Section 3.2 Solutions by Direct Integration
As seen in the
previous example, if a differential equation can be written in the form:
\begin{equation*}
\frac{d}{dx}[y] = f(x)\text{,}
\end{equation*}
then the solution can be found by integrating both sides. This principle extends to more complex equations of the form:
\begin{equation*}
\frac{d}{dx}\left[g(x,y)\right] = f(x)\text{.}
\end{equation*}
Here, \(f(x)\) is a placeholder that indicates that only \(x\) variables can appear on the right-hand side of this equation. Similarly, \(g(x,y)\) indicates that only \(x\) or \(y\) variables can appear inside the derivative.
This leads us to our first method for solving first-order differential equations—direct integration.
Method 1. Direct Integration.
The general solution of a differential equation in the form
\begin{equation}
\frac{d}{dx}\left[g(x,y)\right] = f(x)\text{,}\tag{2}
\end{equation}
can be determined by integrating both sides with respect to \(x\text{.}\)
Consider the following examples to see how the same approach applies to these more interesting problems.
Example 31.
\(\ \ \) Find the particular solution to the differential equation
\begin{equation*}
\frac{d}{dx}\left[5x \cdot y\right] = \frac{1}{x^2} , \quad y\left(1\right) = -4\text{.}
\end{equation*}
Solution.
To solve for \(y\text{,}\) we need to eliminate the derivative by integrating both sides as before.
\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 \\
y \amp = \frac{1}{5x}\Big(-\frac{1}{x} + \overset{=\ c}{\overbrace{c_2 - c_1}}\Big) \\
y \amp = \frac{1}{5x}\left(-\frac{1}{x} + c\right)
\end{align*}
Finally, we use the condition, \(y\left(1\right) = -4\text{,}\) to find the particular solution.
\begin{align*}
-4 \amp = \frac{1}{5}\left(-1 + c\right) \\
-20 \amp = -1 + c \\
c \amp = -19
\end{align*}
Therefore, the particular solution is \(\quad \displaystyle y = \frac{1}{5x}\left(-\frac{1}{x} - 19\right)
\text{.}\)
Example 32.
\(\ \ \) Compute the general solution of the differential equation
\begin{equation*}
\frac{d}{dx}\Big[y\sin(2x)\Big] = \cos x
\end{equation*}
Solution.
To solve the differential equation
\begin{equation*}
\frac{d}{dx}\Big[y\sin(2x)\Big] = \cos x\text{,}
\end{equation*}
we integrate both sides with respect to \(x\) to get
\begin{align*}
\int \frac{d}{dx}\Big[y\sin(2x)\Big]dx \amp = \int \cos x\ dx\\
y\sin(2x) + c_1 \amp = \sin x + c_2
\end{align*}
Finally, we isolate \(y\) and combine constants.
\begin{align*}
y\sin(2x) \amp = \sin x + \underset{=\ c}{\underbrace{c_2 - c_1}} \\
y \amp = \frac{\sin x + c}{\sin(2x)}
\end{align*}
Reading Questions Check your Understanding
1. Solving a differential equation by direct integration involves computing a derivative.
Solving a differential equation by direct integration involves computing a derivative.- True
Incorrect, direct integration involves integrating both sides of the equation.
- False
Correct! Direct integration involves integrating both sides of the equation.
2. Direct integration could be used to solve the equation \(\ds\frac{d}{dx}\left[y^2 + x^3\right] = \sqrt{x}\).
3. Why shouldn’t direct integration be applied to \(\ds\frac{d}{dx}\left[\frac{x}{y^2}\right] = \sin(x+y)\).
Why shouldn’t direct integration be applied to the equation
\begin{equation*}
\frac{d}{dx}\left[\frac{x}{y^2}\right] = \sin(x+y)\text{.}
\end{equation*}
- The equation is not linear.
Incorrect, direct integration can handle this.
- The \(y\) term is squared.
Incorrect, direct integration can handle this.
- The \(y\) term cannot be in the denominator.
Incorrect, direct integration can handle this.
- The right-hand side contains \(y\text{.}\)
Correct! Direct integration only works when the right-hand side contains only \(x\text{.}\)
4. In the equation \(\ds\frac{d}{dx}\left[5x \cdot y\right] = \frac{1}{x^2}\text{,}\) what is the first step in solving for \(y\text{?}\)
In the differential equation
\begin{equation*}
\frac{d}{dx}\left[5x \cdot y\right] = \frac{1}{x^2}\text{,}
\end{equation*}
what is the first step in solving for \(y\text{?}\)
- Integrate both sides with respect to \(x\text{.}\)
- Correct! Integrating both sides is the first step in solving for \(y\text{.}\)
- Differentiate both sides with respect to \(x\text{.}\)
- Incorrect. Differentiating would not help solve the equation; integration is the correct approach.
- Factor the expression \(5x \cdot y\text{.}\)
- Incorrect. Factoring is not necessary here; integration is the correct step.
- Substitute a new variable for \(5x \cdot y\text{.}\)
- Incorrect. Substitution is not needed; direct integration is the correct step.
5. Solve \(\ \ds y'= e^{2x} - 5x\ \) using direct integration.
Complete each step below to solve the differential equation
\begin{equation*}
y'= e^{2x} - 4x.
\end{equation*}
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