I am differentiating a first order DE. I want to use the substitution y = ux and the differential $dy = udx +xdu$. I thought $dy = u + xdu$ because $dx = x$. I thought I am differentiating with respect to x.
I would appreciate any feedback. thank you
On
You wrote : " I thought $dy = u + xdu$ because $dx = x$. "
This is a big mistake because $dx$ is an infinitesimal, that is "something infinitely small" and $x$ is a finite value, hence not infinitely small in general (except when $x=0$).
Thus $dx=x$ is a non-sens.
May be, you can better understand this on the "physical" sens :
$dx$ is then understood as an infinitely small variation of $x$. So, one cannot confuse $x$ with it's variation $dx$.
$du$ is understood as an infinitely small variation of $u$.
$dy$ is understood as an infinitely small variation of $y$.
The rule says that the variation of $y=$($u$ multiplied by $x$) is the sum of two infinitesimal variations :
$$dy=u\,dx+x\,du$$
How can this rule be understood on the "physical" sens ?
When $u$ is increased to $(u+du)$ and $x$ is increased to $(x+dx)$ the product $y=ux$ increases to $$y+dy=(u+du)(x+dx)$$ $$y+dy=ux+u\,dx+x\,du+du\,dx$$ $$y+dy=y+u\,dx+x\,du+du\,dx$$ $$dy=u\,dx+x\,du+du\,dx$$ $du\,dx$ is the product of two infinitesimals, which order of magnitude is infinitely smaller that the infinitesimals themselves. Thus, it is negligible. $$dy=u\,dx+x\,du$$
On
Derivatives are obtained from differentials
$$dy = udx +xdu$$
by direct division as if they are algebraic ... numerator/denominator in a fraction.
$$\dfrac{dy}{dx} = u + x \dfrac{\,du}{dx}$$
Derivatives are divided quantities or quotients of infinitesimally small (called differentials) bits when the function has a slope that varies.
$$y=\underbrace{ux}_{\color{red}{\text {prodct of 2 function}s}}$$ Differentiate $$\frac {dy}{dx}=x\frac {du}{dx}+u=u'x+u$$ Multiply by dx $$\frac {dy}{dx}=x\frac {du}{dx}+u$$ $$\frac {dy}{dx}dx=x\frac {du}{dx}dx+udx$$ $${dy}=x {du}+udx$$
Like when you differntiate any product of two functions $$(fg)'=f'g+fg'$$