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

[seen]

could some1 help me with the working of this question. it's for homework and i would help if i knew how to come to the answer in the back of the book.

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

 

[HallsofIvy]

Surely your textbook (not to mention your teacher!) has talked about "separating variables"!

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

"Separate" that into differential form with y on one side of the equation and x on the other:
(7- 4y)dy= 2dx.
Integrating: 7y-2y²= 2x+ k.

Now think of that as the quadratic equation 2y²- 7y+ (2x+k)= 0 and solve for y using the quadratic formula. (The "C" in Ay²+ By+ C is 2x+k.)

 

[seen]

thank you very much

this is a great forum

very responsive