- #1

- 2

- 0

it says to find the general solution (y=...)

given y'(x) = 2/(7-4y)

...

how do u get to this:

y = 7/4 +/- √(k-x)

that is: seven over four plus or minus the squareroot of k minus x

k being another constant

please help if u can

thanx

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter seen
- Start date

- #1

- 2

- 0

it says to find the general solution (y=...)

given y'(x) = 2/(7-4y)

...

how do u get to this:

y = 7/4 +/- √(k-x)

that is: seven over four plus or minus the squareroot of k minus x

k being another constant

please help if u can

thanx

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 967

y'= 2/(7- 4y) is the same as [itex]\frac{dy}{dx}= \frac{2}{7-4y}[/itex]

"Separate" that into differential form with y on one side of the equation and x on the other:

(7- 4y)dy= 2dx.

Integrating: 7y-2y

Now think of that as the quadratic equation 2y

- #3

- 2

- 0

thank you very much

this is a great forum

very responsive

this is a great forum

very responsive

Share:

- Replies
- 1

- Views
- 2K

- Replies
- 2

- Views
- 8K