Bernoulli Differential Equations

[TranscendArcu]

Suppose I have a Bernoulli differential equation; that is, an equation of the form: $y' + p(x)y = g(x) y^n$. Supposing that I let $n=1$, the equation is linear. Can I solve it by constructing an integrating factor? That is, can I observe:

$y' + p(x)y = g(x) y$
$→ y' + y[(p + g)(x)] = 0$. I would then have,
$Ω(x) = e^{\int (p + g)(x) dx}$ and multiplying through,
$Ω(x)y' + Ω(x)y[(p + g)(x)] = 0$
$→ (Ω(x)y)' = 0 → Ω(x)y = 0$

But, this seems to be leading me to the conclusion that y = 0. Is that right or have I done something wrong? Is it possible to solve a Bernoulli equation with n=1 by constructing an integrating factor?

 

[Ray Vickson]

Suppose I have a Bernoulli differential equation; that is, an equation of the form: $y' + p(x)y = g(x) y^n$. Supposing that I let $n=1$, the equation is linear. Can I solve it by constructing an integrating factor? That is, can I observe:

$y' + p(x)y = g(x) y$
$→ y' + y[(p + g)(x)] = 0$. I would then have,
$Ω(x) = e^{\int (p + g)(x) dx}$ and multiplying through,
$Ω(x)y' + Ω(x)y[(p + g)(x)] = 0$
$→ (Ω(x)y)' = 0 → Ω(x)y = 0$

But, this seems to be leading me to the conclusion that y = 0. Is that right or have I done something wrong? Is it possible to solve a Bernoulli equation with n=1 by constructing an integrating factor?

F'(x) = 0 does not imply F(x) = 0; it implies F(x) = constant.

RGV

 

[TranscendArcu]

Oh! So that must mean I have $Ω(x)y = c$. This implies $y = \frac{c}{Ω(x)}$. Not so?