Question

HARD time understanding why an equation is not seperable.

As an example, if the integral of Y dx = xy + c, then why does the derivative of xy = y + x dy/dx ?

Also... this textbook says that dy/dx=y + sinx is not seperable. If I take the integral of both sides with respect to dx I get :
integral ( dy/dx * dx) = integral (y + sinx) dx which becomes
y=yx -cos(x).

Im pretty sure my math for the integration is right, yet like the integral of y = yx, it just doesnt reintegrate correctly. Reintegrating becomes :
dy/dx=y+xdy/dx+sin(x)

What am I missing ? Its driving me crazyyyy

Answer 1

You ca can't integrate respect to x bouth sides of dy/dx=y + sinx, because y is actualy y(x), and you , at this point at least, have no idea how this function looks like.

Answer 2

separable means that you can separate it into F(x) dx = G(y) dy

Answer 3

Y dx = xy + c
^^
you cant factor this into (x) and (y) parts that readily

Answer 4

ok I am willing to accept the fact that... You cant integrate it because you dont know what y really is, it could be x^100e^y or whatever crazy.

but then how come wolfram alpha says : integral of y dx = yx, but the derivative of yx is not y?

Answer 5

you need to notate that a little better ...

Answer 6

amistre : i know the RULE is both have to be seperated... the thing is I want to know why. I dont want to just know the rule.

Answer 7

becasue in order to separate it up and apply a separation method ... you have to be able to separate it.

Answer 8

\(\int ydx=yx+\phi(y)\), that's how it should be

Answer 9

wow. so anytime you integrate y with respect to x... that means its not + c but its + a function ?

Answer 10

there might be some blurring between PDEs and ODEs

Answer 11

to clear the things, jst follow what @amistre64 said befor:
separable means that you can separate it into F(x) dx = G(y) dy

Answer 12

well. I can move forward that way, but not knowing what im doing wrong here, to me, messes up my whole foundation. I really appretiate both of your help though.

Answer 13

\[\int\limits y(x)~ dx = Y(x) + C\]
which is what an ordinary differential equation tends to represent

Answer 14

yes, just:
\(\int ydx=xy+C(y)\)

Answer 15

this is in case if you treat y as just another variable

Answer 16

\[\frac{d}{dx}[xy(x)]\to y(x)+x\frac{dy}{dx}\]or
\[g(x,y(x),\frac{dy}{dx})=y(x)+x\frac{dy}{dx}\]
and thats what you are lookinng at when integrating

Answer 17

\[g(x,y(x),\frac{dy}{dx})=y(x)+x\frac{dy}{dx}\]
\[g(x,y(x),\frac{dy}{dx})-y(x)=x\frac{dy}{dx}\]
\[\left[g (x,y(x),\frac{dy}{dx})-y(x)\right]dx~=x~dy\]

Answer 18

that just doesnt look like a very nice "separation" to me

Answer 19

is it because the\[ \int\limits\limits g(x,y(x),\frac{dy}{dx}) dx = \int\limits \frac{d}{dx}xy(x)dx - \int\limits xy(x)\] and that last integral would be with respect to nothing... so thats why it wouldnt work?

Answer 20

that - should be an = sign. basically im saying, it doesnt work because the d/dx will cancel with another dx and then youd have an intergral with respect to nothing.

Answer 21

it would prolly be best to simplify it as:

\[\left[g (x,y(x),\frac{dy}{dx})-y(x)\right]dx~=x~dy\]
\[since:~g(x,y(x),\frac{dy}{dx})=\]
\[\left[y(x)+x\frac{dy}{dx}-y(x)\right]dx~=x~dy\]
\[\left[x\frac{dy}{dx}\right]dx~=x~dy\]
\[x~dy~=x~dy\]
which seems rather trivial: but\[\int x~dy=xy+f(y)\]

Answer 22

let f(y) = 0 to get back to your inital setup if need be