伯努利方程求解推导验证：请核对以下步骤是否正确

Hey there, let's walk through your Bernoulli equation solution step by step—you're on the right track here! Let's verify each part of your derivation:

1. Rewriting the original equation #

Starting with your initial equation:
$$y' +y(x^2-1) + xy^6 =0$$
You rearranged it correctly to isolate the non-linear term, setting up the standard Bernoulli form:
$$y' + y(x^2-1) = -xy^6 $$

2. Substitution for Bernoulli conversion #

Your substitution choice is perfect for this type of equation (Bernoulli equations use $v = y^{1-n}$, and here $n=6$, so $1-n=-5$):
$$v = y^{-5}, v' = -5y^{-6}y'$$
Solving for $y'$ gives:
$$y' = \frac{-y^{6}v'}{5}$$

3. Substituting into the equation #

When you plugged $y'$ back into the rearranged equation:
$$\frac{-y^{6}v'}{5} + y(x^2-1) =-xy^6$$
You divided both sides by $-y^6$ (valid as long as $y \neq 0$, which is a trivial solution we can note separately) to convert it to a linear ODE:
$$v' -5v(x^2-1) = 5x$$
This conversion is completely correct—you’ve successfully turned the non-linear Bernoulli equation into a linear first-order ODE in terms of $v$.

4. Calculating the integrating factor #

Your integrating factor calculation is spot-on too. For a linear ODE of the form $v' + P(x)v = Q(x)$, the integrating factor is $r(x) = e^{\int P(x)dx}$. Here, $P(x) = -5(x^2-1)$, so:
$$r(x) = e^{-5\int (x^2-1)dx} = e^{{-5\left(\frac{x}3}{3}-x\right)}$$

Quick note on next steps #

Multiplying both sides by this integrating factor will make the left-hand side the derivative of $r(x)v$, so you can integrate both sides to solve for $v$, then substitute back $v = y^{-5}$ to get the final solution for $y$.

Great work so far—your derivation is solid!

内容的提问来源于stack exchange，提问作者Stuy