Initial value problem of $y' = \sqrt{|y|}(y+1)$

Question

i'm trying to determinate the solution of the intial value problem

$$y' = \sqrt{|y|}(y+1)$$

my solution was as follow

applying substitution as follow

let $u^{2} = y$ and $dy = 2u\ du$

$$2 \int du = \int u^{2} + 1\ dx $$

but i feel it's wrong

Answer 1

$$y' = \sqrt{|y|}(y+1)$$

$$ \frac{dy}{dx} = \sqrt{|y|}(y+1) \\ \equiv \int \frac{1}{\sqrt{y} (y + 1)} dy = \int 1\ dx \\ \equiv \int \frac{2u}{u(u^2+1)} du = x + c \\ \equiv 2 \int \frac{1}{u^2+1} du = x + c \\ \equiv \tan^{-1}{\sqrt{y}} = \frac{1}{2}(x + c) \\ \equiv \tan{(\tan^{-1}{\sqrt{y}})} = \tan{(\frac{x}{2}+c)} \\ \equiv \sqrt{y} = \tan{(\frac{x}{2}+c)} \\ \equiv y = \tan^2{(\frac{x}{2}+c)} $$

Answer 2

You haven't yet established a relationship between $u$ and $x$ so you can't integrate the right hand side.

Things get a lot easier if you write things in Leibniz (differential) notation from the start. $$\frac{dy}{dx}=\sqrt{|y|}(y+1)$$ $$\frac{dy}{\sqrt{|y|}(y+1)}=dx$$ $$\int\frac{|y|^{-0.5}dy}{y+1}=\int dx$$ Now you can let $u^2=y$ and $dy=2u du$ $$\int\frac{|u^2|^{-0.5} 2u du}{u^2+1}=\int dx$$ $$\int\frac{2 du}{u^2+1}=x$$

and that ought to be in your table of integrals