I think we are losing solutions when doing Bernoulli Differential Equations. For example, take:

$$y^2+2y= \frac{\mathrm{d}y}{\mathrm{d}t}$$

If we would take $u=y^{-1}$, we can transform the equation to $$u^{-2}+2u^{-1}= -u^{-2}\cdot\frac{\mathrm{d}u}{\mathrm{d}t}$$

$$-2u-1= \frac{\mathrm{d}u}{\mathrm{d}t}$$

This is an easy solvable differential equation, which is exactly the point of the Bernoulli substitution. However, when substituting $u=y^{-1}$, we assume that $y$ has no zeroes.

One solution we are clearly losing is $y(t)=0$ for all $t \in \mathbb R$, but how can we know for sure that we aren't losing other solutions?


1 Answer 1


Technically, you are correct. However:

In reality you do not lose any solution because locally, for every $x_0, y_0$, there exists precisely one function $y$ such that $y^2 + 2y = y'$ and $y(x_0) = y_0$. This is due to the Picard–Lindelöf theorem.

This means that if $y_0=0$, then the solution is indeed $y(x)=0$.

If $y_0\neq 0$, then for some $\epsilon$, you know that $y(x)\neq 0$ for all $x\in (x-\epsilon, x+\epsilon$.

So you only lose a solution if the starting condition is such that you didn't really need to bother with the Bernoulli process at all.

  • $\begingroup$ +1, but can you give a reference or proof for your first statement (i.e. that there is only one $y$ such that $y^2+2y=y'$ and $y(x_0)=y_0$)? $\endgroup$
    – wythagoras
    Oct 15, 2015 at 18:41
  • $\begingroup$ @wythagoras Added. $\endgroup$
    – 5xum
    Oct 15, 2015 at 18:48

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