# Help with basic integration?

1. Sep 13, 2009

### 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
$$h(t) = \frac{16}{15625}*t^{2}-\frac{8}{125}*t*\sqrt{10}+10$$

2. Sep 13, 2009

### Bohrok

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

3. Sep 13, 2009

### Gregg

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

4. Sep 13, 2009

Thanks all.

Got it.