Bernoulli Differential Equations

[TranscendArcu]

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

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

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: [itex]y' + p(x)y = g(x) y^n[/itex]. Supposing that I let [itex]n=1[/itex], the equation is linear. Can I solve it by constructing an integrating factor? That is, can I observe:

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

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 [itex]Ω(x)y = c[/itex]. This implies [itex] y = \frac{c}{Ω(x)}[/itex]. Not so?