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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 $\forall x: y(x)=0$

3 Answers 3

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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 $y = u^3$.

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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 $C=-1/4$ $y=\left(\frac{e^{4x/3}-1}4\right)^3 e^{-x}$

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    @Babak: You are right about [Bernoulli Equation](http://en.wikipedia.org/wiki/Bernoulli_differential_equation).2012-06-26
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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?

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    Yes, as Zarrax pointed out, this trick works for Bernoulli equations.2012-06-26