Section 3.1 Antiderivatives
When you find an antiderivative, you’re actually solving a basic differential equation. For example, consider finding the general antiderivative of \(x^2\text{.}\) The calculus approach would compute the integral
\begin{equation*}
\int x^2\ dx = \frac13 x^3 + c\text{.}
\end{equation*}
The differential equations approach, on the other hand, seeks all functions whose derivative is \(x^2\text{.}\) This means solving any of the following equivalent equations for \(y\text{:}\)
\begin{equation}
\frac{d}{dx}[y] = x^2, \quad \frac{dy}{dx} = x^2, \quad \text{or} \quad y^\prime = x^2\tag{1}
\end{equation}
To solve for \(y\text{,}\) we integrate both sides with respect to \(x\text{,}\) like so:
\begin{align*}
\int\frac{d}{dx}[y]\ dx \amp = \int x^2\ dx \\
y + c_1 \amp = \frac13 x^3 + c_2 \\
y \amp = \frac13 x^3 + c_2-c_1 \quad \implies \quad y = \frac13 x^3 + c
\end{align*}
where \(c = c_2-c_1\) is a constant. Although this method might seem excessive for simple problems, it illustrates the core concept of isolating the dependent variable and expressing it in terms of the independent variable, \(x\text{.}\)
Note 29. Combining Constants is very common in differential equations.
Example 30.
\(\ \ \) Solve the initial-value problem
\begin{equation*}
2y' - 4\sin x = 2, \quad y(0) = 5 \text{.}
\end{equation*}
Solution.
Start by isolating the derivative, \(y'\text{,}\) on one side of the equation
\begin{align*}
y' \amp = 1 + 2 \sin x
\end{align*}
Integrate both sides with respect to \(x\) to leave us with \(y\) on the left side
\begin{align*}
\int y'\ dx \amp = \int \left(1 + 2 \sin x\right) \ 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*}
Finally, apply the initial condition \(y(0) = 5\) to find \(c\)
\begin{align*}
y(0) \amp = 5 \\
(0) - 2 \cos(0) + c \amp = 5 \\
0 - 2 + c \amp = 5 \\
c \amp = 7
\end{align*}
Thus, the solution to the differential equation is \(\quad y = x - 2 \cos x + 7 \text{.}\)
Reading Questions Check your Understanding
1. We can solve \(\ds \frac{dy}{dx} = x^3 - 7\) for \(y\) by differentiating both sides with respect to \(x\).
- True
- Incorrect, we integrate both sides with respect to \(x\text{.}\)
- False
- Correct!
We can solve \(\ds \frac{dy}{dx} = x^3 - 7\) for \(y\) by differentiating both sides with respect to \(x\text{.}\)
2. The equation \(\ds\frac{dy}{dx} = \ln(3x+1)\) implies that the solution, \(y\text{,}\) is the antiderivative of \(\ln(3x+1)\).
-
The equation \(\ds\frac{dy}{dx} = \ln(3x+1)\) implies that the solution, \(y\text{,}\) is the antiderivative of \(\ln(3x+1)\text{.}\)
- True
- Correct, integrating both sides gives\begin{equation*} y = \int \ln(3x+1)\ dx \quad \leftarrow \text{anti-derivative of } \ln(3x+1)\text{.} \end{equation*}
- False
- Incorrect.
3. Combining constants is a common practice in differential equations..
-
Combining constants is a common practice in differential equations.
- True
- Correct!
- False
- Incorrect, see the footnote above.
4. What is the process to solve for \(y\) in the equation \(\ds\frac{dy}{dx} = \tan(2x)\text{?}\)
-
What is the process to solve for \(y\) in the equation \(\ds\frac{dy}{dx} = \tan(2x)\text{?}\)
- Differentiating both sides with respect to \(x\text{.}\)
- Incorrect, we integrate both sides with respect to \(x\text{.}\)
- Integrating both sides with respect to \(x\text{.}\)
- Correct!
- Integrating both sides with respect to \(y\text{.}\)
- Incorrect, this is not part of the process.
- Multiplying both sides by \(dx\text{.}\)
- Incorrect, this is not part of the process.
5. How do we start solving the differential equation \(\ds\frac13 y'+ 7x + x^2 = 1\text{?}\)
-
How do we start solving the differential equation \(\ds\frac13 y'+ 7x + x^2 = 1\text{?}\)
- By isolating the derivative, \(y'\text{.}\)
- Correct! Isolate \(y'\) first, then integrate both sides.
- By Integrating both sides with respect to \(x\text{.}\)
- Incorrect. While you could do this, we suggest isolating the derivative first.
- Differentiate both sides with respect to \(x\text{.}\)
- Incorrect, we would like to remove derivatives, not add more.
- Convert \(y'\) to \(dy/dx\text{.}\)
- Incorrect, the notation for the derivative does not matter.
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