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

-

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 $y = u^3$.

-

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

-
How did you think about the beginning ? – Belgi Jun 26 '12 at 15:34
Well, if you see $y'+y$, you see that you can get $y'e^x+ye^x=(ye^x)'$; similarly from $y-y'$ you can get $ye^{-x}-y'e^{-x}$. It's easy if you have seen the trick at least once. This is basic idea for obtaining the general solution of homogeneous first-order linear equations, as given e.g. here. (Perhaps someone can give a better reference.) – Martin Sleziak Jun 26 '12 at 15:42
I see, in the reference: What about the constants from the integration, should I add them or does the "$+C$" in the solution takes care of that ? – Belgi Jun 26 '12 at 15:51
@MartinSleziak: Is this equation a Bernoulli Equation which could be solved by taking $w=y^{1-\frac{2}{2}}$? – Babak S. Jun 26 '12 at 15:55
@Babak: You are right about Bernoulli Equation. – Martin Sleziak Jun 26 '12 at 15:59

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?

-
Can the LHS be more general than $y'+y$ and this will still work ? – Belgi Jun 26 '12 at 15:35
Yes, as Zarrax pointed out, this trick works for Bernoulli equations. – Siminore Jun 26 '12 at 16:48