I am trying to solve a non-exact differential equation
$\frac{s-b}{s}ds – \frac{b(2\beta-b)}{8{\beta}^2}db =0$
where $\beta >0$ is a constant. Boundary conditions are given by $b(s=0)=0$.
A reformulation of the above equation using $s =\phi(b)$ yields
$\frac{d ln(F(\phi(b))}{db} = \frac{g(b)}{\phi(b)-b}$.
I am interested in finding an equation for $\phi(b)$.
How should I go about it? Thank you very much for your help.
Best Answer
I have tried the integrating factors of the form $P(s,b)$ and $e^{P(s,b)}$, where $P(s,b)$ is an arbitrary polynomial of $s$ and $b$. It is relatively easy to show that neither satisfies the requirements. To shorten the writeup, I will multiply your problem by $s$ and introduce two constants $A$ and $B$. So, ignoring the case $s=0$, original problem is equivalent to
$$(s-b)ds + (Abs + Bb^2s)db = 0$$
We will be looking for solutions $F(s,b)$ that form a full derivative, namely
$$dF(s,b) = \frac{\partial F}{\partial s}ds + \frac{\partial F}{\partial b}db = 0$$
Example 1: $\mu(s,b) = P(s,b)$, a finite degree polynomial
Multiplying the original equation by $\mu$ and comparing it to the full derivative we obtain
$$\frac{\partial F}{\partial s} = \mu(s-b) \;\; \mathrm{and} \;\;\; \frac{\partial F}{\partial b} = \mu (Abs + Bb^2s)$$
We will integrate both equations to obtain $F$, and then equate them to get an expression for $\mu$. We get
$$\int_s \mu(s,b)(s-b)ds + K_1(b) = \int_b \mu(s,b) (Abs + Bb^2s)db + K_2(s)$$
where the integration constants $K_1$ and $K_2$ may depend on the variables that were not integrated over. We observe that all polynomials on the LHS of the equation will be lower order than on the RHS. For example if the degree of $\mu$ is 10, the degree of the LHS is 12 and RHS is 14. So there will always be a term of the form $s^ib^j$ that will not cancel out.
Example 2: $\mu(s,b) = e^{P(s,b)}$
First step is the same, as before. For the 2nd step it is a bit easier to differentiate than to integrate. We will assume that $F$ well-behaves, namely
$$\frac{\partial^2 F}{\partial b \partial s} = \frac{\partial^2 F}{\partial s \partial b}$$
Substituting the first order partial derivatives, we get
$$\frac{\partial}{\partial b} [\mu(s-b)] = \frac{\partial}{\partial s} [\mu(Abs + Bb^2s)]$$
Simplifying, and substituting $\mu = e^P$, we get
$$e^P \bigl[\frac{\partial P}{\partial b}(s-b) - 1 \bigr] = e^P \bigl[(Abs + Bb^2s)\frac{\partial P}{\partial s} + (Ab + Bb^2) \bigr]$$
Cancelling out $e^P$ and moving last term to the LHS, we get
$$\frac{\partial P}{\partial b}(s-b) - 1 - (Ab + Bb^2) = (Abs + Bb^2s)\frac{\partial P}{\partial s} $$
As you can see, the very same problem persists. The polynomial degree of mixed terms of RHS is greater than that of LHS, so equality can not hold.
Conclusion:
I do not think that there is a good strategy for guessing integrating factors if standard choices do not work. Depending on application, series expansion or numerical solution may be of interest
Tough luck...