First-order differential equations

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Introduction

We consider the general first-order differential equation:


\tau \frac{dy(t)}{d t} + y(t) = x(t)


The general solution is given by:


y(t) = y_0 e^{-(t-t_0)/\tau} + \frac{e^{-t/\tau}}{\tau} \int_{t_0}^{t} x(t') e^{t'/\tau} dt'


where y0 = y(t = t0). An equivalent form is:


y(t) = y_0 e^{-(t-t_0)/\tau} + \frac{e^{-(t-t_0)/\tau}}{\tau} \int_{t_0}^{t} x(t') e^{(t'-t_0)/\tau} dt'


To obtain the general solution, begin with the first order differential equation:


\tau \frac{dy(t)}{d t} + y(t) = x(t)


Divide both sides by \tau \,\!:


\frac{dy(t)}{d t} + \frac{1}{\tau} y(t) = \frac{1}{\tau} x(t)


Multiply the LHS by the integrating factor e − t / τ:


e^{-t/\tau} \frac{d}{dt}\left[ {e^{t/\tau} y(t)} \right] = \frac{1}{\tau} x(t)


Simplify:


\frac{d}{dt} \left[ {e^{t/\tau} y(t)} \right]= \frac{1}{\tau} x(t) e^{t/\tau}


Integrate both sides:


e^{t/\tau}y(t)-e^{t_0/\tau}y(t_0)= \frac{1}{\tau} \int_{t_0}^{t} x(t') e^{t'/\tau} dt'


Solve for y(t):


y(t) = y_0 e^{-(t-t_0)/\tau} + \frac{e^{-t/\tau}}{\tau} \int_{t_0}^{t} x(t') e^{t'/\tau} dt'
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Example Solutions of First Order Differential Equations

Consider:

\frac{dy(t)}{d t}= x(t)

\int_{t_0}^{t}\frac{dy dt}{dt}=\int_{t_0}^{t} {x(t')dt'}

The general solution is given as:

y(t)=y_0+\int_{t_0}^{t} {x(t')dt'}


Now Consider:

\frac{dy(t)}{dt}=-ay(t)

The solution steps are as follows:

\frac{1}{y(t)}\frac{dy(t)}{dt}=-a

\frac{d}{dt}{[ln(y(t))]}=-a

The general solution is given as:

y(t)=y_{0}e^{-a(t-t_{0})}


Now Consider:

\frac{dy(t)}{dt}=-ay(t)+x(t)

The solution steps are as follows:

1. Separate y(t) and x(t) terms


\frac{dy(t)}{dt}+ay(t)=x(t)


2. Multiply LHS by "integrating factor"


e^{-at}\frac{d(e^{at}y(t))}{dt}=x(t)


3. Divide both sides by e − at


\frac{d(e^{at}y(t))}{dt}=e^{at}x(t)


4. Multiply both sides by dt and integrate


e^{at}y(t)-e^{at_0}y_0=\int_{t_0}^{t} x(t') e^{at'}dt'


The general solution is given as:

y(t)=y_{0}e^{-a(t-t_{0})}+e^{-at} \int_{t_0}^{t} x(t') e^{a(t'-t_{0})}dt'

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Example

x(t) = \Theta(t-t_0) \,\!

y_0 = 0 \,\!


That is, x(t) = 1 \,\! for t > t_0 \,\! and x(t) = 0 \,\! otherwise.

The solution is

y(t) = e^{-(t-t_0)/\tau}

since the integral equals 1. The quantity \tau \,\! can be seen to be the time constant whereby y(t) \,\! drops to 1/e \,\! of its original value

For more information, check out the wikipedia page on the same topic... [[1]]

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