A simple differential equation

John O' Meara
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The depth of water x at any time t in a tank is given by
A\frac{dx}{dt}= B-c\sqrt{x}\\,
where A, b and C are constants. Initially x=0; show that C^2t=2AB\ln{\frac{B}{B_C\sqrt{x}}}-2AC\sqrt{x}\\
To solve this let x= y^2\\.
Then \int \frac{A}{B-C\sqrt{x}} = \int dt = \int \frac{2Ay}{B-Cy}dy \\.
Let u=B-Cy, therefore y=\frac{B-u}{c}\\, therefore the integral is
-2A \int \frac{B-u}{C^2u} du = \frac{-2A}{C^2} \int \frac{B-u}{u} du \\.
which = \frac{-2AB}{C^2} \int \frac{1}{u} du + \frac{2A}{C^2} \int du \\.
Therefore we have \frac{-2AB}{C^2}\ln(u) + \frac{2A}{C^2}u \\. We have \frac{-2AB}{C^2} \ln{B_Cy} + \frac{2A}{C^2}(B-Cy) \\.
Which implies C^2t = -AB\ln{\frac{B}{C\sqrt{x}}} + 2A(B-C\sqrt{x})\\.
As you can see I get a different answer. I integrated by parts also to get a different result. Please help. Thanks.
 
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I tried solving it dextercioby's way I used y instead of his t, and I got the wrong answer, I got the integral = \frac{ -AB}{2C} \int dy + \frac{AB}{2C} \int y dy \\. Where \sqrt{x} = \frac{B-y}{C} \\ and dx=-dy \frac{B-y}{2C} \\. Thanks for helping.
 
Sorry I wrote down the wrong answer to dextercioby's method of solving this. What I actually got was the following: t = \frac{-AB}{2C} \int \frac{1}{y}dy + \frac{A}{2C} \int dy. Which is not the right answer.
 
I was thinking someone might have a second opinion on what I did originally, i.e., #1. Thanks for the help.
 
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