∂L/∂u the derivative of L with respect to u? If it is then that implies you have at least 7 functions (but only 6 equations)

Is sgn(v) the sign() function applied to v? If it is, then what value is sgn(0) ? Some places define sgn(0) as NaN and some define it as 0 and some define it as 1.

Your definition of x' starts with 1/m2[EXPRESSION] . Are we to take that as (1/m2)*(EXPRESSION), or are we to take that as 1/(m2*(EXPRESSION)) ?

As you have some variables that appear to be constructed of more than one character, we cannot tell whether cv is a single constant or if it is instead c*v .

Is 2∂L/∂uv 2*v*diff(L(t),u(t)) ? Or is it 2*diff(L(t),u(t)*v(t)) ?

Is ∂^2 L intended to be diff(L(t),t,t) ?

Please rewrite your expressions in full operator notation, using * for all multiplications, and converting all characters such as μ and ∂ to characters that are valid in MATLAB expressions, and converting all function calls to their MATLAB equivalents. Use () when necessary, but do not use [] or else we will interpret that portion as being the construction of a vector.

If ∂ is the differential operator, then since u is really u(t), a function rather than a variable, ∂L/∂u would have to be the differential of one function with respect to a second function. In such a case, you would not have ODE: you would have PDE, it seems to me.

[Davis Tee]

Hi Mr. Roberson, I sent you a email about the questions u asked.

Sorry for had provided unclear info, let me write a proper script and i send u it later.

Thanks have a great day ahead.