Thread: Cannot Differentiate ydx?

Hi,

In class today, we worked out the following question. However, I don't understand why the teacher said we can't differentiate ydx because it will be equal to zero, as well as where the 'd' disappeared to (please see my comments in red below)

Question: Solve for the following differential equation for the original equation: $\displaystyle \frac{dy}{dx} = \frac{-y}{4+x}$

$\displaystyle \frac{dy}{dx} = \frac{-y}{4+x}$

$\displaystyle \int (4 + x)dy = \int -ydx$

$\displaystyle \int 4dy + xdy = \int -ydx$ (you cannot differentiate ydx, ydx = 0)

$\displaystyle \int 4y + xy = 0 + C$ (where did the d go? Why not $\displaystyle 4xdy + x^2dy$)?

$\displaystyle 4y + xy + C = 0$

Originally Posted by sparky

Hi,

In class today, we worked out the following question. However, I don't understand why the teacher said we can't differentiate ydx because it will be equal to zero, as well as where the 'd' disappeared to (please see my comments in red below)

Question: Solve for the following differential equation for the original equation: $\displaystyle \frac{dy}{dx} = \frac{-y}{4+x}$

$\displaystyle \frac{dy}{dx} = \frac{-y}{4+x}$

$\displaystyle \int (4 + x)dy = \int -ydx$

$\displaystyle \int 4dy + xdy = \int -ydx$ (you cannot differentiate ydx, ydx = 0)

$\displaystyle \int 4y + xy = 0 + C$ (where did the d go? Why not $\displaystyle 4xdy + x^2dy$)?

$\displaystyle 4y + xy + C = 0$

I would argue it this way:
$\displaystyle \displaystyle \frac{dy}{dx} = -\frac{y}{x + 4}$

$\displaystyle \displaystyle (x + 4) dy = -y dx$

$\displaystyle \displaystyle x dy + 4 dy = -y dx$

$\displaystyle \displaystyle (x dy + y dx) + 4 dy = 0$

$\displaystyle \displaystyle d(xy) + 4 dy = 0$

$\displaystyle \displaystyle \int d(xy) + 4 \int dy = C$

xy + 4y = C

etc.

-Dan

To separate the variables, you need the $\displaystyle \displaystyle y$ terms to be with the $\displaystyle \displaystyle dy$ and the $\displaystyle \displaystyle x$ terms to be with the $\displaystyle \displaystyle dx$.

So this should really be (if you're going to use the lazy notation...)

$\displaystyle \displaystyle \frac{dy}{dx} = -\frac{y}{4+x}$

$\displaystyle \displaystyle \frac{dy}{y} = -\frac{dx}{4+x}$

$\displaystyle \displaystyle \int{\frac{dy}{y}} = -\int{\frac{dx}{4+x}}$...


Personally, I prefer using the "reverse chain rule" approach (you end up with the same integral but is more mathematically correct, since it doesn't treat $\displaystyle \displaystyle \frac{dy}{dx}$ as a fraction)...

$\displaystyle \displaystyle \frac{dy}{dx} = -\frac{y}{4+x}$

$\displaystyle \displaystyle \frac{1}{y}\,\frac{dy}{dx} = -\frac{1}{4+x}$

$\displaystyle \displaystyle \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = -\int{\frac{1}{4+x}\,dx}$

$\displaystyle \displaystyle \int{\frac{1}{y}\,dy} = -\int{\frac{1}{4+x}\,dx}$, which gives us the same integrals.

Go from here...

Originally Posted by Prove It

Personally, I prefer using the "reverse chain rule" approach (you end up with the same integral but is more mathematically correct, since it doesn't treat $\displaystyle \displaystyle \frac{dy}{dx}$ as a fraction)...

Not that it matters in the end (let's not be dogmatic)... but, you might seem, at least, to be cancelling a fraction when you do this:

$\displaystyle \displaystyle \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = -\int{\frac{1}{4+x}\,dx}$

$\displaystyle \displaystyle \int{\frac{1}{y}\,dy} = -\int{\frac{1}{4+x}\,dx}$, which gives us the same integrals.

One way to 'correct' such an impression is of course by replacing d with w.r.t., ... or else... ooh, I don't know, how about a diagram -



i.e.



... where (key in spoiler)

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

What's topsquark doing then? If I'm reading it right, using the fraction notation to get us up onto the integral level via the product rule, which in this case at least avoids having to go in and out of logs. Which is useful. So, re-writing this

Originally Posted by topsquark

$\displaystyle \displaystyle (x + 4) dy = -y dx$

$\displaystyle \displaystyle x dy + 4 dy = -y dx$

$\displaystyle \displaystyle (x dy + y dx) + 4 dy = 0$

as

$\displaystyle \displaystyle (x + 4) \frac{dy}{dx} = -y$

$\displaystyle \displaystyle x \frac{dy}{dx} + 4 \frac{dy}{dx} = -y$

$\displaystyle \displaystyle x \frac{dy}{dx} + y + 4 \frac{dy}{dx} = 0$


we can present his version like this...



i.e.



... where

Spoiler:

... is the product rule straight continuous lines differentiating downwards (integrating up) with respect to t.

I have to admit that all that didn't enable me to decide whether Sparky's teacher was doing either of these things or something else.

Originally Posted by sparky

$\displaystyle \int 4dy + xdy = \int -ydx$ (you cannot differentiate ydx, ydx = 0)

Cannot integrate? Or were they referring to an integration by parts / product rule type maneouvre? Could be, because the next line is where topsquark and the second pair of diagrams (above) arrived at. Apart from

$\displaystyle \int 4y + xy = 0 + C$ (where did the d go?)?

I think they just left an integral sign in there by mistake, after integrating.

(Why not $\displaystyle 4xdy + x^2dy$)?

I don't follow. Are you integrating with respect to x, perhaps?

Anyway, I hope this helps, or is of interest.

__________________________________________

Don't integrate - balloontegrate!

Balloon Calculus: standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!