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Handout Lesson 5, First-order Linear Differential Equations, Part I
This lesson is based on Section 1.5 of your textbook by Edwards, Penney, and Calvis.
What is a first-order linear differential equation?
Definition 30 .
A differential equation is first-order linear if it can be written in the form
\begin{equation*}
A(x)\frac{dy}{dx}+B(x)y+C(x)=0
\end{equation*}
The standard form for a first-order linear differential equation is
\begin{equation*}
\frac{dy}{dx}+P(x)y=Q(x)
\end{equation*}
Example 31 . Identifying types of differential equations.
Determine if the differential equation is separable, first-order linear, both, or neither.
\(\displaystyle \frac{dy}{dx}+y=2\)
\(\displaystyle \frac{dy}{dx}=xy^2\)
\(\displaystyle \cos(x)\frac{dy}{dx} + 2xy=x^3+2\)
\(\displaystyle \frac{dy}{dx}+\cos(y)y^3=x+y\)
Solving first-order linear differential equations. I am going to demonstrate the algorithm for solving first-order linear differential equation
in standard form by example.
Example 32 . A first look at solving linear differential equations.
(based on number 1 from Section 1.5 of your textbook by Edwards, et.al.)
Solve
\begin{equation*}
\frac{dy}{dx}+y=2
\end{equation*}
In the last example,
\(e^x\) is called an
. When we multiplied both sides of the differenital equation by the integrating factor, then it became possible to integrate both sides of the differential equation.
The obvious question is
“How do you find an integrating factor for a first-order linear differential equation?”
Example 33 . Finding the integrating factor.
(based on number 4 from Section 1.5 of your textbook by Edwards, et.al.)
Solve
\begin{equation*}
\frac{dy}{dx}-2xy=e^{x^2}
\end{equation*}
Now let’s summarize what we have learned.
Algorithm for solving first-order linear differential equations in standard form.
Given: a first-order linear differential equation.
Put the equation in standard form.
\begin{gather}
\frac{dy}{dx}+P(x)y=Q(x)\tag{†}
\end{gather}
Calculated the integrating factor.
\begin{equation*}
\rho(x)=e^{\int P(x) \, dx}
\end{equation*}
Do NOT \(\int P(x) \, dx\)
Multiply both sides of
(†) by
\(\rho(x)\text{.}\)
\begin{equation*}
e^{\int P(x) \, dx} \frac{dy}{dx}+P(x)e^{\int P(x) \, dx}y
=Q(x)e^{\int P(x) \, dx}
\end{equation*}
which yields
\begin{align*}
\frac{d}{dx}\biggr[ye^{\int P(x) \, dx}\biggr]
& = & Q(x)e^{\int P(x) \, dx} \\
\frac{d}{dx}\biggr[y\rho(x) \biggr]
& = & Q(x)\rho(x) \\
\end{align*}
Integrate both sides of the equation to find
\begin{equation*}
y\rho(x)
=\int Q(x)\rho(x) \, dx +C
\end{equation*}
So
\begin{equation*}
y=\frac{\int Q(x)\rho(x) \, dx+C}{\rho(x)}
\end{equation*}
Theorem 34 .
The solutions of the standard form first-order linear differential equation
\begin{equation*}
\frac{dy}{dx}+P(x)y=Q(x)
\end{equation*}
are given by
\begin{equation*}
y=\frac{\int Q(x)\rho(x) \, dx +C}{\rho(x)}
\end{equation*}
where \(\rho(x)=\intfact\text{.}\)
Example 35 . Solving first-order linear equations.
(Based on Excercise 1.4.1 on T. Bazett’s web site) Solve
\begin{equation*}
e^{x^2}\frac{dy}{dx}+2e^{x^2}xy=e^x
\end{equation*}
Example 36 . Solving first-order linear equations.
(Based on Excercise 1.4.4 on T. Bazett’s web site) Solve
\begin{equation*}
\sec(x)\frac{dy}{dx}+y=1
\end{equation*}
