Separable Differential Equation

  • Thread starter crm07149
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  • #1
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Homework Statement


The volume V of water in a particular container is related to the depth h of the water by the equation dV/dh below. If V = 0 when h = 0, find V when h = 4.


Homework Equations


dV/dh = 16 sqrt(4-(h-2)2)

V(0) = 0
V(4) = ?


The Attempt at a Solution



I have not gotten the ODE in a form that I can integrate using separation of variables. I have gotten to the point dV/dh = 16 sqrt(-h2+4h) but this doesn't help much.
 
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Answers and Replies

  • #2
jks
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Unless you miswrote the equation, you don't need to use separation of variables in this problem, since the right side is an equation of 1 variable. But you can do it like this:

dV/dh = √(4-(h-2)²)

move the dh over
∫dV = ∫√(4-(h-2)²) dh

V = (integrate that expression with respect to h) + C
 
  • #3
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The problem was from my textbook under the "separable equations" section so I listed it as that.

Integrating the expression √(4-(h-2)²) is what I was having trouble with. I tried u substitution, integration by parts, and other methods for awhile. Am I missing something obvious to integrate the expression?
 
  • #4
35,236
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I would try a trig substitution. To make things a little simpler, I would make an ordinary substitution u = h - 2, and then the trig substitution.

I usually draw a right triangle, with t being one of the acute angles, hypotenuse 2 and opposite (to angle t) side u. So the substitution here is sin t = u/2.

Don't forget that you'll need to undo two substitutions when you finally evaluate your antiderivative.
 

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