# How can i solve a differential equation like this one?

My Problem is: this given differential equation $$x^3+y^3+x^2y-xy^2y^{\prime}=0$$ $$(x\neq 0,\ y\neq 0)$$

My Approach was: i had the idea to bring it in this form:

$$x^3+y^3+x^2y-xy^2y^{\prime}=0$$ $$x^3+y^3+x^2y=xy^2y^{\prime}$$ $$\frac{x^3}{xy^2}+\frac{y^3}{xy^2}+\frac{x^2y}{xy^2}=\frac{xy^2y^{\prime}}{xy^2}$$ $$\frac{x^3}{xy^2}+\frac{y^3}{xy^2}+\frac{x^2y}{xy^2}=y^{\prime}$$ $$\frac{x^2}{y^2}+\frac{y}{x}+\frac{x}{y}=y^{\prime}$$

But this is the point where i get stuck. It seems the expression is gettin more and more complex, and is not leading to any solution... how can i solve this?

Hint: Make the change of variable $y=zx$. You will get a separable equation. You can do that from the point you reached, but it is quite a bit simpler to start all over again. For general information (which you will not need in this case) search for homogeneous differential equation.
More: Let $y=zx$. Then $y'=z+xz'$. Substituting in the equation, we get $$x^3+z^3x^3 +x^3z-x^3 z^2(z+xz')=0.$$ Divide through by $x^3$. There is some nice cancellation, and we end up with $$xz^2z'=1+z.$$ This is a separable equation, which I expect you can handle. There is a complication in that we will end up with an implicit solution.
• The system does not like long strings of comments. I will delete most of mine, suggest you do the same. We want to integrate $\frac{z^2}{z+1}$. Either make change of variable $w=z+1$, or just divide $z^2$ by $z+1$. We get $z-1+\frac{1}{z+1}$. Integrate. We get $\frac{z^2}{2}-z +\ln(z+1)+C$. Jun 10 '13 at 17:52
• so the solution is: $$\frac{z^2}{2}-z +\ln(z+1)+C = log(x)$$ I am confused, what can i do with this term? Jun 10 '13 at 18:01
• If you have an initial condition, you can evaluate $C$. Whether you do or not, replace $z$ by $y/x$. Now you have an equation linking $y$ and $x$. This gives the solution(s) implicitly. You will not be able to solve for $y$ explicitly in terms of $x$. That's usually considered OK. Many DE only yield implicit solutions. Jun 10 '13 at 18:09
• okay.. we'll have: $y=x\cdot z$ that's why: $\frac{y}{x}=z$. No we insert and get: $$\frac{y^2}{2x^2}-\frac{y}{x}+ln(\frac{y}{x}+1)+ C = log(x)$$ $$\frac{y^2}{2x^2}-\frac{y}{x}+ln(\frac{y}{x}+1)-log(x) + C = 0$$ ? please say i am right, this time. Jun 10 '13 at 18:18
• Looks right. There are many ways to further manipulate, but nothing that gives $y$ explicitly in terms of $x$. Jun 10 '13 at 18:21
Since you have expressed y' in terms of a function of $\frac{y}{x}$, you can do a substitution: set v = $y\over x$, and $\frac{dy}{dx} = v + x\frac{dv}{dx}$, and you get $$v + x\frac{dv}{dx} = v^{-2} + v^{-1} + v \longrightarrow x\frac{dv}{dx} = v^{-2} + v^{-1}.$$Separate the equation:$$\frac{dv}{v^{-2} + v^{-1}} = \frac{dx}{x}$$ Integrate both sides: $$\ln\left(v+1\right)+\frac{\left(v-2\right)v}{2}+C = \ln x$$ Now solve for v, undo the substitution, and solve for y.