I am trying to solve the following Bernoulli's equation:
$\frac{dx}{dt}=a(t)x-b(t)x^2$. with $x(0)=x_0$ and $a(t)>0$, $b(t)>0$.
On the way to solution, I make the substitution $u=1/x$ implying $u'=-x^{-2}\frac{dx}{dt}$ to transform it to the following equation in $u$.
$u'+a(t)u-b(t)=0$, which is a linear first order ODE, can be solved via Integrating factor: $e^{\int a(t)dt}$.
Leading to the following:
$\frac{d}{dt}\left[e^{\int a(t)dt)}*u\right]=b(t)e^{\int a(t)dt}$
$\implies u=\frac{1}{e^{\int a(t)dt}}\int b(t)e^{\int a(\tau)d\tau}dt$
Lets call $\int a(t)dt=\alpha(t)$. $\implies u=\frac{1}{e^\alpha}\int b(t)e^{\alpha}dt$.
Integrate by parts:
$\int b(t)e^{\alpha(t)}dt=b\int e^{\alpha(t)}dt-\int b(t)\left[\int e^{\alpha(\tau)}d\tau\right]dt+c$
which means:
$u=\frac{1}{e^{\alpha(t)}}\left[b\int e^{\alpha(t)}dt-\int b(t)\left[\int e^{\alpha(\tau)}d\tau\right]dt \right]$, this leads us to the solution $x(t)$ by substituting back $u=x^{-1}$ and $\alpha=\int a(t) dt$ i.e.,
$x(t)=\frac{e^{\int a(\tau)d\tau}}{\frac{1}{x_0}+b\int e^{\alpha(t)}dt-\int b(t)\left[\int e^{\alpha(\tau)}d\tau\right]dt}$. where $c=1/x_0$.
However, the solution that I needed to come to is the following:
$x(t)=\frac{e^{\int a(\tau)d\tau}}{\frac{1}{x_0}+\int b(t)e^{\int a(\tau)d\tau}dt}$.
So I have one extra term in my solution $b(t)\int e^{\int a dt} $, that emerges from the 'uv' term in integration by parts. I would appreciate if someone can point out where I am doing a mistake?
Best Answer
You already arrive at the solution formula $$ u=e^{-\alpha}\int b(t)e^{\alpha}dt $$ that you write in undetermined integrals. Add integration bounds $$ u(t)=e^{-\alpha(t)}\left(u_0+\int_{t_0}^t b(s)e^{\alpha(s)}ds\right) $$ and reverse the definition of $u$ $$ x(t)=\frac{e^{\alpha(t)}}{x_0^{-1}+\int_{t_0}^t b(s)e^{\alpha(s)}ds} $$ and you get exactly the reference solution formula. No partial integration required.
Of course, if the original also has this mix of integration constant and undetermined integrals, then you are right in questioning this reference solution formula.