2.6: First Order Linear Differential Equations
- Page ID
- 381
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
\[ y' + p(x)y = g(x) \]
Before we come up with the general solution we will work out the specific example
\[ y' + \frac{2}{x y} = \ln \, x. \]
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
\[ (my)' = my' + m'y. \]
This leads us to multiplying both sides of the equation by m which is called an integrating factor.
\[ m y' + m \frac{2}{x y} = m \ln\, x. \]
We now search for a \(m\) with
\[ m' = m \frac{2}{x} \]
or
\[ \frac{dm}{m} = \frac{2}{x} \; dx. \]
Integrating both sides, produces
\[ \ln\, m = 2\ln\, x = \ln(x^2) \]
or \( m = x^2 \) by exponentiating both sides.
Going back to the original differential equation and multiplying both sides by \( x^2 \), we get
\[ x^2y' + 2xy = x^2 \ln\, x . \]
Using the product rule in reverse gives
\[ (x^2y)' = x^2 \ln \, x . \]
Now integrate both sides. Note that the integral of the derivative is the original. Integrate by parts to get
\[ \int x^2 \, \ln \, x \, dx. \]
- \(u = \ln x\) and \(dv = x^2 \,dx\)
- \(du = \dfrac{1}{x} dx \) and \( v = \dfrac{1}{3} x^3\)
\[ = \dfrac{x^3 \ln\, x}{3} - \int \dfrac{x^2}{3} \, dx = \dfrac{x^3 \ln \, x}{3} - \dfrac{x^3}{9} +C \]
Hence
\[ x^2y = \frac{1}{3} x^3 \ln x - \frac{1}{9} x^3 + C. \]
Divide by \(x^2\)
\[ y = \dfrac{1}{3} x \ln \, x - \dfrac{1}{9} x + \dfrac{C}{x^2} . \]
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
\[ y' + p(t)y = g(t). \]
We multiply both sides by \(m\) to get
\[ my' + mp(x)y = mg(x). \]
We now search for an \(m\) with
\[ m' = mp(x) \]
or
\[ \dfrac{dm}{m} = p(x) \, dx . \]
Integrating both sides, produces
\[ \text{ln} \, m = \int p(x)\, dx \]
exponentiating both sides
\[ m = e^{\int p(x)\, dx}. \]
Going back to the original differential equation and multiplying both sides by \(m\), we get
\[ my' + mp(x)y = mg(x) \]
\[ (my)' = mg(x) \]
\[ my = \int \mu \, g(x) \, dx. \]
Solving for \(y\) gives
\[ y = e^{-\int p(x)\, dx}\int g(x) e^{\int p(x) \, dx } \, dx .\]