A simple differential equation

[John O' Meara]

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.

 

[dextercioby]

Just make $B-C\sqrt{x} =t$

 

[John O' Meara]

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.

 

[John O' Meara]

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.

 

[John O' Meara]

I was thinking someone might have a second opinion on what I did originally, i.e., #1. Thanks for the help.