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?  

$y^3~dy+(x+y^2)~dx=0$
$y^3\dfrac{dy}{dx}=-x-y^2$
Let $u=y^2$ ,
Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$
$\therefore\dfrac{y^2}{2}\dfrac{du}{dx}=-x-y^2$
$\dfrac{du}{dx}=-\dfrac{2x}{y^2}-2$
$\dfrac{du}{dx}=-\dfrac{2x}{u}-2$
Luckily this becomes a first-order homogeneous ODE.
