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Intro to Differential Equations#

  • See chapters 1-5 of text

  • A differential equation is an equation containing one or more detivatives of an unknown function.
    A) Ordinary Differential Equation (ODE)
    \(\rightarrow\) Unknown depends on only one variable\

    (88)#\[\begin{align} \frac{dy}{dx} = x \rightarrow y=y(x)\\ \frac{dC}{dt} = -kC^2 \rightarrow C = C(t) \end{align}\]

    B) Partial Differential Equation (PDE)
    \(\rightarrow\) Unknown function depends on two or more variables

    (89)#\[\begin{align} \frac{\partial y}{\partial t} = k \frac{\partial y}{\partial x} \rightarrow y=y(t,x) \end{align}\]

Notation#

  • For an unknown function y(x)

    1. First derivative of \(y\): \(\frac{dy}{dx} \equiv y'\)

    2. Second derivative: \(\frac{d^2y}{dx^2}\equiv y''\)

    3. Third derivative: \(\frac{d^3y}{dx^3}\equiv y'''\)

Examples of ODEs:#

  • \(y'=cos(x)\)

  • \(y'' + 4y = 0\)

  • \(x^2y'''y' + 2e^xy''=x^2+2\)

“Order” of ODE: the order of the highest order derivative
\(\rightarrow\) above equations are \(1^{st}\), \(2^{nd}\) and \(3^{rd}\) order respectively.

First Order Differential Equations#

(90)#\[\begin{align} F(x,y,y')=0 && \text{or} && y'=f(x, y) \end{align}\]

will have a general solution

(91)#\[\begin{align} y=h(x) && \rightarrow \text{explicit}\\ H(x,y) = 0 && \rightarrow \text{implicit} \end{align}\]

Linear differential equations#

A linear ODE is one where we can write it in the form $\(a_n(t)y^{(n)}(t)+a_{n-1}(t)y^{(n-1)}(t)+\dots+a_1(t)y'(t)+a_0(t)y(t)=g(t)\)\( where \)y^n(t)$ is the nth derivative of y. If we can get it in this form, it is linear. If not, non-linear.

Examples:#

  • \(y'=\cos(t)\)

    • Linear!

  • \(y''+4y=0\)

    • Linear!

  • \(x^2y'''y'+2e^xy''=x^2+2\)

    • Nonlinear because of the \(y'''y'\) term

  • \(\exp(y)+y''=4x\)

    • Nonlinear because of the \(\exp(y)\) term

  • \(\exp(x)+y''=3x\)

    • Linear

  • \(y+y''=3x\)

    • Linear

Classification of Differential Equations#

  • We just discussed two types of differential equations:

    • Partial Differential Equations (PDE’s)

    • Ordinary Differential Equations (ODE’s)

  • We can also describe the order of the differential equation

    • 1st order

    • 2nd order

    • etc

  • Linearity of differential equation

    • Linear

    • Non-linear

As we progress in the class we will add more classifications.

Separable \(1^\circ \) ODEs#

\(\rightarrow\) should already know these
\(\rightarrow\) Equation is separable if it can be written as:

(92)#\[\begin{align} g(y)y'=f(x) \end{align}\]

\(\rightarrow\) since \(y'=\frac{dy}{dx}\), then

(93)#\[\begin{align} g(y)dy = f(x)dx && \text{separate all $y$'s from all $x$'s} \end{align}\]

Simple integration solution

(94)#\[\begin{align} \int g(y)dy = \int f(x) dx + \underline{c} && \text{don't forget the constant!} \end{align}\]

Example separable first order ODE#

\(y'=2x\)

(95)#\[\begin{align} \frac{dy}{dx} = 2x\\ dy = 2x dx && \text{separate}\\ \int dy =\int 2x dx + c && \text{integrate}\\ y(x) = x^2 + c && \text{general solution} \end{align}\]

We need more information to determine \(c\). Once \(c\) is known \(\rightarrow\) particular solution.
e.g. if we know \(y(x=0)=2\), then

(96)#\[\begin{align} 2&=0^2+c\\ c&=2 \end{align}\]

and the particular solution is \(y(x) = x^2 + 2\)

  • Separable \(1^\circ\) ODE’s arise frequantly in engineering problems

Example separable first order ODE#

\(\underline{Ex}\): There is a pond just north of Pittsburgh where people like to swim in summer. Unfortunately on Memorial Day (May 25), the pond is polluted with a substance \(T\) at a concentration of 100 ppm. There is a flow rate of fresh water into the pond that equals the rate at which water leaches into the soil (pond volume of \(10^4 m^3\) doesn’t change). Swimming will be safe only once \(C_T < 5\) ppm. Will swimmers be back in business before Labor Day (Sept. 5)?

\(\rightarrow\) Solve by performing a mass balance on what? (the pollutant)

(97)#\[\begin{align} \text{Accumulation = In - Out + Gen - Consumption} \end{align}\]

Inlet term is zero because only fresh water enters. Generation and Consumption are zero as well because there’s no reaction.
\(\rightarrow\) Unlike linear systems, this problem is interested in what happens before steady state.
\(\rightarrow\) There is an element of time involved

(98)#\[\begin{align} \frac{dM}{dt} = \frac{d(C \cdot V)}{dt} = -FC \end{align}\]

Check units,

(99)#\[\begin{align} \frac{kg}{s} [=] \frac{\frac{kg}{L} L}{s} [=] \frac{L}{s} \cdot \frac{kg}{L} \end{align}\]

\(\rightarrow\) since \(V\) is constant, we can pull it out

(100)#\[\begin{align} V\frac{dC}{dt} = -FC && \text{$1^\circ$ ODE that models our system}\\ \frac{dC}{C} = -\frac{F}{V} dt && \text{separate}\\ \int \frac{dC}{C} = -\frac{F}{V} \int dt + \alpha && \text{integrate}\\ \ln C = -\frac{F}{V} t + \alpha\\ C(t) = \exp\left[-\frac{F}{V}t + \alpha\right] && \text{General solution} \end{align}\]

\(\rightarrow\) to find \(\alpha\), we need to know \(C(t)\) at some time \(\rightarrow\) will yield our particular solution
\(\rightarrow\) we know \(C(t=0)=C_0=100ppm\)

(101)#\[\begin{align} \therefore C(t=0) = C_0 &= \exp\left[-\frac{F}{V} \cdot 0 + \alpha\right]\\ C_0 &= \exp(\alpha)\\ \alpha &= \ln C_0 && \rightarrow \text{can plug into general solution}\\ \end{align}\]
(102)#\[\begin{align} C(t) &= \exp\left[-\frac{F}{V}t + lnC_0 \right]\\ &=\exp\left[-\frac{F}{V}t\right]\cdot \exp[lnC_0]\\ C(t) &= C_0 \exp \left[ -\frac{F}{V}t\right] && \text{particular solution} \end{align}\]

For our problem: \(C_0 =100ppm\), \(V=10^4 m^3\), \(F=100 m^3\)/day

(103)#\[\begin{align} C(t) &= 100 ppm \cdot \exp\left(\frac{100 m^3/day}{10^4m^3}t \right)\\ C(t) & = 100 \exp(-0.01t) && \text{where $C[=]ppm, t[=]days$} \end{align}\]

\(\rightarrow\) Can swimmers swim on Labor Day? Let’s determine how long it will take for \(C(t)\) to drop to 5 ppm.

(104)#\[\begin{align} 5=100\exp(-0.01t)\\ ln\left(\frac{5}{100} \right) = -0.01t\\ t = 300 days && \rightarrow \text{Can't swim until next summer :(} \end{align}\]

Transient Batch Reactor#

\begin{equation} \ce{CH_3COOC_2H5 + H_2O -> CH_3COOH + C_2H_5OH}\\ \text{ethyl acetate}\hspace{3cm} \text{acetic acid} \hspace{0.8cm} \text{ethanol} \end{equation}

\(\rightarrow\) Develop a model for the concentration of ethyl acetate in the reactor as a function of time. The reaction consumes E.A. at a rate proportional to the instantaneous concentration of E.A.
\(\rightarrow\) Assume the reactor is charged with \(C_0\).
Step 1: What is changing with time?
Concentration of E.A. \(\rightarrow\) mass is conserved.
MB on E.A: Accum = In - Out + Gen - Consump
The reactor has no inlet and outlet. E.A. isn’t being produced by the reaction. Hence, Generation term is also zero.

(105)#\[\begin{align} \frac{dM}{dt} = -kC \end{align}\]

We were told Consump \(\propto C_{EA}\). Given \(C[=]\frac{mass}{vol} \implies k[=]\frac{vol}{time}\) to match LHS units of \(\frac{mass}{time}\)

(106)#\[\begin{align} V\frac{dC}{dt} = -kC\\ \frac{dC}{C} = -\frac{k}{V} dt && \text{separate}\\ \ln C = -\frac{k}{V}t + \beta\\ C(t) = \exp\left(-\frac{k}{V}t \right) \cdot \exp(\beta) \end{align}\]

Let \(\exp(\beta) = \alpha\),

(107)#\[\begin{align} C(t) = \alpha \cdot \exp\left(-\frac{k}{V}t\right) && \text{general solution} \end{align}\]

\(\rightarrow\) We know that \(C(t=0) = C_0\)

(108)#\[\begin{align} \therefore C(t=0) = C_0 = \alpha \cdot \exp(0) \implies C_0 = \alpha \end{align}\]

then,

(109)#\[\begin{align} C(t) = C_0 \cdot \exp\left(-\frac{k}{V} t\right) && \text{particular solution} \end{align}\]

  • What if, instead of knowing the initial concentration, we were told that ater 10 minutes, conc. = \(C_1\)?
    \(\rightarrow\) only \(\alpha\) will change

(110)#\[\begin{align} C(t=10) = C_1 = \alpha \cdot \exp\left(\frac{-10k}{V} \right) \implies \alpha=\frac{C_1}{\exp\left(\frac{-10k}{V} \right)} \end{align}\]

and

(111)#\[\begin{align} C(t) = \frac{ C_1 \exp\left(-\frac{k}{V} t \right)}{\exp\left(-\frac{10k}{V} t \right)} =C_1 \exp\left[\frac{k}{V} (10 - t)\right] \end{align}\]

The basic batch reactor pops up often.
\(\rightarrow\) Redo the problem with different kinetics:
What if E.A. is consumed at a rate proportional to the square of the instantaneous concentration of E.A.?
Now: Accumulation = - Consumption

(112)#\[\begin{align} V\frac{dC}{dt} = -kC^2 && \text{where now $k[=]\frac{vol^2}{mass \cdot time}$}\\ \frac{dC}{C^2} = -\frac{k}{V}dt\\ -\frac{1}{C} = -\frac{k}{V}t + \alpha\\ \frac{1}{C} = \frac{k}{V}t + \alpha && \text{OK to change sign of $\alpha$ since it is arbitrary}\\ C(t) = \frac{1}{\frac{k}{V}t+ \alpha} && \text{general solution} \end{align}\]

Given \(C(0) = C_0\), then \(C(t=0) = 1/\alpha \implies \alpha = \frac{1}{C_0}\)

(113)#\[\begin{align} C(t) = \frac{1}{\frac{k}{V}t+ \frac{1}{C_0}}\\ C(t) = \frac{C_0}{\frac{k}{V}C_0t+ 1} && \text{particular solution} \end{align}\]

Reality checks#

  • Consider the possible steady states of the system.

  • What do you know? What will be a reasonable answer?

  • If you’re worried about your solution, plug it in and check if it’s a solution