Help with basic integration?

[EtherealMonkey]

This is the integral that I am having trouble with
\frac{dh}{dt} = -0.064*\sqrt{h}

I hope you will kindly nudge me to the point where I have made a mistake
I did this problem by hand (and can scan and present if anyone is interested.)
Anyway, I used Maple to check my work and found that I must have made a mistake during integration because I am missing a term.

Here is my work:

\int \frac{1}{\sqrt{h}} dh = -0.064 \int dt

Using integration by substitution:

u = \sqrt{h}

du = \frac{1}{\sqrt{h}} dh

\int \frac{1}{\sqrt{h}} dh \rightarrow 2\int \frac{u}{u} du \rightarrow 2*\sqrt{h}

After integration by substitution and the simple integration on the RHS:
2*\sqrt{h} = -0.064*t+C_{1}

Putting the equation in General form:
\sqrt{h}+0.032*t=C_{1}

The problem is occurring before this point (I believe)
h = -0.001024*t^{2}+C_{1}

This was an initial value problem with h(0) = 10;
But, the problem occurs before this point (as I stated above).
Because, the general solution of the IVP in Maple includes a single term in t, where the solution I found below - does not.

Using h(0)=10;
h = -0.001024*t^{2} + 10;

The answer I found with Maple (for my own reference, I suppose
h(t) = \frac{16}{15625}*t^{2}-\frac{8}{125}*t*\sqrt{10}+10

[Bohrok]

After integration by substitution and the simple integration on the RHS:
2*\sqrt{h} = -0.064*t+C_{1}

Putting the equation in General form:
\sqrt{h}+0.032*t=C_{1}

The problem is occurring before this point (I believe)
h = -0.001024*t^{2}+C_{1}

After you get √h by itself in the middle equation, square both sides entirely.

[Gregg]

h=(\frac{16}{125})^2(t+C)^2

[EtherealMonkey]

Thanks all.

Got it.