Transforming ODEs into exact equations. I want some examples of ODEs that can only be solved by transforming them into exact equations. They shouldn't be solvable with; Direct integration, separation of variables, manipulating a reverse product rule, homogenous linear equation method, Bernoulli's equation, Laplace transform. What's the general formula for these kinds of equations?
This is an example DE that was solved by transforming it to an exact and taking it from there. Definitely not the best approach, as you can manipulate a reverse product rule or even simpler solve it as a separable one. What I want is a DE that can ONLY be solved analytically by transforming it in an exact one and has no other possible methods. Anyway, here goes;
$(x+3)y'= x y(x) \iff  
-(x y(x))+(x+3)y'(x)= 0$
Let $R(x, y)=-xy$ and $S(x, y)  =  x+3$.
$R(x, y)_y  =  -x \not= 1  =  S(x, y)_x$
To transform it into exact;
$(m(x) R(x, y))_y  =  (m(x) S(x, y))_x    \iff
-(x m(x))  =  m(x)_x (x+3)+m(x) \iff    \frac{m(x)_x}{m(x)}  =  -\frac{x+1}{x+3} \iff \ln(m(x))  =  -x+2 ln(x+3) \iff m(x)  =  \frac{(x+3)^2}{e^x} \iff -\frac{x (x+3)^2 y(x)}{e^x}+(x+3)^3 \frac{y'}{e^x}  =  0$
Let $P(x, y)  =  -e^{-x} x (x+3)^2 y$ and $Q(x, y)  =  e^{-x} (x+3)^3$.
This is an exact equation; $P(x, y)_y  =  -e^{-x} x (x+3)^2  =  Q(x, y)_x$.
Let $f(x, y)$ such that $f(x, y)_x  =  P(x, y)$  and $f(x, y)_y  =  Q(x, y)$
Then, the solution will be given by $f(x, y)  =  c_1$
Ergo $f(x, y)  =   \int -\frac{y x (x+3)^2}{e^x} dx  =  \frac{y (x+3)^3}{e^x}+g(y)$, $g(y)$ is a function of $y$ that we could have lost due to partial differentiation.
And $f(x, y)_y  =  \frac{y (x+3)^3}{e^x}+g(y)_y  =     \frac {(x+3)^3}{e^x}+g(y)_y$
Substituting into $f(x, y)_y  =  Q(x, y)$:
$\frac{(x+3)^3}{e^x}+g(y)_y  =  \frac{(x+3)^3}{e^x}  \iff
g(y)_y  =  0 \iff g(y)  =   \int 0 dy  =  0$
Thus $\frac{y (x+3)^3}{e^x}  =  c_1$
$y(x) = \frac{c_1 e^x}{(x+3)^3}$.
 A: These equations are called equations which demand an integrating multiplier. They are solved by searching integrating multiplier and transforming to total DE (exact, as you call). There is no general form of these equations, but, if you want, you can come up with any cumbersome nonlinear function $F(x,y)$ and a multiplier (this is only one of many possible) $m(x)=x^{\alpha}, \alpha>1$ or $m(y)=y^{\alpha}, \alpha>1$ and obtain an equation in a following way: consider a one-parametric family of curves
$$F(x,y)=C ,$$
multiply by $m(x)$(for instance)
$$m(x)F(x,y)=C ,$$
and differentiate it:
$$\frac{\partial m F}{\partial x}dx+\frac{\partial m F}{\partial y}dy=0 $$
$$\left(x^{\alpha -1}F+x^{\alpha}\frac{\partial F}{\partial x}\right)dx+x^{\alpha}\frac{\partial F}{\partial y}dy.$$
After dividing by $x^{\alpha-1}$ we obtain very cumbersome equation, it can be transformed to what you want, and if $F(x,y)$ is very cumbersome, this equation can't be solved with another method with probability almost 1. Everything depends here on your imagination.
