Differential equation- where is my mistake?

"Water flows out of a tank such that the depth \(h\) metres of water in the tank falls at a rate which is proportional to the square root of \( h \) "

Show that the general solution of the differential equation may be written as \( h = (c-kt)^2 \)

However, I'm getting something different. I have a 2 in there somewhere. Bear with me as i figure out how to show you what I did with tex, but please show me what I'm doing wrong! Thanks in advance, Fudge.

 

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so, \( \frac{dh}{dt} = -k\sqrt h\)

and \( \frac{1}{\sqrt h} dh = -k dt \)

intergrating both sides, i get \( 2 \sqrt h = -kt + c \)

if only this 2 wasn't there, it would be perfect. where did i go wrong?

 

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aaah, i had that idea... i thought there was something i was missing but alright, thank you!