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Goal is to find general solution by reducing the following ODE to separable using a substitution such as $u = y/x$ or $v = xy$ or $v = y - x$ or something similar. Here is the ODE to solve:

$$xy' = (y - x)^3 + y$$

Here is as far as I could get.

$$v = y - x$$ $$y = v + x$$ $$y' = v' + 1 $$ $$x(v' + 1) = v^3 + (v + x)$$ $$xv' + x = v^3 + v + x$$ $$xv' = v^3 + v$$ $$\int\frac{dv}{v(v^2 + 1)} = \int\frac{dx}{x}$$

Although I was able to separate the variables this does not appear to be what the text intended. I am also having difficulty integrating left side for $v$.

Question: Is there a better substitution in which we end up with an easier integral to solve? If so how could we set it up?

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The integral on the left is solved by partial fraction.

$$ \frac {1}{v(v^2+1)}=\frac {A}{v} +\frac {(Bv+C)}{v^2+1} $$

Your substitution is fine and your over all work is correct.

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  • $\begingroup$ Of course I should have known that. It's been years since college and I am returning this fall so I need to review ODE's. Thank you. $\endgroup$ May 15, 2019 at 2:19

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