Use the substitution $u=y'$ to change $xy''=y'+(y')^3$ into a Bernoulli equation and solve.

I can get most of the way through this, but I'm stuck on the last integral, so I'd like to verify that what I've done is correct and then get assistance on this last integral.

If $u=y'$ then $\frac{du}{dx}=y''$, so we have $x \frac{du}{dx}=u+u^3$. Transforming this to a Bernoulli equation, we have $$\frac{du}{dx}-\frac{1}{x}\cdot u=\frac{1}{x}\cdot u^3$$

We then use the substitution $v=u^{1-3}=u^{-2}$ or $u=v^{-1/2}$

Now, $\frac{dv}{dx}=-2u^{-3}\cdot u'=-2u^{-3}\frac{du}{dx}\Rightarrow\frac{-1}{2}u^3\cdot\frac{dv}{dx}=\frac{du}{dx}$. Making the appropriate substitutions and simplifying, we get $$\frac{dv}{dx}+2\cdot\frac{1}{x}\cdot v=-2\cdot \frac{1}{x}$$

The integrating factor is $e^{\int\frac{2}{x}dx}=e^{2\ln x}=x^2$. Thus, we get $\int\frac{d}{dx}[x^2v]\cdot dx=\int-2x\cdot dx$, which gives us $x^2v=-x^2+c$ or $$v=-1+cx^{-2}$$

Substituting $u=v^{-1/2}$ we have $$u=\frac{1}{\sqrt{-1+cx^{-2}}}$$

Since $y'=u$ we must integrate both sides of the equation with respect to $x$ so $$y=\int\frac{1}{\sqrt{-1+cx^{-2}}}dx$$

This is the integral that I am not sure how to do. Thank you in advance for any assistance.


2 Answers 2


$$y=\int\frac{1}{\sqrt{-1+cx^{-2}}}dx$$ is a trivial integral. Multiply by $\frac{x}{x}$ to obtain


Clearly, differentiating $\sqrt{c-x^2}$ gives $-\frac{x}{\sqrt{c-x^2}}$, therefore $y = -\sqrt{c - x^2} + K$


To solve $xy'' = y' + (y')^3$, first let $u = y'$, so that $xu' = u + u^3$

Writing $u' = \frac{du}{dx}$ readily reveals the seperable nature of this equation, and it can be found that

$\frac{1}{u(1+u^2)} \mathrm{du} = \frac{1}{x}\mathrm{dx}$

To solve this, integrate both sides. Partial fractions will suffice to integrate the LHS. You will arive at:

$\log(u) - \frac{1}{2}\log(u^2+1) = \log(ax)$ (the constant can be written as $\log(a)$)

Simplifying, and then exponentiating,

$\frac{u}{\sqrt{u^2+1}} = ax$

$\frac{u^2}{u^2+1} = 1 - \frac{1}{u^2+1} = a^2x^2$

$u = \frac{dy}{dx} = \frac{\pm ax}{\sqrt{1-a^2x^2}}$

Integrating, you find

$y = \int\frac{\pm ax}{\sqrt{1-a^2x^2}}\mathrm{dx} = \mp \sqrt{1-a^2x^2}+ K$

  • $\begingroup$ I do realize this equation is separable and can solve it that way, but the prompt of the question specified rewriting it as a Bernoulli equation and solving it that way. $\endgroup$
    – mmmmmm
    Aug 6, 2021 at 18:59
  • $\begingroup$ @mmmmmm See my new answer. The integral you posed is a trivial one. $\endgroup$
    – egglog
    Aug 6, 2021 at 19:07

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