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)?
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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