Proving a solution of a Bernoulli type equation Prove that
\begin{equation}
y(x) = \sqrt{\dfrac{3x}{2x + 3c}}
\end{equation}
is a solution of
\begin{equation}
\dfrac{dy}{dx} + \dfrac{y}{2x} = -\frac{y^3}{3x}
\end{equation}
All the math to resolve this differential equation is already done. The exercise simply asks to prove the solution.
I start by pointing out that it has the form
\begin{equation}
\dfrac{dy}{dx} + P(x)y = Q(x)y^3
\end{equation}
where
\begin{equation}
P(x) = \dfrac{1}{2x}, \qquad Q(x) = -\frac{1}{3x}
\end{equation}
Rewriting y(x) as
\begin{equation}
y(x) = (3x)^{\frac{1}{2}} (2x + 3c)^{-\frac{1}{2}}
\end{equation}
Getting rid of that square root, I'll need it later on to simplify things
\begin{equation}
[y(x)]^2 = 3x(2x + 3c)^{-1}
\end{equation}
Calculating dy/dx
\begin{align}
\dfrac{dy}{dx} &= \frac{1}{2}(3x)^{-\frac{1}{2}}(3)(2x + 3c)^{-\frac{1}{2}} + \left(-\dfrac{1}{2}\right)(2x + 3c)^{-\frac{3}{2}}(2)(3x)^{\frac{1}{2}} \\
&= \frac{3}{2}(3x)^{-\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} - (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{3}{2}} \\
&= (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= y \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= \dfrac{y}{2x} - y(2x + 3c)^{-1} \\
&= \dfrac{y}{2x} - y\left(\dfrac{y^2}{3x}\right) \\
&= \dfrac{y}{2x} - \dfrac{y^3}{3x}
\end{align}
Finally
\begin{align}
\dfrac{dy}{dx} + P(x)y &= \dfrac{y}{2x} - \dfrac{y^3}{3x} + \dfrac{y}{2x} \\
&= \dfrac{y}{x} - \dfrac{y^3}{3x}
\end{align}
which obviously isn't the same as equation 2. I don't know where I screwed up.
 A: Your work seems fine and we obtain
$$\dfrac{dy}{dx} =\dfrac{y}{2x} - \dfrac{y^3}{3x}$$
which should be the correct differential equation.
I don't understand why you have added the term $P(x)y$ in the last step.
A: You could probably get an easier calculation by going over the logarithmic derivative,
$$
\log(y(x))=\frac12(\log(3x)-\log(2x+3c))
\\~\\
\implies
\frac{y'(x)}{y(x)}=\frac12\left(\frac1x-\frac2{2x+3c}\right)
=\frac1{2x}-\frac1{3x}y(x)^2
$$
The last step is obtained by doing the minimum to eliminate the constant $c$.
As you can see, this again confirms your result.

As to the original equation itself, it solves as
$$
(y^{-2})'=-2y^{-3}y'=\frac{y^{-2}}x+\frac2{3x}\\
\left(\frac{y^{-2}}x\right)'=\frac2{3x^2}\implies \frac{y^{-2}}x=-\frac2{3x}+c\\
y(x)=\pm\sqrt{\frac3{3cx-2}}
$$
which is quite different from your solution formula.
A: @kira1985 you should not add the p(x).y on both side, here you have to find the answer .
But you are using the answer it self to solve the problem .As the answer is unknown to have to just find the value of the P(x) and Q(x).
which is
/ = /2 − (^3)/3
P(x) = - 1/(2x)
Q(x) = - 1/(3x)
Is the write answer for the given question, please check your question once more.
