Question

is it possible to turn differential equation general solution into differential equation?

Answer 1

differentiate ?

Answer 2

just differentiate?

Answer 3

lol @saifoo.khan shut up :p

@UnkleRhaukus can you provide an example?

Answer 4

yep
the arbitrary constant will go

Answer 5

i sure can give me a second

Answer 6

let's say i have \[x^2y + x + c = 0\]
dow do i turn that into a differential equation?

Answer 7

let's just use my example :S

Answer 8

loooooooooool. @lgbasallote

Answer 9

i think the original equation was
\[(2xy + 1)dx + (2xy)dy = 0\]
or something like that...how do i produce it?

Answer 10

\[(x^2-y^2)\text dx+(2xy)\text d y=0\]
\[\frac{\text d y}{\text dx}=\frac{y^2-x^2}{2xy}\]\[\frac{\text d y}{\text dx}=\frac{\left(\frac yx\right)^2-1}{2\left(\frac yx\right)}\]
\[\text{let } \frac y x = v\]\[y=vx\]
\[\frac{\text d y }{\text d x}=v+x\frac{\text d v}{\text dx}\]\[v+x\frac{\text d v}{\text dx}=\frac{v^2-1}{2v}\]\[x\frac{\text d v}{\text dx}=\frac{v^2-1}{2v}-v\]\[x\frac{\text d v}{\text dx}=\frac{v^2-1-2v^2}{2v}\]\[x\frac{\text d v}{\text dx}=-\frac{1+v^2}{2v}\]\[\frac{2v}{1+v^2}\text dv=-\frac{\text d x}{ x}\]\[\int\frac{2v}{1+v^2}\text dv=\int-\frac{\text d x}{ x}\]\[\ln(1+v^2)=-\ln x + c\]\[\ln(1+v^2)=\ln e^c-\ln x \]\[\ln(1+v^2)=\ln \frac{k}{x}\]\[1+v^2=\frac kx\]\[1+ \left(\frac yx\right)^2 =\frac kx\]
\[x^2+y^2=kx\]

Answer 11

\[x+\frac {y^2}x=k\]
\[\frac{\text d}{\text dx}\left(x+\frac{y^2}{x}\right)=\frac{\text d}{\text dx}\left(k\right)\]\[1-\frac{y^2}{x^2}+2\frac yx\frac{\text dy}{\text dx}=0\]\[1+2\frac yx\frac{\text dy}{\text dx}=\frac{y^2}{x^2}\]\[x^2+2xy\frac{\text dy}{\text dx}=y^2\]\[2xy\frac{\text d y }{\text dx}=y^2-x^2\]
\[\frac{\text d y}{\text dx}=\frac{y^2-x^2}{2xy}\]

Answer 12

that...was....TOTALLY WICKED!!!`

Answer 13

awesome...just awesome :O

Answer 14

you used product rule on the derivative of y^2/x right?

Answer 15

\[x^2y + x + c = 0\]
\[y=cx^{-2}-x^{-1}\]
\[\frac{\text dy}{\text dx}=\frac{\text d}{\text dx}\left(cx^{-2}-x^{-1}\right)\]

Answer 16

hmm what's the derivative of c/x^2?

Answer 17

-2c/x^3?

Answer 18

but since -2C is still a constant it's just c? so c/x^3?

Answer 19

then -1/x becomes 1/x^2

Answer 20

thats right i did use the product rule for \(\frac{\text d}{\text dx}\left(\frac{y^2}{x}\right)\)

Answer 21

i dont think this last one is making sense...

Answer 22

ahh sowhat next

Answer 23

hmm the constant didn't do away like it wasn't ment to

Answer 24

hmm i need to sleep...ill get back to this in the morning -_- lol

Answer 25

maybe this way is better

\[x^2y+x+c=0\]\[x^2y+x=-c\]
\[\frac{\text d}{\text dx}\left(x^2y+x\right) =0\]

Answer 26

using the product rule\[2xy+x^2\frac{\text dy}{\text dx}+1=0\]

Answer 27

\[x^2\frac{\text dy}{\text dx}=-(1+2xy)\]
\[\frac{\text dy}{\text dx}=\frac{-(1+2xy)}{x^2}\]
\[{x^2}{\text dy}=-{(1+2xy)}\text dx\]
\[{(2xy+1)}\text dx+(x^2){\text dy}=0\]

Answer 28

http://www.myspacehippo.com/files/comments/congratulations/MShippo37915.jpg

Answer 29

thank you

Answer 30

lol nice @UnkleRhaukus i just gained new respect for you <tips hat>