How rigorous is multiplying both sides of an eqaution for the differential of a function?

69 Views Asked by At

I have to solve this equation:

$$ -C_0 f + \frac{1}{2}f^2 +\frac{d^2 f}{d X^2}=A $$ where $C_0$ and $A$ are two real nonzero constant; $f:\mathcal{R}\to \mathcal{R}$ I have seen that the person who solved this equation made the following operation: he multiplyed both sides for $\frac{df}{dX} $ and then simplified the $dX$ in order to integrate in $df$:

$$ -C_0 f \frac{df}{dX}+ \frac{1}{2}f^2\frac{df}{dX} +\frac{d^2 f}{d X^2}\frac{df}{dX}=C_{\text{int}}\frac{df}{dX} $$

then

$$ -C_0 f df+ \frac{1}{2}f^2 df +\frac{d^2 f}{d X^2}df=C_{\text{int}} df $$

and then he integrated in df. Colud anyone explain me why this is rigorous. (My problem is that I cannot understand how it is possible to threat differential quantities as non zero number)

1

There are 1 best solutions below

0
On BEST ANSWER

When you see that notation, think in terms of integrals: $$ \int f(g)\,dg = \int f(g(t))\frac{dg}{dt}\,dt $$