Question

Hi everyone! Using u=y-2x+3, Solve
dy/dx=2+(root(y-2x+3))...Anyone? Kinda lost here :o)

Answer 1

I get du/dx = -2dy/dx and -1/2du/dx=dy/dx...I have tried a bunch of stuff but can't figure it out.

Answer 2

@jim_thompson5910

Answer 3

@ganeshie8

Answer 4

u=y-2x+3

y is a function of x here, so the derivative of y with respect to x will not be 0

Answer 5

\[u = y-2x+3\]
\[\implies \frac{du}{dx} = \frac{dy}{dx} - 2\]

Answer 6

how do you do that? I thought you take the partial multiplied by dy/dx?

Answer 7

I got -2dy/dx

Answer 8

what am I missing here?

Answer 9

@SinginDaCalc2Blues i have tried to solve it for you, if you could get it.

Answer 10

What do you mean?

Answer 11

i attached my solution http://assets.openstudy.com/updates/attachments/54db04aee4b0ffc24cc39693-ribhu-1423640318915-photo164.jpg

Answer 12

okay, let me take a look! :O)

Answer 13

sorry i messed it in the last step @SinginDaCalc2Blues

Answer 14

I messed the integration, the final answer would be \[2\sqrt{y-2x+3}\] = x+c

Answer 15

@SinginDaCalc2Blues just see it

Answer 16

It looks like it may be wrong...I'm pretty sure the integral of 1/root(u) = 2root(u), so I think the final solution should be x - 2root(y-2x+3) + c =0

Can you check your work to make sure?

Answer 17

yeah thats what i wrote after correcting myself and i did admit that i messed up the integration.

Answer 18

Oh, I didn't see your post. Thank you very much for the help! :o)

Answer 19

u got my handwriting clearly?? lol

Answer 20

yes thanks