5.3 First Order Linear Differential Equations

Subsection 5.3.1 Homogeneous DEs

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

Definition 5.21. First Order Homogeneous Linear DE.

A first order homogeneous linear differential equation is one of the form \(\ds y' + p(t)y=0\) or equivalently \(\ds y' = -p(t)y\text{.}\)

We have already seen a first order homogeneous linear differential equation, namely the simple growth and decay model \(y'=ky\text{.}\)

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

\begin{align*} y' \amp = -p(t)y\\ \int {1\over y}\,dy \amp = \int -p(t)\,dt\\ \ln|y| \amp = P(t)+C\\ y\amp = \pm\,e^{P(t)+C}\\ y\amp = Ae^{P(t)}\text{,} \end{align*}

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

Example 5.22. Solving an IVP I.

Solve the initial value problem

\begin{equation*} \ds y' + y\cos t =0\text{,} \end{equation*}

subject to

1. \(y(0)=1/2\)

2. \(y(2)=1/2\)

Solution

We start with

\begin{equation*} P(t)=\int -\cos t\,dt = -\sin t\text{,} \end{equation*}

so the general solution to the differential equation is

\begin{equation*} y=Ae^{-\sin t}\text{.} \end{equation*}

1. To compute the constant coefficient \(A\text{,}\) we substitute:

   \begin{equation*} \frac{1}{2} = Ae^{-\sin 0} = A\text{,} \end{equation*}

   so the solutions is

   \begin{equation*} y = \frac{1}{2} e^{-\sin t}\text{.} \end{equation*}

2. To compute the constant coefficient \(A\text{,}\) we substitute:

   \begin{align*} \frac{1}{2} \amp = Ae^{-\sin 2}\\ A \amp = \frac{1}{2}e^{\sin 2} \end{align*}

   so the solution is

   \begin{equation*} y = {1\over 2}e^{\sin 2}e^{-\sin t}\text{.} \end{equation*}

Example 5.23. Solving an IVP II.

Solve the initial value problem \(ty'+3y=0\text{,}\) \(y(1)=2\text{,}\) assuming \(t>0\text{.}\)

Solution

We write the equation in standard form: \(y'+3y/t=0\text{.}\) Then

\begin{equation*} P(t)=\int -{3\over t}\,dt=-3\ln t \end{equation*}

and

\begin{equation*} y=Ae^{-3\ln t}=At^{-3}\text{.} \end{equation*}

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