Differential equation with a problem of definition domain

Question

What are the real functions $y$ satisfying $3x y'(x)+y(x) = \frac{1}{y^2(x)}$ , I tried to put $z$ a new real function such that $z(x) = y^3(x)$ , and i found that the solution $$ z(x) = \begin{cases} 1 & \text{if x=0;}\\ \frac{\lambda}{x^3}+1 ;\lambda \in \mathbb{R} & \text{ otherwise.} \end{cases} $$ But I'm having a domain of definiton problem of $y$ ! Can you help me?

Answer 1

Hint: your equation is separabel and can be written in the form $$\frac{y'(x)}{\frac{1}{3}\left(\frac{1}{y(x)^2}-y(x)\right)}=\frac{1}{x}$$

Answer 2

I think your answer is correct(you just left to compare $z$ with $y$)

Here is another way:

Rewrite $3x y'(x)+y(x) = \frac{1}{y^2(x)}$ as $$\frac{y^2(x)}{1+y^3(x)}y'(x)=\frac1{3x}\implies \frac{y^2}{1-y^3}dy=\frac1{3x}dx\\\implies \int\frac{y^2}{1-y^3}dy=\int\frac1{3x}dx$$

L.H.S:$$\int\frac{y^2}{1-y^3}dy;\;u=1-y^3, -\frac{du}3=y^2dy\\\rightarrow-\int\frac{1}{3u}du=-\frac13\ln u+c=-\frac13\ln(1-y^3)+c$$

R.H.S:$$\int\frac1{3x}dx=\frac13\ln x+k$$ Thus:$$\frac13\ln x+K_0=-\frac13\ln(1-y^3)\implies -\ln x+K_1=\ln(1-y^3)\\\implies \ln(e^{K_1}/x)=\ln(1-y^3)\implies K_2/x=1-y^3\\\implies 1-K_2/x=y^3\implies y=\sqrt[3]{1+\frac\lambda x}$$

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Note that we have trivial solution of $y=1$.

Edit, like Isham pointed out, this is not a trivial solution, but the only solution because of $x=0\implies \lambda=0$

Answer 3

$$3x y'(x)+y(x) = \frac{1}{y^2(x)}$$ Multiply both side by $y^2 (y\neq 0)$ $$3x y'(x)y^2+y^3 = 1$$ $$x (y^3)'+y^3 = 1$$ $$xz'+z = 1$$ $$ zx= x+K$$ for $x \neq 0$ $$ y^3= 1+K/x$$ $$ y= \sqrt[3]{1+K/x}$$ For $x=0$ $$y^3(0) -1 =0 $$ $$zx=x+K=0 \implies K=0 \implies y^3-1=0$$

If the interval include $x=0$ then $y=1$ is the solution. Otherwise $ y= \sqrt[3]{1+K/x}$