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Linear 1st-order ODEs

Integrating Factors - Recap

\(\rightarrow\) Functions that make inexact \(1^\circ\) ODEs exact (can’t always find them).
\(\rightarrow\) Two possibilities in this class: \(F(x)\) or \(F(y)\)

• if \(F(x)\), then

(148)\[\begin{align} \frac{1}{F}\frac{dF}{dx} = \frac{1}{Q}\left[\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}\right] = R(x) \end{align}\]

and \(F(x) = \exp[\int R(x)dx]\)

• if \(F(y)\), then

(149)\[\begin{align} \frac{1}{F}\frac{dF}{dy} = \frac{1}{P}\left[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right] = R(y) \end{align}\]

and \(F(y) = \exp[\int R(y)dy]\)

\(\rightarrow\) Once you find your I.F., follow the steps for solving an exact equation:

1. Write in form \(M(x)dx + N(y)dy = 0\)

2. Check for exactness

3. If exact find \(u(x,y)\)

4. Find implicit general solution from \(u(x,y)=c\)

Linear \(1^\circ\) ODEs

A \(1^\circ\) differential equation is linear if it can be written:

(150)\[\begin{align} y' + p(x) y = r(x) \end{align}\]

\(\rightarrow\) it is linear in \(y\) and \(y'\)
\(\rightarrow\) \(p\) and \(r\) may be any function of \(x\) only

• Two cases to solve:

1. Homogeneous Linear ODE

(151)\[\begin{align} r(x) = 0 && \implies \text{can solve by separating} \end{align}\]

1. Non-homogeneous Linear ODE

(152)\[\begin{align} r(x) \neq 0 && \implies \text{can solve as an exact equation with a simple integrating factor} \end{align}\]

Homogeneous Case

(153)\[\begin{align} y' &+ p(x)y = 0\\ \frac{dy}{dx} &= -p(x) y\\ \frac{dy}{y} &= -p(x)dx\\ \ln y &= -\int p(x)dx + \beta\\ y(x) &= \alpha \exp\left[- \int p(x) dx\right] \end{align}\]

Non-Homogeneous Case

(154)\[\begin{align} y' &+ p(x)y = r(x)\\ \frac{dy}{dx} &= -p(x)y + r(x)\\ \end{align}\]

1. Write in the form

(155)\[\begin{align} [p(x)y - r(x) ]dx + 1 dy = 0 \end{align}\]

1. Check for exactness:

(156)\[\begin{align} \frac{\partial}{\partial y}P(x,y) &= \frac{\partial}{\partial x} Q(x,y)\\ p(x) & \neq 0 && \implies \text{not exact} \end{align}\]

1. Find an integrating factor. Try \(F(x,y)=F(x)\)

(157)\[\begin{align} \frac{1}{F}\frac{dF(x)}{dx} &= \frac{1}{Q} \left[\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right]\\ &= \frac{1}{1}(p(x) - 0)\\ &= p(x)\\ \therefore F(x) &= \exp\left(\int p(x)dx\right) && \implies \text{this will always be the I.F. for a linear ODE} \end{align}\]

• Once you recognize that you have a linear \(1^\circ\) ODE, you know that you can solve it as an exact equation using the I.F. \(F(x) = \exp\left( \int p(x)dx \right)\)

• Therefore since we will always have the same I.F., let’s solve the non-homogeneous linear \(1^\circ\) ODE to obtain a general solution \(\implies\) won’t have to derive every time.

(158)\[\begin{align} F(x)y' + F(x) p(x) y = F(x) r(x)\\ \exp\left[ \int p(x) dx\right] y' + \exp \left[ \int p(x) dx\right] p(x) y = \exp\left[\int p(x)dx \right] r(x)\\ \frac{d}{dx} \left( \exp\left[\int p(x) dx \right] y \right) = \exp\left[\int p(x)dx \right] r(x) && \text{(chain rule)} \end{align}\]

• Note that the RHS is dependent only on x, so we can simply integrate

(159)\[\begin{align} \exp \left[\int p(x)dx \right] y = \int \exp\left[\int p(x)dx \right] r(x) dx + c\\ y(x) = \exp \left[-\int p(x)dx \right]\left(\int \exp \left[\int p(x) dx \right] r(x)dx + c \right)\\ \end{align}\]

(160)\[\begin{align} y(x) = e^{-\int p dx} \left[\int e^{\int p dx} r dx + c \right] && \rightarrow \text{general solution for linear $1^\circ$ ODE} \end{align}\]

• If homogeneous, then \(r=0\) and \(y(x) = ce^{-\int p dx}\) as we found before.

Example

\(\frac{dy}{dx} = (x+1)^2 - y\)

(161)\[\begin{align} y' + y = (x+1)^2 && \rightarrow \text{non-homogeneous linear $1^\circ$ ODE} \end{align}\]

\(\therefore\) we know we can solve by turning it into an exact equation using an I.F. of the form

(162)\[\begin{align} F(x) = \exp\left(\int p(x)dx \right) && \text{where $p(x) = 1$ (the coeff. of y)}\\ F(x) = \exp\left(\int 1 dx \right) = e^x \end{align}\]

• Multiply our ODE by our I.F.:\

(163)\[\begin{align} e^xy' + e^x y &= e^x(x+1)^2\\ \frac{d}{dx}(e^xy) &= e^x(x+1)^2\\ e^xy &= \int e^x (x+1)^2 dx\\ e^xy &= e^x (x^2+1) + c && \text{integration by parts, often worst step. See this calc in next section}\\ y(x) &= (x^2+1) + ce^{-x} && \text{general solution} \end{align}\]

• Or we could have found this from our fully generalized solution

(164)\[\begin{align} y(x) &= e^{-\int p dx} \left[\int e^{\int p dx} r dx + c\right], && \text{where $p(x) = 1$ and $r(x) = (x+1)^2$}\\ y(x) &= e^{-x}\left[\int e^x (x+1)^2 dx + c \right]\\ &= (x^2 + 1) + ce^{-x} \end{align}\]

Integrating by Parts (Refresher)

Method to integrate product of 2 terms

(165)\[\begin{align} \int u \cdot \frac{dv}{dx} dx = u \cdot v - \int v \cdot \frac{du}{dx} dx \end{align}\]

Example

Consider \(\int e^x (x+1)^2 dx\)

\(\rightarrow\) easiest to set

(166)\[\begin{align} u = (x+1)^2 = x^2 + 2x + 1 && \text{and} && \frac{dv}{dx} = e^x \end{align}\]

then,

(167)\[\begin{align} \frac{du}{dx} = 2x + 2 && \text{and} && v = e^x \end{align}\]

(168)\[\begin{align} \int e^x (x+1)^2 dx = (x+1)^2 e^x - \int e^x (2x+2) dx && \text{need IBP again!}\\ u = 2x + 2 \implies \frac{du}{dx} = 2\\ \frac{dv}{dx} = e^x \implies v = e^x \end{align}\]

(169)\[\begin{align} \text{Integral} &= (x+1)^2 e^x - \left[(2x+2)e^x - \int e^x \cdot 2 dx \right]\\ &= (x+1)^2 e^x - (2x+2)e^x + 2e^x\\ &= e^x (x^2 + 2x + 1 - 2x -2 + 2)\\ \int e^x (x+1)^2 dx &= e^x (x^2+1) \end{align}\]

Recap of linear \(1^\circ ODEs:\)

(170)\[\begin{align} y' + p(x)y &= r(x)\\ r(x) &= 0 \implies \text{homogeneous and separable}\\ r(x) &\neq 0 \implies \text{non-homogeneous}\\ \text{I.F.} = F(x) &= \exp\left[\int p(x) dx \right] \end{align}\]

• Chain rule provides a simple solution:

(171)\[\begin{align} h = \int p(x) dx\\ y(x) = e^{-h}\left[\int e^h r(x) dx + c \right] && \text{general solution} \end{align}\]

Integral can be difficult to solve

Chemical Engineering Example

• Ethyl acetate decomposition

(172)\[\begin{equation} \ce{CH_3COOC_2H5 + H_2O -> CH_3COOH + C_2H_5OH}\\ \text{(A)} \hspace{2cm} \text{$\rightarrow$first order kinetics} \end{equation}\]

CSTR : Continuous Stirred Tank Reactor

• At steady state, each drop of feed is immediately converted to the composition of product stream.

• What is concentration profile of A before steady state?
  Mass balance on A:

(173)\[\begin{align} \text{Accum = In - Out + Gen - Consump}\\ V\frac{dC}{dt} = FC_{in} - F C(t) - kV C(t)\\ \frac{dC}{dt} = \frac{F}{V}C_{in} - \frac{1}{V}(F+kV)C \end{align}\]

\(\implies\) We can solve 2 ways:

1. Separate

2. Treat as linear equation

1. Separate

(174)\[\begin{align} \int \frac{dC}{\frac{FC_{in}}{V} - \frac{(F+kV)}{V}C} = \int dt \end{align}\]

Useful integral:

(175)\[\begin{align} \int(ax + b)^{-1}dx = \frac{1}{a}\ln (ax + b) + c \end{align}\]

Take \(a=-\frac{F+kV}{V}\) and \(b=\frac{FC_0}{V}\)

(176)\[\begin{align} \frac{-V}{(F+kV)}\ln \left[ \frac{FC_{in}}{V} - \frac{(F+kV)}{V}C\right] &= t + \alpha\\ \ln \left[ \frac{FC_{in}}{V} - \frac{(F+kV)}{V}C\right] &= -\frac{(F+kV)}{V}t + \frac{-(F+kV)}{V}\alpha \end{align}\]

Take the last term as \(\beta\):

(177)\[\begin{align} \frac{FC_{in}}{V} - \frac{(F+kV)}{V}C &= \exp \left[\frac{-(F+kV)t}{V} + \beta \right]\\ &=\phi \exp \left[\frac{-(F+kV)}{V}t\right]\\ \end{align}\]

(178)\[\begin{align} C(t) &= \frac{V}{(F+kV)}\left(\frac{FC_{in}}{V} - \phi \exp \left[\frac{-(F+kV)}{V} t \right] \right)\\ C(t) &= \frac{FC_{in}}{F+kV} - \gamma \exp \left[\frac{-(F+kV)}{V} t \right] && \text{General Solution} \end{align}\]

• Find \(\gamma\) with I.C. \(C(t=0) = C_0\) which is what the CSTR is charged with; doesn’t have to be equal to \(C_{in}\)

(179)\[\begin{align} C(0) = C_0 &= \frac{FC_{in}}{F+kV} - \gamma \exp(0)\\ \gamma &= \frac{FC_{in}}{F+kV} - C_0 \end{align}\]

(180)\[\begin{align} C(t) &= \frac{FC_{in}}{F+kV} - \left(\frac{FC_{in}}{F+kV}-C_0 \right) \exp \left[-\frac{(F+kV)}{V}t \right]\\ C(t) &= \frac{FC_{in}}{F+kV}\left(1 - \exp \left[-\frac{(F+kV)}{V}t \right] \right) + C_0 \exp \left[-\frac{(F+kV)}{V}t \right] && \text{specific solution} \end{align}\]

Check : Does \(C(0) = C_o\)? (yes)
Does \(C(t)\) as \(T\rightarrow \infty\) = \(\frac{F}{(F+kV)}C_{in}\)? (yes!)