I'm trying to understand what is wrong with this solution, since I'm not getting the same answer in

dikcijom2k 2022-07-02 Answered
I'm trying to understand what is wrong with this solution, since I'm not getting the same answer in Matlab
y x y = x y 3 / 2 , y ( 1 ) = 4
y x y = x y 3 / 2 d y d x = x ( y + y 3 / 2 ) d y ( y + y 3 / 2 ) = x d x 2 ln 1 + y y = x 2 2 + c y = 1 ( 1 e x 2 4 + c ) 2 4 = 1 ( 1 e 1 4 + c ) 2 c = 0.655 y = 1 ( 1 e x 2 4 0.655 ) 2
This is what I'm getting in Matlab
( ( tanh ( x 2 8 + atanh ( 3 ) 1 8 ) + 1 ) 2 4 ( tanh ( x 2 8 + atanh ( 5 ) + 1 8 ) 1 ) 2 4 )
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Answers (2)

Nathen Austin
Answered 2022-07-03 Author has 14 answers
I think
y = 1 ( 1 e x 2 4 + c ) 2
instead of what you wrote. y cannot be negative anyway, as y is present in the equations.

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cooloicons62
Answered 2022-07-04 Author has 4 answers
If you distribute the constant in the exponent to be also a factor, then you get to
1 + y 1 / 2 = C e 1 x 2 4
and from that to C = 3 2 , so that in the end
y ( x ) = ( 3 2 e 1 x 2 4 1 ) 2
One can now express the exponential in various ways in terms of hyperbolic functions, for instance one has that coth ( u ) + 1 = e u sinh ( u ) = 2 1 e 2 u , which has a resemblance to parts of the solution.

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