Solving First Order Non Linear Differential Equation I was trying to solve the following question:
$\frac { dy }{ dx } ({ x }^{ 2 }{ y }^{ 3 }+xy)=1$
I took xy common and made the substitution
$x{ y }^{ 2 }=t$
On solving and simplifying after substitution I got
$\frac { dt }{ dx } -\frac { t }{ x } =\frac { 2 }{ t+1 } $.
Now I'm stuck up here. 
I don't know what to do next.
Neither do I know what is first order non linear differential equation is nor do i know how to solve it.
(I typed the entire equation on wolframalpha and it showed it is a first order non linear differential equation).
Any suggestions?
 A: The following equation can be written as 
${ x }^{ 2 }{ y }^{ 3 }+xy=\frac { dx }{ dy } $
Dividing LHS and RHS by ${ x }^{ 2 }$
We get
$\frac { 1 }{ { x }^{ 2 } } \frac { dx }{ dy } -\frac { y }{ x } ={ y }^{ 3 }$
It's Bernouli Differential Equation.
Put  $\frac { 1 }{ x } =u$
Hence on differentiating with respect to y we get
$-\frac { 1 }{ { x }^{ 2 } } \frac { dx }{ dy } =\frac { du }{ dy } $
Putting the value in the original equation
$\frac { du }{ dy } +uy=-{ y }^{ 3 }$
Now this differential equation can be easily solved.
A: $$\frac{dy}{dx}(x^2y^3+xy)=1$$
$$(x^2y^3+xy)=\frac{dx}{dy}$$
$$x^2y^3+xy=x'$$
it's Bernoulli differential equation put $x=uv$ then $x'=u'v+uv'$
$$u'v+uv'-uvy=u^2v^2y^3$$
$$u'v+u(v'-vy)=u^2v^2y^3$$
$$v'-vy=0$$
$$v'=vy$$
$$\frac{dv}{v}=ydy$$
$$ln|v|=\frac{1}{2}y^2$$
$$v=e^{\frac{1}{2}y^2}$$
$$u'\cdot e^{\frac{1}{2}y^2}+u\cdot 0=u^2(e^{\frac{1}{2}y^2})^2y^3$$
$$u'=u^2\cdot e^{\frac{1}{2}y^2}y^3$$
$$\frac{du}{u^2}= e^{\frac{1}{2}y^2}y^3dy$$
$$-\frac{1}{u}=e^{\frac{1}{2}y^2}(y-2)+c$$
$$u=-\frac{1}{e^{\frac{1}{2}y^2}(y-2)+c}$$
$$x=uv=-\frac{e^{\frac{1}{2}y^2}}{e^{\frac{1}{2}y^2}(y-2)+c}$$
A: Of cource the first integral is 
$$
{ x }^{ 2 }{ y }^{ 3 }+xy-x=C.
$$
It is a solution.
