Solved: "Value of differential equation at non existent points..."

acceloq3c

Answered question

2022-04-22

Value of differential equation at non existent points in the original function.

y^{2} + y = x^{3}

Kathleen Keller

Beginner2022-04-23Added 20 answers

Step 1
Generally speaking, the equation

f (y) = g (x)

leads to multiple branches, where if both

f (\cdot)

and

g (\cdot)

are differentiable, then a derivation of y w.r.t. x can be defined as

y^{'} f^{'} (y) = g^{'} (x)

.
The latter derivative makes sense as long as the points lie on the branches of

f (y) = g (x)

and is insignificant elsewhere.
Also note that for example in your case, y can be defined as a function of x on each branch (

y = - \frac{1}{2} + \sqrt{x^{3} + \frac{1}{4}}

on one branch and

y = - \frac{1}{2} - \sqrt{x^{3} + \frac{1}{4}}

on another), and whether or not this function can be explicitly found, y in

y^{'} = \frac{3 x^{2}}{2 y + 1}

should be replaced by some h(x), for which

y^{'} = \frac{3 x^{2}}{2 h (x) + 1} .

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