Skip to main content
Contents Index Calc Dark Mode Prev Next Profile \(\newcommand\DLGray{\color{Gray}}
\newcommand\DLO{\color{BurntOrange}}
\newcommand\DLRa{\color{WildStrawberry}}
\newcommand\DLGa{\color{Green}}
\newcommand\DLGb{\color{PineGreen}}
\newcommand\DLBa{\color{RoyalBlue}}
\newcommand\DLBb{\color{Cerulean}}
\newcommand\ds{\displaystyle}
\newcommand\ddx{\frac{d}{dx}}
\newcommand\os{\overset}
\newcommand\us{\underset}
\newcommand\ob{\overbrace}
\newcommand\obt{\overbracket}
\newcommand\ub{\underbrace}
\newcommand\ubt{\underbracket}
\newcommand\ul{\underline}
\newcommand\laplacesym{\mathscr{L}}
\newcommand\lap[1]{\laplacesym\left\{#1\right\}}
\newcommand\ilap[1]{\laplacesym^{-1}\left\{#1\right\}}
\newcommand\tikznode[3][]
{\tikz[remember picture,baseline=(#2.base)]
\node[minimum size=0pt,inner sep=0pt,#1](#2){#3};
}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\newcommand{\sfrac}[2]{{#1}/{#2}}
\)
Section 4.2 Solutions by Direct Integration
In the previous section, we saw how finding the antiderivative of \(x^2\) can be rephrased as solving the differential equation
\begin{equation*}
y'(x) = x^2.
\end{equation*}
📝: Link to Algebra.
Just as taking the square root of both sides of the equation
\begin{equation*}
x^2 = 9
\end{equation*}
helps isolate \(x\text{,}\) integrating both sides of
\begin{equation*}
y' = x^2
\end{equation*}
helps isolate the unknown function, \(y\text{.}\)
To find
\(y\text{,}\) we would like a way to “remove” the derivative from
\(y'\text{.}\) The fundamental theorem of calculus tells us that this is possible by integrating both sides with respect to
\(x\text{.}\)
Subsection Solving \(y' = f(x)\)
Let’s use integration to solve the following equation:
\begin{equation*}
2y' - 4\sin x = 2\text{.}
\end{equation*}
We begin by isolating \(y'\text{:}\)
\begin{equation*}
2y' = 2 + 4\sin x \quad \Rightarrow \quad y' = \frac12(2 + 4\sin x) \quad \Rightarrow \quad y' = 1 + 2\sin x\text{.}
\end{equation*}
Now we integrate both sides with respect to \(x\) to leave us with \(y\text{:}\)
\begin{align*}
\int y'\ dx \amp = \int (1 + 2\sin x)\ dx \\
y + c_1 \amp = x - 2 \cos x + c_2 \\
y \amp = x - 2 \cos x + \boxed{c_2 - c_1}\leftarrow c \\
y(x) \amp = x - 2 \cos x + c
\end{align*}
The constants of integration are combined into a single constant
\(c\text{.}\) Merging constants like this is a standard way to simplify the solution.
Subsection Solving \([g(x,y)]^{\prime} = f(x)\)
Let’s extend this idea to solve the equation:
\begin{equation*}
x^2\frac{d}{dx}\left[5x \cdot y\right] - 1 = 0\text{.}
\end{equation*}
Currently, \(y\) is trapped inside a derivative, but perhaps integration could release it. Start by isolating this derivative:
\begin{equation*}
\frac{d}{dx}\left[5x \cdot y\right] = \frac{1}{x^2}.
\end{equation*}
To “remove” this derivative, we must integrate with respect to \(x\text{:}\)
\begin{align*}
\int \frac{d}{dx}\left[5x \cdot y\right]\ dx \amp = \int x^{-2}\ dx \\
5x \cdot y + c_1 \amp = -x^{-1} + c_2
\end{align*}
With \(y\) now free, we can isolate it to obtain the solution:
\begin{align*}
5x y \amp = -\frac{1}{x} + c_2 - c_1 \\
y \amp = \frac{1}{5x}\left(-\frac{1}{x} + c\right) \quad \leftarrow \ \text{general solution}
\end{align*}
where \(c = c_2-c_1\) is the combined arbitrary constant.
Subsection Direct Integration Approach
Direct Integration.
If a differential equation can be expressed in one of the forms:
\begin{equation}
\frac{dy}{dx} = f(x) \quad\text{or}\quad \frac{d}{dx}\left[g(x,y)\right] = f(x)\text{,}\tag{4.1}
\end{equation}
then the general solution can be found by
integrating both sides with respect to \(x\text{,}\) and
solving for \(y\text{.}\)
🌌 Example 37 .
Find the general solution to the differential equation
\begin{equation*}
\frac{1}{\cos x}\frac{d}{dx}\Big[y\sin(2x)\Big] - 1 = \sec x
\end{equation*}
Solution .
First, we get the derivative by itself on the left-hand side
📝: ☝️ Recall the Identity.
\begin{equation*}
\sec x = \frac{1}{\cos x}
\end{equation*}
\begin{align*}
\frac{d}{dx}\Big[ y \sin(2x) \Big] \amp = \cos x\left( 1 + \frac{1}{\cos x} \right)\\
\frac{d}{dx}
\Big[
\us{\textcolor{BurntOrange}{g(x,y)}}{\us{\textcolor{BurntOrange}{\uparrow}}{\ul{y\sin(2x)}}}
\Big]
\amp = \us{\textcolor{BurntOrange}{f(x)}}{\us{\textcolor{BurntOrange}{\uparrow}}{\ul{\cos x + 1}}}
\end{align*}
We now integrate both sides with respect to \(x\text{,}\) thereby eliminating \(y\) from the derivative and allowing us to isolate the general solution.
\begin{align*}
\int \frac{d}{dx}\Big[y\sin(2x)\Big]dx \amp = \int \left(\cos x + 1\right) \ dx\\
y\sin(2x) + c_1 \amp = \sin x + x + c_2\\
y \amp = \frac{\sin x + x + c}{\sin(2x)}\qquad (c = c_2 - c_1)\text{.}
\end{align*}
🌌 Example 38 .
Show why direct integration doesn’t apply to the equation:
\begin{equation*}
\frac{dy}{dx} - y = x
\end{equation*}
Solution .
Isolating the derivative gives:
\begin{equation*}
\frac{dy}{dx} = x + y.
\end{equation*}
Since the derivative is with respect to \(x\text{,}\) we must integrate both sides with respect to \(x\) as well:
\begin{align*}
\int\frac{dy}{dx}\ dx \amp = \int (x + y)\ dx\\
y \amp = \int x\ dx + \int y\ dx\\
y \amp = \frac12 x^2 + \int y\ dx \quad \leftarrow\text{stuck}\text{.}
\end{align*}
The integral left behind cannot be computed because we are integrating with respect to
\(x\text{,}\) not
\(y\text{.}\) We know that
\(y\) is a function of
\(x\text{,}\) but we don’t know which one (if we did then we would already have the solution!)
Direct integration is a simple method for solving differential equations, provided the structure permits it. If your equation can be written as
\begin{equation*}
\frac{d}{dx}[x\ \text{and}\ y\ \text{terms}] = \text{only}\ x\ \text{terms}\text{,}
\end{equation*}
you’re in business. Just integrate both sides and solve for \(y\text{.}\)
Checkpoint 39 .
(a) How could you Find \(y\text{?}\)
How could you solve for \(y\) in the equation
\begin{equation*}
\frac12\frac{dy}{dx} - \tan(2x) = x\text{?}
\end{equation*}
Isolate \(\frac{dy}{dx}\) & integrate both sides with respect to \(x\text{.}\)
Correct!
Differentiate both sides with respect to \(x\text{.}\)
Incorrect, differentiating both sides only puts another derivative on \(\frac{dy}{dx}\text{.}\)
Isolate \(\frac{dy}{dx}\) & integrate both sides with respect to \(y\text{.}\)
Incorrect, the integration is not with respect to \(y\text{.}\)
Find the antiderivative of \(\tan(2x)\text{.}\)
Incorrect, the solution is the antiderivative of \(2\tan(2x) + 2x\text{,}\) not just \(\tan(2x)\text{.}\)
(b) Solution ⇄ Antiderivative.
The solution to the differential equation
\begin{equation*}
\frac13 y' - 7x + x^2 = 1
\end{equation*}
is the integral of which function?
\(\quad y\)
Incorrect. \(y\) is the solution to the differential equation.
\(\quad 21x - 3x^2 + 1\)
Incorrect, perhaps check your algebra.
\(\quad 7x - x^2 - 1\)
Incorrect, perhaps check your algebra.
\(\quad 21x - 3x^2 + 3\)
Correct! Isolating \(y'\) gives
\begin{equation*}
y' = 21x - 3x^2 + 3\text{,}
\end{equation*}
so the solution is \(\int (21x - 3x^2 + 3)\ dx\text{.}\)
(c) Does Direct Integration Apply?
Direct integration could be used to solve the equation
\begin{equation*}
\frac{d}{dx}\left[y^2 + x^3\right] = \sqrt{x}\text{.}
\end{equation*}
True
Correct!
False
Incorrect. This equation is in the form
(4.1) .
(d) Why Doesn’t Direct Integration Apply Here?
Give the reason direct integration cannot be applied to the equation
\begin{equation*}
\frac{d}{dx}\left[\frac{x}{y^2}\right] = \sin(x+y)\text{.}
\end{equation*}
There is a fraction in the derivative.
Incorrect, direct integration doesn’t care about fractions.
The \(y\) term is squared.
Incorrect, direct integration can handle this.
There is a sine term on the right side of the equation.
Incorrect, the sine is not the issue here.
The right-hand side contains \(y\text{.}\)
Correct! Direct integration only works when the right-hand side contains only the independent variable, in this case \(x\text{.}\)
(e) Combining Constants.
Combining constants is a common practice in differential equations.
True
Correct! Combining constants is an easy way to simplify a solution and is a standard practice in differential equations.
False
Incorrect, revisit the examples above.
You have attempted
of
activities on this page.
