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?