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7.2: First-order differential equations

  • Page ID
    22404
  • 2.1 Introduction

    We consider the general first-order differential equation:

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

    The general solution is given by:

    \[y(t)=y_{0} e^{-\left(t-t_{0}\right) / \tau}+\frac{e^{-\left(t-t_{0}\right) / \tau}}{\tau} \int_{t_{0}}^{t} x\left(t^{\prime}\right) e^{\left(t^{\prime}-t_{0}\right) / \tau} d t^{\prime}\]

    where y0 = y(t = t0). Note that t' is used to be distinguished from the upper limit t of the integral.

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

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

    Divide both sides by tau \,\!:

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

    Rewrite the LHS in condensed form using the integrating factor et / τ:

    Notice how a chain differentiation will return the LHS to the previous form

    Simplify:

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

    Integrate both sides:

    \[e^{t / \tau} y(t)-e^{t_{0} / \tau} y\left(t_{0}\right)=\frac{1}{\tau} \int_{t_{0}}^{t} x\left(t^{\prime}\right) e^{t^{\prime} / \tau} d t^{\prime}\]

    Solve for y(t):

    \[y(t)=y_{0} e^{-\left(t-t_{0}\right) / \tau}+\frac{e^{-t / \tau}}{\tau} \int_{t_{0}}^{t} x\left(t^{\prime}\right) e^{t^{\prime} / \tau} d t^{\prime}\]

    2.2 Example Solutions of First Order Differential Equations

    Consider:

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

    Multiplying both sides by dt gives:

    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)

    Dividing both sides by y(t) gives:

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

    which can be rewritten as:

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

    Multiplying both sides by dt, integrating, and setting both sides of the equation as exponents to the exponential function gives the general solution:

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

    Now Consider:

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

    The detailed solution steps are as follows:

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

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

    2. Rewrite the LHS in condensed form using the "integrating factor" eat

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

    Notice how a chain differentiation will return the LHS to the form written in step 1

    3. Divide both sides by eat

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

    4. Multiply both sides by dt and integrate

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

    The general solution is given as:

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

    2.3 Step Function Simplification

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

    For (t) = \Theta(t-t_0) \,\!, which is the step function and _0 = 0 \,\!

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

    The previously derived general solution:

    \[y(t)=y_{0} e^{-\left(t-t_{0}\right) / \tau}+\frac{e^{-\left(t-t_{0}\right) / \tau}}{\tau} \int_{t_{0}}^{t} x\left(t^{\prime}\right) e^{\left(t^{\prime}-t_{0}\right) / \tau} d t^{\prime}\]

    reduces to:

    image-601.png, since the integral equals 1.

    The quantity tau \,\!can be seen to be the time constant whereby (t) \,\!drops to of its original value

    2.4 Sample Problem

    Problem Statement:

    Consider the differential equation frac{dy(t)}{d t} = -0.5 y(t)+ x(t)

    where (t)= 2+ 0.01 t \,\!

    Assuming _0 = 0 \,\!, what is the behavior of (t)\,\!?

    Solution

    General solution was derived previously as:

    \[y(t)=y_{0} e^{-a\left(t-t_{0}\right)}+e^{-a t} \int_{t_{0}}^{t} x\left(t^{\prime}\right) e^{a\left(t^{\prime}-t_{0}\right)} d t^{\prime}\]

    For  = 0.5 \,\!and _0 = 0 \,\!, while setting _0 = 0 \,\!, the solution reduces to:

    \[y(t)=e^{-0.5 t} \int_{0}^{t}\left(2+0.01 t^{\prime}\right) e^{0.5\left(t^{\prime}\right)} d t^{\prime}\]

    The following link may be referred to for integral tables: S.O.S. Math

    Simplifying the solution gives the following:

    \[y(t)=\frac{\left.\left(e^{-0.5 t}\right)\left((x+198)\left(e^{0.5 t}\right)-198\right)\right)}{50}\]

    Plotting this function displays the following behavior:

    lot.jpg

    As can be seen clearly from the graph, initially the systemic response shows more exponential characteristics. However, as time progresses, linear behavior dominates.