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I'm currently trying to solve this particular ODE:

$$(x^3 + xy^2)y' = y^3$$

but am having trouble finding the way to do so. I'm currently going through the textbook Advanced Engineering Mathematics 10e (Kreyszig) and the methods that I've studied are separating variables, using integrating factors, and using the Bernoulli equation. I've tried all three with this particular ODE and have been unsuccessful.

My initial thought was to change it into its differential form:

$$-y^3dx + (x^3 + xy^2)dy = 0$$

so that we can find the integrating factor to solve it. However, I've observed that finding the integrating factor isn't quite working out the way that the textbook describes (i.e. integrating factor $F$ should be of one variable rather than two) and am puzzled as to how to proceed from here.

Any tips or advice on how to solve this would be greatly appreciated. Thank you.

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The terms that can combine to the derivative of a monomial are $-y^3\,dx+xy^2\,dy=x^2y^2\,d(y/x)$. This suggests to insert $y=xu$ into the ODE to get $$ x^3(1+u^2)(xu'+u)=x^3u^3\implies x(1+u^2)u'+u=0 $$ which is now separable. You can then, if you want, reconstruct the integrating factor that produces the integrable form in one step. This is more confirmation than solution method.


Trying to match the list of "classical methods", a change of dependency, $x(y)$ instead of $y(x)$, and $x'(y)=1/y'(x)$, gives the equation $$ y^3x'=y^2x+x^3, $$ which is now a Bernoulli equation. Set $u=x^{-2}$ to get the linear DE $$ y^3u'(y)=-2y^2u-2\\~\\ (y^2u(y))'=-\frac{2}y\implies \frac{y^2}{x^2}=-2\ln|y|+c $$

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  • $\begingroup$ Hello. Thanks for the answer. Is this substitution method something that I should have learned in a previous calculus course? There isn't any explanation of it in the textbook, and so I'm wondering if the author took it that the reader will "obviously" know this. $\endgroup$ – Seankala Oct 13 '19 at 8:03
  • $\begingroup$ There is some intuition in recognizing homogeneous (in the non-linear sense) ODE. Miraculously, text-book examples reduce nicely after substitution. // Obviously, these examples/exercises are constructed to be solvable, so you can always try to find traces of the building blocks to reverse-engineer the equation. Having terms with the same degree structure is such a hint. // Any contrived method to reduce the PDE of the integrating factor to an ODE relies on the same constructedness of the task. $\endgroup$ – Lutz Lehmann Oct 13 '19 at 8:10
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The given DE is homogeneous.

So, let $x = ky$. We have $$x' = yk'+k$$ $$x'y^3 = x^3+xy^2\to (yk'+k)y^3 =k^3y^3 + ky^3$$ $$\implies yk'+k = k^3+k$$ $$\implies \frac{dk}{k^3} = \frac{dy}{y}$$ $$\implies \ln y = \frac{k^{-2}}{-2} + c$$ $$\implies \boxed{y = c'e^{y^2\over -2x^2}}$$

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This seems to be a general first order differential equation of homogeneous type (unfortunately, not the typical homogeneous). That is is an ODE of the form: $$\frac{dy}{dx}=F\bigg(\frac{y}{x}\bigg)$$

How can we see it: $$(x^3 + xy^2)y' = y^3\Longrightarrow y'=\frac{y^3}{x^3+xy^2}\Longrightarrow y'=\frac{\frac{y^3}{x^3}}{\frac{x^3+xy^2}{x^3}}$$

Now set $v=\frac{y}{x}$ implies $xv=y\Longrightarrow y'=v+xv'$ we get the following separable ODE of the function $v(x)$:

$$v+xv'=\frac{v^3}{1+v^2}$$

(the differential is $\frac{dv}{dx}$ just to be clear).

An Important thing to note is that because the nature of the substitution we made, the solution can't always be written explicitly in the form $y=f(x)$ do don't worry if you can't reach this form.

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We will solve instead for $\frac{dx}{dy}$ then

$$ \frac{dx}{dy} = \frac{x^3(y)}{y^3}+\frac{x(y)}{y} $$

having as solution

$$ x = \pm\frac{y}{\sqrt{C_0-2\ln y}} $$

after that we can solve for the inverse.

NOTE

$$ \frac{dx}{dy} = \frac{x^3(y)}{y^3}+\frac{x(y)}{y} $$

can be easily solved with the substitution

$$ z = \frac yx $$

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