How to solve $y'+y=e^{x}y^{\frac{2}{3}}$? I am given the ODE $y'+y=e^{x}y^{\frac{2}{3}}$ and the initial condition $y(0)=0$.
I don't have an idea on how to to this, but the solution in the book starts with letting 
$u=y^{\frac{1}{3}}$, with this I got to $u'=\frac{1}{3}u + \frac{1}{3}e^x$ and from there I don't know how to continue.
How can I continue and how did the book got to $u=y^{\frac{1}{3}}$ in the first place ?
Edit: I am not looking for the solution $\forall x: y(x)=0$
 A: There is one obvious solution $y(x)=0$.
$$y'+y=e^{x}y^{2/3}$$
$$y'e^x+ye^x=e^{2x}y^{2/3}$$
$$(ye^x)'=e^{2x}y^{2/3}$$
Substitute $u(x)=y(x)e^x$
$$u'=u^{2/3}e^{4x/3}$$
$$u^{-2/3}u'=e^{4x/3}$$
We have divided by $u^{2/3}$, so this is ok only for points such that $u(x)\ne0$.
$$\frac13u^{-2/3}u'=\frac13e^{4x/3}$$
$$(u^{1/3})'=(e^{4x/3}/4)'$$
$$u^{1/3}=e^{4x/3}/4+C$$
$$u=(C+e^{4x/3}/4)^3$$
$$y=(C+e^{4x/3}/4)^3e^{-x}$$
To get $y=0$, we need $C=-1/4$
$$y=\left(\frac{e^{4x/3}-1}4\right)^3 e^{-x}$$
A: The equation $u'=\frac{1}{3}u+\frac{1}{3}e^x$ is a linear differential equation of the first order. Its solutions can be immediately written by a general formula. You can find plenty of references on these equations.
For the second question, consider $y'+y=a(x)y^\alpha$, where $\alpha$ is a given number. Can you find a change of unknown like $u=y^\beta$, where $\beta$ depends on $\alpha$, that reduces the equation to a simpler one?
A: The idea is that in differential equations of the form $y' + g(x) y = f(x)y^{\alpha}\,$ $(\alpha \neq 1)$, the Bernoulli equations, you can do the variable change $y = u^{\beta}$, and your equation becomes
$$\beta u' u^{\beta -1} + g(x)u^{\beta} = f(x)u^{\beta \alpha}$$
Multiplying through by ${1 \over \beta}u^{1 - \beta}$, this is the same as
$$u' + {1 \over \beta} g(x)u = {1 \over \beta}f(x)u^{\beta \alpha + 1 - \beta}$$
So the idea is that if you choose $\beta$ so that $\beta \alpha + 1 - \beta = 0$, in other words, $\beta = {1 \over 1 - \alpha}$, your equation becomes first-order linear and you can solve it using first-order linear methods. In this case, $\alpha = \frac{2}{3}$, so $\beta = 3$ and you make the variable change $y = u^3$.
