Differential Equation Question

Differential equation question?

Lake Erie has a volume of roughly 100 cubic miles, and its equal inflow and outflow rates are 40 cubic miles per year. At year t = 0, a certain pollutant has a volume concentration of 0.05%, but after that the concentration of pollutant flowing into the lake drops to 0.01%. Answer the following questions, assuming the pollutant leaving the lake is well mixed with lake water.

(a) What is the differential equation satisfied by the volume V (in cubic miles) of pollutant in the lake?

(b) What is the volume V (in cubic miles) of pollutant in the lake at time t?

(c) How long (in years) will it take to reduce the pollutant concentration to 0.02% in volume?

Basic Differential Equations Question?

To verify whether y is a solution you need to figure out y'.
y' = (e^t²)'∫e^(-s²)ds + (e^t²)[∫e^(-s²)ds]' + (e^t²)'
y' = 2te^t²∫e^(-s²)ds + (e^t²)[∫e^(-s²)ds]' + 2te^t²
The derivative of an integral is just the function of the integral, so to finish off y'

y' = 2te^t²∫e^(-s²)ds + (e^t²)[e^(-t²)] + 2te^t²
y' = 2te^t²∫e^(-s²)ds + 1 + 2te^t²

Now plug y' into the differential equation

2te^t²∫e^(-s²)ds + 1 + 2te^t² -2t[e^(t²) ∫e^(-s²)ds+e^(t²) ?= 1

2te^t²∫e^(-s²)ds + 1 + 2te^t² -2te^(t²)∫e^(-s²)ds+-2te^(t²) ?=1
{2te^t²∫e^(-s²)ds -2te^(t²)∫e^(-s²)ds} + {2te^t²-2te^(t²)} +1 ?=1
Then 1 = 1

Differential Equations Questions?

Still working on the 1st one, but the 2nd one can be put into the form:

dy/dx + P(x)y = Q(x)

And then an integrating factor can be found to solve. I found that it's more convenient to use y as the independent variable, so all the variables in the above form will be reversed:

y*dx = (2x - 2y^4)*dy
dx/dy = 2x/y -2y^3

dx/dy - 2x/y = -2y^3 ==> P(y) = -2/y

So the integrating factor is 1/y^2

Multiplying by 1/y^2 yields:

1/y^2*dx/dy - 2x/y^3 = -2y Here's where that factor shows it's usefulness, the left side becomes:

d/dy (x/y^2) = -2y Now integrate:

x/y^2 = -y^2 + C, The typical arbitrary constant

x = -y^4 + Cy^2

Of course now getting it into terms of y will be a pain, but I think this is how it's supposed to be solved.

Ok now the 1st one can be solved the same way, but we'll have to first make a substitution.

dy/dx =2y/x - 6y^2
dy/dx -2y/x = -6y^2 , Multiply by 1/y^2

1/y^2*dy/dx -2/xy = -6 , Let v = -1/y , then we have:

dv/dx + 2v/x = -6 , The integrating factor is x^2

x^2*dv/dx +2vx = -6x^2

d/dx(vx^2) = -6x^2 , Intergrate:

vx^2 = -2x^3 + C ==> v = -2x + C/x^2

So, y = 1/(2x - C/x^2)

Calculus...Differential Equation Question?!?!?

The differential equation is question is:

dy/dt = k*y

This is a separable equation that can be rewritten as:

dy/y = k dt

This integrates to:

ln(y) - ln(c) = k*t

where ln(c) is the constant of integration.

Rearranging this, we have that:

ln(y/c) = k*t

y/c = exp(k*t)

y(t) = c*exp(k*t)

dy/dt = k*c*exp(k*t)

You are told that when t = 3, y = 2, and dy/dt = 4 so:

2 = c*exp(k*3)
and
4 = k*c*exp(k*3)

Divide the second equation by the first to get:

4/2 = k = 2

Use this value in the first equation to solve for c:
2 = c*exp(2*3)

c = 2*exp(-6)

The equation for y(t) is now:

y(t) = 2*exp(-6) * exp(2*t) = 2*exp(2t - 6)

When t= 6, y(6) = 2*exp(12-6) = 2*exp(6)

Differential equations circuit question?

Did you want Laplace (s-domain) or time-domain analysis?

Writing Kirchoff's Voltage Law around the circuit:
E = L di/dt + Ri +1/C ∫ i dt
= L di/dt + Ri +1/C q(t)

or in Laplace s-domain
E = sLi + Ri + 1/sC i = (R + sL + 1/sC) i

Worded differential equations question?

"...The rate of increase of the mass, x kg, of X is proportional to the product of the masses of unreacted A and B present at time t minutes...."
That means, that the time derivative of x at an arbitrary time t is given by.
dx/dt = k·a·b
Where a,b are the masses of A and B in solution in kg at that time. k is a proportionality factor to be determined later.

The change of a,b and x are linked to each other by the stochiometry of the reaction. Since 4kg of X are formed per 1kg of A and 3kg of B which have reacted away, you can relate the change of the masses in kg as:
Δa/-1 = Δb/-3 = Δx/4
Take the difference between the initial amount and the of each chemical:
(a-a₀)/-1 = (b-b₀)/-3 = (x-x₀)/4

Using relations above you can express the masses of two chemicals as function of the mass of the third. Here we want to express a and b as function of x. So choose relation between a and x and relation between b and x and solve them for a and b respectively:
(a-a₀)/-1 = (x-x₀)/4
=>
a = a₀ - (1/4)·(x-x₀)
and
(b-b₀)/-3 = (x-x₀)/4
=>
b= b₀ -(3/4)·(x-x₀)

The initial masses are a₀=2, b₀=3 and x₀=0. So.
a = 2 - (1/4)·x
b= 3 -(3/4)·x

Insert this to rate equation
dx/dt = k · (2 - (1/4)·x) · (3 - (3/4)·x)
or
dx/dt = k' · (8 - x) · (12 - 3·x)
where k' = k/16

Solving this 1st order DE you'll find a solution with one additional constant. So you need two boundary conditions to evaluate the two constants. The conditions are:
(i) at t=0 x=0
(ii) at t=1 [min] x=1


Edit:
1.
The squares are probably erroneously displayed Unicode characters. I used two different
(i)
subscript 0 as is
(a-a₀)/-1 = (b-b₀)/-3 = (x-x₀)/4
denoting initial masses
(ii)
middot as multiplication sign as in
dx/dt = k · (2 - (1/4)·x) · (3 - (3/4)·x)
it is the same as
dx/dt = k *(2 - (1/4)*x) * (3 - (3/4)*x)

2.
The change of reactant masses is negative, while the change of product mass positive. Thus to relate the changes you have to provide a negative sign either for reactants or for products.
I used to set stoichiometric factors of reactants negative, because it is useful convention for some other calculations like evaluation of Δ H or equilbrium constants.

Solve a Differential Equation question correctly and Earn Best Answer!?

a)
-dQ/dt = kQ
dQ/dt = -kQ
dQ/Q= -kdt

Integrate

Integral from Q_0 to Q(t) of ( dQ/Q) = Integral from t_0 to t of ( -ktdt)
ln(Q (t)) - ln(Q_0)= Integral from t_0 to t of ( -kdt)
ln(Q (t)) - ln(Q_0) = -kt - -kt_0

assuming Q_0 starts at time t_0=0

ln(Q (t)) - ln(Q_0) = -kt
ln(Q (t)) = ln(Q_0) - kt
exp(ln(Q (t))) = exp(ln(Q_0) - kt)
Q (t) = exp(ln(Q_0)) * exp( - kt)
Q (t) = Q_0 * exp( - kt)

b)
From our boundary condition
Q(3hr)/Q_0 = 25%
Q (t) = Q_0 * exp( - kt)
Q(3hr)/Q_0 = exp( - k(3hr)) = 0.25
-k(3hr) = ln(0.25)
k = -ln(0.25)/3hr = 0.462 / hr

Now solving with Q_0 = 300mg, t = 6hr
Q( 6hr ) = (300mg)*exp(-(0.462/hr)*6hr)
Q( 6hr ) = 18.75mg

Hope that helps, let me know if you have any questions about my notation.