Question

The point (1,4) lies on the graph of an equation y=f(x) for which dy/dx=4 sqrt(xy) where x>= 0 and y >= 0. When x = 0 the value of y is:

Answer 1

I was thinking of doing a differential equation setup, is this the right approach?

Answer 2

Something like this: \[\int\limits_{}^{}dy \div \sqrt{y} = \int\limits_{}^{}\sqrt{x}dx\]

Answer 3

you are missing 4 but close

Answer 4

\[\int\limits\limits_{}^{}dy \div \sqrt{y} = \int\limits\limits_{}^{}4 \sqrt{x}dx\]

Answer 5

Have you integrated yet?

Answer 6

you are suppose to find y by evaluating the above
then find the constant by using the point (!,4)

then you can find y for whatever x they want where x>0 of course

Answer 7

Thanks @freckles !! How could I forget something as silly as that 4...and to think I had even put the 4 onto my paper.. I completely forgot to include it in the integration. Thank you very much

Answer 8

what did you get for C?

Answer 9

I got C = 2/3 but I'm starting to see this answer is wrong..

Answer 10

yep it is a bit off