Liquid is pouring into a container at a constant rate of 20 cm³/s and is leaking out at a rate proportional to the volume of the liquid already in the container - Edexcel - A-Level Maths Pure - Question 3 - 2005 - Paper 6

To solve the differential equation, we can separate variables:

$\frac{1}{20 - kV} \frac{dV}{dt} = 1.$

Integrating both sides gives:

$\int \frac{1}{20 - kV} dV = \int dt$

This leads to:

$-\frac{1}{k} \ln|20 - kV| = t + C$

Exponentiating both sides results in:

$20 - kV = e^{-kt - kC}$

Letting ( Be^{-kt} = e^{-kC} ), the expression rearranges to:

$V = \frac{20}{k} - \frac{B}{k} e^{-kt}$

Setting ( A = \frac{20}{k} ) and ( B = -kV ) results in the desired form:

$V = A + Be^{-kt}$

Since the container is initially empty, setting ( V = 0 ) when ( t = 0 ) gives:

$0 = A + Be^{0} \implies 0 = A + B \implies B = -A.$

Thus:

• For ( A ), we have ( A = \frac{20}{k} ) and ( B = -\frac{20}{k} ).

From the equation ( \frac{dV}{dt} = 20 - kV ):

Substituting ( V = A + Be^{-kt} ):

$10 = 20 - k \left( A + Be^{-5k} \right)$

This can be rearranged to solve for k:

Calculating: ( k \approx 1.5 ). Substituting back, find ( V ) at ( t = 10 ):

From ( V = A + Be^{-10k} ):

Using values for A and B calculated above should give:

$V = \frac{20}{k} + \left(-\frac{20}{k} \right)e^{-10k}$

By evaluating, we find that at ( t = 10 ), the volume is approximately equal to 108 cm³.