2.6: First Order Linear Differential Equations
- Page ID
- 381
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\ket}[1]{\left| #1 \right>}\)
\(\newcommand{\bra}[1]{\left< #1 \right|}\)
\(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
\(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
\(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
\(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
\(\newcommand{\expect}[1]{\langle #1 \rangle}\)
\(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)
In this section we will concentrate on first order linear differential equations. Recall that this means that only a first derivative appears in the differential equation and that the equation is linear. The general first order linear differential equation has the form
\[ y' + p(x)y = g(x) \nonumber \]
Before we come up with the general solution we will work out the specific example
\[ y' + \frac{2}{x y} = \ln \, x. \nonumber \]
The strategy for solving this is to realize that the left hand side looks a little like the product rule for differentiation. The product rule is
\[ (my)' = my' + m'y. \nonumber \]
This leads us to multiplying both sides of the equation by m which is called an integrating factor.
\[ m y' + m \frac{2}{x y} = m \ln\, x. \nonumber \]
We now search for a \(m\) with
\[ m' = m \frac{2}{x} \nonumber \]
or
\[ \frac{dm}{m} = \frac{2}{x} \; dx. \nonumber \]
Integrating both sides, produces
\[ \ln\, m = 2\ln\, x = \ln(x^2) \nonumber \]
or \( m = x^2 \) by exponentiating both sides.
Going back to the original differential equation and multiplying both sides by \( x^2 \), we get
\[ x^2y' + 2xy = x^2 \ln\, x . \nonumber \]
Using the product rule in reverse gives
\[ (x^2y)' = x^2 \ln \, x . \nonumber \]
Now integrate both sides. Note that the integral of the derivative is the original. Integrate by parts to get
\[ \int x^2 \, \ln \, x \, dx. \nonumber \]
- \(u = \ln x\) and \(dv = x^2 \,dx\)
- \(du = \dfrac{1}{x} dx \) and \( v = \dfrac{1}{3} x^3\)
\[ = \dfrac{x^3 \ln\, x}{3} - \int \dfrac{x^2}{3} \, dx = \dfrac{x^3 \ln \, x}{3} - \dfrac{x^3}{9} +C \nonumber \]
Hence
\[ x^2y = \frac{1}{3} x^3 \ln x - \frac{1}{9} x^3 + C. \nonumber \]
Divide by \(x^2\)
\[ y = \dfrac{1}{3} x \ln \, x - \dfrac{1}{9} x + \dfrac{C}{x^2} . \nonumber \]
Notice that when \(C\) is nonzero, the solutions are undefined at \(x = 0\). Also given an initial value with \(x\) positive, there will be no solution for negative \(x\). Now we will derive the general solution to first order linear differential equations.
Consider
\[ y' + p(t)y = g(t). \nonumber \]
We multiply both sides by \(m\) to get
\[ my' + mp(x)y = mg(x). \nonumber \]
We now search for an \(m\) with
\[ m' = mp(x) \nonumber \]
or
\[ \dfrac{dm}{m} = p(x) \, dx . \nonumber \]
Integrating both sides, produces
\[ \text{ln} \, m = \int p(x)\, dx \nonumber \]
exponentiating both sides
\[ m = e^{\int p(x)\, dx}. \nonumber \]
Going back to the original differential equation and multiplying both sides by \(m\), we get
\[ my' + mp(x)y = mg(x) \nonumber \]
\[ (my)' = mg(x) \nonumber \]
\[ my = \int \mu \, g(x) \, dx. \nonumber \]
Solving for \(y\) gives
\[ y = e^{-\int p(x)\, dx}\int g(x) e^{\int p(x) \, dx } \, dx . \nonumber \]