(Differential Equations) General and singular solutions

[namegoeshere]

Homework Statement

Find the general solution and any singular solutions to (2xy^3+4x)y'=x^2y^2+y^2.

Homework Equations

The Attempt at a Solution

2x(y^3+2)y'=y^2(x^2+1)
\int\frac{y^3+2}{y^2}\,dy=\int\frac{x^2+1}{2x}\,dx
\frac{y^3-4}{2y}=\frac{x^2+2\ln x}{4}+C
​

Is this correct?
To find the singular solution, do I set y'=0 and see if its solutions fit into the general solution? i.e.

y'=\frac{x^2y^2+y^2}{2x^3+4x}=0 \Rightarrow y=0​

Which in this example does not fit into the general solution since the left side becomes \frac{(0)^3-4}{2(0)}. Therefore y=0 is a singular solution?

 

[LCKurtz]

Homework Statement

Find the general solution and any singular solutions to (2xy^3+4x)y'=x^2y^2+y^2.

Homework Equations

The Attempt at a Solution

2x(y^3+2)y'=y^2(x^2+1)
\int\frac{y^3+2}{y^2}\,dy=\int\frac{x^2+1}{2x}\,dx
\frac{y^3-4}{2y}=\frac{x^2+2\ln x}{4}+C
​

Is this correct?

That looks OK, at least for $x>0$.

To find the singular solution, do I set y'=0 and see if its solutions fit into the general solution? i.e.

y'=\frac{x^2y^2+y^2}{2x^3+4x}=0 \Rightarrow y=0​

Which in this example does not fit into the general solution since the left side becomes \frac{(0)^3-4}{2(0)}. Therefore y=0 is a singular solution?

In your solution, you divided by $y^2$. That rules out finding a potential solution of $y=0$. So you plug that into the original equation and see if it works. You don't set $y'=0$ to do it.

 

[namegoeshere]

That looks OK, at least for $x>0$.
In your solution, you divided by $y^2$. That rules out finding a potential solution of $y=0$. So you plug that into the original equation and see if it works. You don't set $y'=0$ to do it.

Ah, I get it now. Based my singular solution attempt off the one given example we had, and admittedly what I tried didn't make intuitive sense to me. Thanks. :)