Differential equation : h d^2h/dx^2 + (dh/dx)^2 – dh/dx * tan(ax) + c – h * sec^2(ax) * a = 0
Boundary conditions: h(x=0)=h0 and h(x=L)=h0
Dependent variable: h
Independent variable: x
constants: a,c,L,h0
Method to be used : RK 4th order

please help me

let y = h and z = dh/dx=dy/dx,dz/dx = a * sec^2(ax) + (1/h) * (z * tan(ax) – z^2 – c) confused how to give boundary conditionsDifferential equation : h d^2h/dx^2 + (dh/dx)^2 – dh/dx * tan(ax) + c – h * sec^2(ax) * a = 0
Boundary conditions: h(x=0)=h0 and h(x=L)=h0
Dependent variable: h
Independent variable: x
constants: a,c,L,h0
Method to be used : RK 4th order

please help me

let y = h and z = dh/dx=dy/dx,dz/dx = a * sec^2(ax) + (1/h) * (z * tan(ax) – z^2 – c) confused how to give boundary conditions Differential equation : h d^2h/dx^2 + (dh/dx)^2 – dh/dx * tan(ax) + c – h * sec^2(ax) * a = 0
Boundary conditions: h(x=0)=h0 and h(x=L)=h0
Dependent variable: h
Independent variable: x
constants: a,c,L,h0
Method to be used : RK 4th order

please help me

let y = h and z = dh/dx=dy/dx,dz/dx = a * sec^2(ax) + (1/h) * (z * tan(ax) – z^2 – c) confused how to give boundary conditions rk 4th order method, second order nonlinear differential equation, boundary value problem MATLAB Answers — New Questions

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