Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

17.2: First Order Homogeneous Linear Equations

[ "article:topic", "authorname:guichard", "showtoc:no" ]
  • Page ID
    4842
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    A simple, but important and useful, type of separable equation is the first order homogeneous linear equation:

    Definition: first order homogeneous linear differential equation

    A first order homogeneous linear differential equation is one of the form

    \[\dot y + p(t)y=0\]

    or equivalently

    \[\dot y = -p(t)y.\]

    "Linear'' in this definition indicates that both \(\dot y\) and \(y\) occur to the first power; "homogeneous'' refers to the zero on the right hand side of the first form of the equation.

    Example \(\PageIndex{2}\)

    The equation \(\dot y = 2t(25-y)\) can be written \(\dot y + 2ty= 50t\). This is linear, but not homogeneous. The equation \(\dot y=ky\), or \(\dot y-ky=0\) is linear and  homogeneous, with a particularly simple \(p(t)=-k\).

    Because first order homogeneous linear equations are separable, we can solve them in the usual way:

    $$\eqalign{ \dot y &= -p(t)y\cr \int {1\over y}\,dy &= \int -p(t)\,dt\cr \ln|y| &= P(t)+C\cr y&=\pm\,e^{P(t)}\cr y&=Ae^{P(t)},\cr} $$

    where \(P(t)\) is an anti-derivative of \(-p(t)\). As in previous examples, if we allow \(A=0\) we get the constant solution \(y=0\).

    Example \(\PageIndex{3}\)

    Solve the initial value problems \(\dot y + y\cos t =0\), \(y(0)=1/2\) and \(y(2)=1/2\).

    Solution

    We start with

    $$P(t)=\int -\cos t\,dt = -\sin t,$$

    so the general solution to the differential equation is

    $$y=Ae^{-\sin t}.$$

    To compute \(A\) we substitute:

    $$ {1\over 2} = Ae^{-\sin 0} = A,$$

    so the solutions is

    $$ y = {1\over 2} e^{-\sin t}.$$

    For the second problem,

    $$ \eqalign{{1\over 2} &= Ae^{-\sin 2}\cr A &= {1\over 2}e^{\sin 2}\cr}$$

    so the solution is

    $$ y = {1\over 2}e^{\sin 2}e^{-\sin t}.$$

    Example \(\PageIndex{4}\)

    Solve the initial value problem \(y\dot y+3y=0\), \(y(1)=2\), assuming \(t>0\).

    Solution

    We write the equation in standard form: \(\dot y+3y/t=0\). Then

    $$P(t)=\int -{3\over t}\,dt=-3\ln t$$

    and

    $$ y=Ae^{-3\ln t}=At^{-3}.$$

    Substituting to find \(A\): \(2=A(1)^{-3}=A\), so the solution is \(y=2t^{-3}\).

    Contributors

    David Guichard (Whitman College)

    • Integrated by Justin Marshall.