<s>[INST] Is there a less trickier way to solve this differential equation? Solve the following differential equation:
$$ y^3 dy+(x + y^2)dx = 0$$
Solution:
The following solution uses the substitution $y^2=tx$.
$$y^3\frac{dy}{dx}+x+y^2=0\tag1$$
Differentiating $y^2 = tx$,
$$ 2y\frac {dy}{dx}= t + x\frac{dt}{dx}$$
Now substituting in (1),
$$tx\ (t + x\frac{dt}{dx})+2(x+tx)=0$$
$$t^2x+tx^2\frac{dt}{dx}+ 2x + 2tx=0 $$
$$2tx\ dx + t^2x\ dx + 2x\ dx+tx^2\ dt=0$$
$$x\ dx\ (t^2 + 2t + 2)=-tx^2\  dt$$
$$\int{\frac{1}{x}}dx= \int{\frac{-t}{t^2+2t+2}}dt$$
$$lnx=\int{\frac{-t-1}{t^2+2t+2}}dt + \int{\frac{1}{t^2+2t+2}dt}$$
$$lnx=\int{\frac{-t-1}{(t+1)^2+1}}dt+\int{\frac{1}{(t+1)^2+1}}dt$$
$$lnx=-\frac{1}{2}ln[(t+1)^2+1]+ \arctan(\frac{y^2}{x}+1)$$
$$lnx=-\frac{1}{2}ln[(\frac{y^2}{x}+1)^2+1)]+ \arctan(\frac{y^2}{x}+1)$$
I have the following questions:
1) How did the person who solved this problem get the idea of substituting
$y=tx^2$ ? This substitution doesn't seem to be obvious.
2) Is there any other method (without using the above substitution) to solve this problem, in a less trickier manner?  
 [/INST] The substitution presented does seem somewhat unmotivated.
Given
\begin{equation}
(x+y^2)dx+y^3dy=0
\end{equation}
one would be inclined to first try the substitution $x=uy^2$ so that the $y^2$ might factor out of the first term.
This gives $dx=y^2du+2uy\,dy$ yielding after a bit of algebra the separable equation
\begin{equation}
y^4(u+1)du+y^3(2u^2+2u+1)dy=0
\end{equation}
Separation gives
\begin{equation}
\dfrac{u+1}{2u^2+2u+1}\,du+\dfrac{1}{y}\,dy=0
\end{equation}
which is straightforward, perhaps aided by the substitution $v=2u+1$ with the first integral partitioned into its logarithmic and inverse tangent parts.
</s>