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The following equation is Homogeneous (source:wolfram alpha), and has the answer $\frac{x}{y}+\mathrm{e}^{x^3}=c$ as solved by putting $y=vx$.

$$y\mathrm{d}x-x\mathrm{d}y+3x^2y^2\mathrm{e}^{x^3}\mathrm{d}x=0$$

or $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y + 3x^2y^2\mathrm{e}^{x^3}}{x}$$

As the Equation is not of the form of $f(zx,zy)=z(f(x,y))$ so I couldn't understand it as homogeneous at first glance. It can be solved by Bernoulli and exact forms but its lengthy and complex than homogeneous form.The solution at wolfram alpha stated that the equation can be solved by the homogeneous method.

Can somebody Please Explain me how to spot the Homogeneous equations that do not follow the definition like the above equation

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  • $\begingroup$ I've edited it to use Latex. I may have have butchered the last equation, please fix this as I could not tell what the equation should have been $\endgroup$ – Alex Mathers Jun 25 '15 at 14:48
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An equation of the form $F(x,y)dy + G(x,y)dx + H(x,y) = 0$ is homogeneous (or rather can be solved by the same sort of trick that works for classically homogeneous equations) if for some $\alpha$ and $\beta$ the substitution $$y = v^\beta x^\alpha \\ dy = \beta v ^{\beta-1} x^\alpha + \alpha v^\beta x^{\alpha-1}$$ transforms it into the form $$ M(v)dv = N(x)dx$$

Since the algebra involved in that transformation is usually simple, a good approach is to just try it, and see if you can adjust $\alpha$ and $\beta$ to make terms thjat don't fit that form cancel. (You might first try just $y = vx^\alpha$ to see if that does the trick for some $\alpha$.)

For your equation: $$ y\,dx -x\,dy +3x^2y^2 d^{x^3}dx = 0 $$ the $y = vx^\alpha$ transformation yields $$ v x^\alpha dx - x^{\alpha+1}dv -\alpha v x^\alpha dx +3x^{2+2\alpha}v^2 e^{x^3}dx = 0 $$ This is in the correct form if cancellations cause $v x^\alpha dx -\alpha v x^\alpha dx=0 $, that is, if $\alpha = 1$. So you know this trick will work for the substitution $y=vx$ and we can read off the resulting equation as $$ -x^{2}dv - +3x^{4}v^2 e^{x^3}dx = 0 \implies\\ \frac{dv}{v^2}=3x^2 d^{x^3} dx $$ which is easy to integrate to $$ -\frac{1}{v} = e^{x^3} +C $$ and since $v = \frac{y}{x}$, $$-\frac{x}{y} = e^{x^3} - C$$ which is equivalent to the stated answer.

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