This is problem 20 chapter 1.9 in Differential Equations and Their Applications – Martin Braun (4:th ed).
The Bernoulli differential equation is given by
$$(dy/dt) + a(t)y = b(t)y^{n} \tag{1}.$$
Multiplying both sides by $\mu(t) = \exp(\int a(t) \ dt)$ we can reduce it to
$$\frac{d}{dt}(\mu (t) y) = b(t)\mu (t)y^{n} \tag{2}.$$
My task is to find an appropriate integrating factor to find the general solution of $(1)$ using $(2)$.
The book suggest to divide by a expression of $y$ which gives me the idea of rewriting $(2)$ to
$$-b(t)\mu(t)y + y^{1-n}\frac{d}{dt}(\mu(t)y) = 0 \tag{3}.$$
I know the equation
$$M(t,y)+N(t,y)\frac{dy}{dt} = 0 \tag{4}$$
has a solution in case the partial derivatives satisfy $\partial M / \partial y = \partial N / \partial t$, or, using integration factors has a solution if the expression
$$\frac{\partial M / \partial y – \partial N / \partial t}{N} \tag{5}$$
is a function $R(t)$ (i.e only dependent on $t$, in which case the integrating factor is $c(t) = \exp(\int R(t) \ dt)$).
I am unsure exactly how to transform $(3)$ into $(4)$. If for example I make the substitution $z = \mu(t)y$ then we get
$$-bz(t,y) + z^{1-n}(t,y)\mu^{n-1}(t,y)\frac{dz(t,y)}{dt} = 0 \tag{6}$$
But this seems incorrect as it is not really on the form of $(4)$.
My substitution in $(6)$ does not seem correct. How can I otherwise proceed?
Best Answer
You already have the integrating factor ($\mu(t) = \exp(\int a(t) \ dt)$) and it depends on $a(t)$ then the differential equation becomes separable. $$\frac{d}{dt}(\mu (t) y) = b(t)\mu (t)y^{n}$$ Divide by $y^n \mu ^{n}$: $$\dfrac {1}{(\mu (t) y)^n} \frac{d}{dt} {(\mu (t) y)} = b(t)\mu^{1-n} (t) $$ $$\dfrac {d {(\mu (t) y)} }{(\mu (t) y)^n} = b(t)\mu^{1-n} (t)dt $$ Integrate both sides. $$\int \dfrac {d(\mu (t) y)}{(\mu (t) y)^n} = \int b(t)\mu^{1-n} (t) dt$$