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Solving three non-linear ODEs analytically in one variable

Mathematica Asked on May 30, 2021

Any idea what is the best way to substitute to analytically solve these equations in y(x) ?

3 y'[x] /y[x] - m == g[x];

-2  y''[x]/y[x] - (y'[x]/y[x])^2 == g[x];

-3 ( y''[x]/y[x] + (y'[x]/y[x])^2) == g[x];

My trail:

Here is my idea, I think we can equal any two of them and eliminate g[t],

So for instance in case the first equals the second

DSolve[{-2 (y''[x]/y[x]) - (y'[x]/y[x])^2 == 3 (y'[x]/y[x])^2 - m}, y, x]

Which gives

{{y -> 
   Function[{x}, C[2] Cosh[Sqrt[3/2] Sqrt[m] (x + 2 C[1])]^(1/3)]}}

BUT this is not the same solution when the first equals the third or the second equals the third.

So how to solve the three equations together to give a single solution for y(x)?

One Answer

The general solution of the first equation is

y =.
y = DSolveValue[3 y'[x]/y[x] - m == g[x], y, x]
(*E^ Integrate [(m + g[K[1]])/3,{K[1], 1, x}]*C[1]*)

substituted into the remaining equations gives

{-2 y''[x]/y[x] - (y'[x]/y[x])^2 == g[x],-3 (y''[x]/y[x] + (y'[x]/y[x])^2) == g[x] } // Simplify


(*{m^2 + (3 + 2 m) g[x] + g[x]^2 + 2 Derivative[1][g][x] == 0,2 m^2 + (3 + 4 m) g[x] + 2 g[x]^2 + 3 Derivative[1][g][x] == 0}*)

Only if these equations are true the solution y[x] holds for the three odes!

Answered by Ulrich Neumann on May 30, 2021

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