- #1

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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

- 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

- 41,833

- 961

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

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thank you very much

this is a great forum

very responsive

this is a great forum

very responsive

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