Find the error - initial value problem

Question

Find a local solution for the following initial value problem $$y' = (y+1)^3, \qquad y(0)=0$$

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I have approached this using separation of variables and obtained

$$y=\sqrt{-\frac{1}{2x-1}}-1$$

locally, of course. However, differentiating this yields

$$y'=\left(\frac{1}{-2x+1} \right)^{2.5}$$

which doesn't solve the general differential equation above. What have I done wrong?

Edit: Added the computation of $y'$:

$$y'=\frac{1}{2}\left( -\frac{1}{2x-1} \right)^{-\frac{1}{2}}\cdot\left((-2x+1)^{-1} \right)'=\frac{1}{2}\left( -\frac{1}{2x-1}\right)^{-\frac{1}{2}}\cdot 2\cdot (-2x+1)^{-2}=\frac{1}{(-2x+1)^{2.5}}$$

Answer 1

The constant solution $y=-1$ can be ruled out because of the initial condition. Any other solution is given implicitly by $$ \int \frac{1}{(y+1)^3} dy = \int dx \Leftrightarrow \frac{(y+1)^{-2}}{-2} = x + C $$

since $y(0)=0$, we get $C=-\frac 12$, which means that $$ \frac{1}{(y+1)^2} = -2x+1\Leftrightarrow (y+1)^2 = \frac{1}{-2x+1}\Leftrightarrow y = -1 + \sqrt{\frac{1}{-2x+1}} $$

This function (which you also obtained) solves the differential equation.

Answer 2

Using SymPy:

>>> from sympy import *
>>> x = Symbol('x', real=True)
>>> f = (1 / sqrt(1 - 2 * x)) - 1
>>> diff(f,x)
(1 - 2*x)**(-3/2)

Hence,

$$f ' (x) = \frac{1}{\left(1 - 2 x\right)^{\frac{3}{2}}}$$

while you computed

$$f ' (x) = \frac{1}{\left(1 - 2 x\right)^{\frac{5}{2}}}$$