How to Solve a Differential Equation for Water Flow in a Cylindrical Tank

[vorse]

Homework Statement

Water is pumped into a cylindrical tank with cross section area A at a constant rate k, and
leaks out through a hole of area a in the bottom of the tank at the rate
αa (2gh(t))^1/2
where g is the acceleration due to gravity, h(t) is the depth of water in the tank at time t,
and α is a constant with 0.5 ≤ α ≤ 1.0. It follows that
lim h(t)
t→∞

Homework Equations

A h'(t) = k-αa (2gh(t))^1/2

The Attempt at a Solution

all i got to is [ A/ k-αa (2gh(t))^1/2 ] dh = dt

I can't seem to solve the differential equation to get h(t)

 

[tiny-tim]

Hi vorse!

(shouldn't your "a" should be the same as "A"? oh, and have a square-root: √ )

The LHS is of the form (P + Q√h)dh … so just integrate it.

 

[vorse]

A and a are different; they are both constant however, so it shouldn't matter much in the integration;


"The LHS is of the form (P + Q√h)dh … so just integrate it." what do you mean by this form?


If i let u = P+Q√h
then du = 1/2Q(h)^-1/2 right? and this substitution doesn't work. I don't have anything to substitute for the du; can you clarify a little bit?

 

[Mark44]

A and a are different; they are both constant however, so it shouldn't matter much in the integration;


"The LHS is of the form (P + Q√h)dh … so just integrate it." what do you mean by this form?


If i let u = P+Q√h
then du = 1/2Q(h)^-1/2 right? and this substitution doesn't work. I don't have anything to substitute for the du; can you clarify a little bit?

(P + Q√h)dh = Pdh + Qh^1/2dh
Can't you integrate these two expressions without resorting to a substitution?

 

[vorse]

well, I tried separating the denominator into the form f(x) = A/cx+d +B/dx+e, etc... i think it's called partial separating integration or something, not sure, but that didn't work out. I think I'm here is because I don't know how to integrate the following equation.

 

[tiny-tim]

well, I tried separating the denominator into the form f(x) = A/cx+d +B/dx+e, etc... i think it's called partial separating integration or something, not sure, but that didn't work out. I think I'm here is because I don't know how to integrate the following equation.

What denominator?

I don't see a fraction.

 

[vorse]

all i got to is [ A/ k-αa (2gh(t))^1/2 ] dh = dtsee the A is divided by ( k-αa (2gh(t))^1/2)

so, in a clearer way to write it [A / ( k-αa (2gh(t))^1/2)]dh = dt

btw, what programs are out there where I can type math equations on the comp?