Solving Dy/Dx = 3-x/y+4: Integral of Both Sides

[KingCrimson]

there was this curve where
Dy/Dx = 3-x/y+4
so he cross multiplied the equation
as (3-x)Dx=(y+4)Dy , then he proceeded to integrate the function .
i don't know if this is correct or not , i mean Dy/dx is just a notation , how can you treat as if it was just a normal fraction ?
i thought he treated them as if they were differentials
then he took the integral of both sides , and he treated Dx as if it was just a notation that means integral with respect to X and so with Dy

 

[AlephZero]

You are right, this does look like an abuse of the notation, but that's the way it is usually written. You can write it out more formally:
$$\begin{aligned}\frac{dy}{dx} &= \frac{3-x}{y+4}\\
(y+4)\frac{dy}{dx} &= 3-x\\
\int(y+4)\frac{dy}{dx}dx &= \int(3-x)dx\\
\int(y+4)dy &= \int(3-x)dx
\end{aligned}$$
But life is too short to write out the solution to every problem in all that detail.

 

[KingCrimson]

You are right, this does look like an abuse of the notation, but that's the way it is usually written. You can write it out more formally:
$$\begin{aligned}\frac{dy}{dx} &= \frac{3-x}{y+4}\\
(y+4)\frac{dy}{dx} &= 3-x\\
\int(y+4)\frac{dy}{dx}dx &= \int(3-x)dx\\
\int(y+4)dy &= \int(3-x)dx
\end{aligned}$$
But life is too short to write out the solution to every problem in all that detail.

in 3rd > 4th step
how did (dy/dx)dx become dy ?

 

[DataGG]

in 3rd > 4th step
how did (dy/dx)dx become dy ?

You've $dx/dx$ on the right side. One of the $dx$ goes to the left side and voila

 

[HallsofIvy]

in 3rd > 4th step
how did (dy/dx)dx become dy ?

That's an application of the chain rule. You can also define "differentials":
f'(x)= dy/dx is NOT a fraction but IS the limit of a fraction:
$$\frac{dy}{dx}= \lim_{h\to 0}\frac{f(x+ h)- f(x)}{h}$$
So it has properties like a fraction that can be proven by going back before the limit, using the fraction property, then taking the limit.
We can make use of the fact that the derivative is like a fraction by defining "differentials":
if y= f(x), so that y'= f'(x), define dy= f'(x)dx where dx and dy are "differentials".

For example, if y is a function of x and x is a function of t, we can write the "chain rule" as
$$\frac{dy}{dt}= \left(\frac{dy}{dx}\right)\left(\frac{dx}{dt}\right)$$
That is commonly proved by, as I said, going back before the limits to write
$$\frac{y(x(t)+ h(t))- y(x(t))}{h(t)}\frac{h(t+ j)- h(t)}{j}$$
and cancelling. But with "differential" notation we can think of it as actually cancelling the "dx" terms.

 

[KingCrimson]

That's an application of the chain rule. You can also define "differentials":
f'(x)= dy/dx is NOT a fraction but IS the limit of a fraction:
$$\frac{dy}{dx}= \lim_{h\to 0}\frac{f(x+ h)- f(x)}{h}$$
So it has properties like a fraction that can be proven by going back before the limit, using the fraction property, then taking the limit.
We can make use of the fact that the derivative is like a fraction by defining "differentials":
if y= f(x), so that y'= f'(x), define dy= f'(x)dx where dx and dy are "differentials".

For example, if y is a function of x and x is a function of t, we can write the "chain rule" as
$$\frac{dy}{dt}= \left(\frac{dy}{dx}\right)\left(\frac{dx}{dt}\right)$$
That is commonly proved by, as I said, going back before the limits to write
$$\frac{y(x(t)+ h(t))- y(x(t))}{h(t)}\frac{h(t+ j)- h(t)}{j}$$
and cancelling. But with "differential" notation we can think of it as actually cancelling the "dx" terms.

oh
then the dX in the end of every integral is not then just a sign to know what are you integrating with respect to , it is actually a differential .. hmmm.. thanks :D

 

[paisiello2]

Maybe an infinitesimal quantity would be a better description. A differential would be d/dx.

 

[bcrowell]

dy/dx can be considered to be a ratio of two infinitesimal numbers. This is the way calculus was traditionally understood until about 1890, and Abraham Robinson provided some theoretical justification around 1960. There is a nice freshman calc book by Keisler that takes this approach, and it's free online: http://www.math.wisc.edu/~keisler/calc.html

 

[Mark44]

Maybe an infinitesimal quantity would be a better description. A differential would be d/dx.

No. d/dx is a derivative operator. It is not a differential.