DE-BOOK Direct Integration

Section Direct Integration

Solving a differential equation might sound like a new challenge, but you’ve actually done it before, in calculus! Any time you find an antiderivative, you’re solving a differential equation.

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In this section, we’ll reframe antiderivatives as solutions to differential equations and use this perspective to introduce our first method: direct integration. By the end, you’ll be able to find general and particular solutions to simple differential equations using integration.

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Checkpoint 37. 🤔💭 Review Question: Recall from Calculus...

Suppose you saw the following expression from calculus:

\begin{equation*} \frac{dy}{dx} = x^3 - 7. \end{equation*}

Select all of the true statements.

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\(\quad y\) is the derivative of \(x^3 - 7\)
Saying \(y\) is the derivative of \(x^3 - 7\text{,}\) is the same as \(\left(x^3 - 7\right)' = y\)
\(\quad x^3 - 7\) is the derivative of \(y\)
This one should be the most straight forward.
\(\quad y\) is the antiderivative of \(x^3 - 7\)
Saying \(y\) is the antiderivative of \(x^3 - 7\) is the same thing as saying “when you take the derivative of \(y\) you shoud get \(x^3 - 7\)”
\(\quad x^3 - 7\) is the antiderivative of \(y\)
Saying \(x^3 - 7\) is the antiderivative of \(y\) is the same as saying “when you take the derivative of \(x^3 - 7\) you should get \(y\)”, which is not true.

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📰 Antiderivatives.

You might not realize it, but solving some differential equations is a skill you already have! For instance, computing the antiderivative of \(x^2\) means you’re finding a function \(y\) whose derivative is \(x^2\text{.}\) In other words:

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“Find \(y\) such that the derivative of \(y\) is \(x^2\text{.}\)”
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This is exactly the same as solving the differential equation:

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“Find \(y\) such that \(\ds \frac{dy}{dx} = x^2\text{.}\)”
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To solve this, we integrate both sides with respect to \(x\text{:}\)

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\begin{align*} \int\frac{d}{dx}[y]\ dx \amp = \int x^2\ dx \\ y + c_1 \amp = \frac{1}{3} x^3 + c_2 \\ y \amp = \frac{1}{3} x^3 + (c_2 - c_1) = \frac{1}{3} x^3 + c \end{align*}

The constant \(c = c_2 - c_1\) absorbs the difference between the two integration constants. This is a standard simplification when solving differential equations.

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\begin{equation*} y = \frac{1}{3}x^3 + c \end{equation*}

Checkpoint 38. 📖❓ When it Comes to Constants, Less is More.

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.

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Now let’s look at a slightly more involved example.

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🌌 Example 39. Isolate \(y\).

Find the function, \(y\text{,}\) such that

\begin{equation*} 2y' - 4\sin x = 2\text{.} \end{equation*}

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Solution.

This is another antiderivative problem where we first have to isolate \(y'\) before integrating, like so

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\begin{equation*} y' = 1 + 2 \sin x\text{.} \end{equation*}

This equation now says “The unknown \(y\) is the function whose derivative is \(1 + 2 \sin x\)” or, in other words,

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“The unknown \(y\) is the antiderivative of \(1 + 2 \sin x\)”.
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Instead of writing down the antiderivative as our solution, let’s take the differential equations approach and integrate both sides of the equation with respect to \(x\)

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\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*}

As before, the constants of integration are combined into a single constant \(c\text{.}\) This practice of merging constants is standard in solving differential equations and simplifies the final result.

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Checkpoint 40. 📖❓ Finding \(y\).

Solving for \(y\) in the equation

\begin{equation*} \frac{dy}{dx} = \ln(3x+1) \end{equation*}

amounts to finding the antiderivative of \(\ln(3x+1)\text{.}\)

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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.

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In summary, solving simple differential equations often reduces to finding the antiderivative of a function. By integrating both sides and simplifying constants, you can find the general solution.

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Checkpoint 41. 🤔💭 Check Your Understanding.

(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*}

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Differentiating both sides with respect to \(x\text{.}\)
Incorrect, differentiating both sides only puts another derivative on \(\frac{dy}{dx}\text{.}\)
Isolate \(\frac{dy}{dx}\) and integrating both sides with respect to \(x\text{.}\)
Correct!
Isolate \(\frac{dy}{dx}\) and integrating 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 \(\tan(2x)\text{.}\)

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(b) 🤔💭❓ Solution ⇄ Antiderivative.

The solution to the differential equation

\begin{equation*} \frac13 y'- 7x + x^2 = 1 \end{equation*}

is the antiderivative of which function?

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\(\quad y\)
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 the antiderivative of \(21x - 3x^2 + 3\text{.}\)

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📰 Solving Differential Equations via Direct Integration.

We’ve seen that when you can isolate the derivative of \(y\) in the form:

\begin{equation*} \frac{d}{dx}[y] = f(x) \end{equation*}

you can solve the equation by integrating both sides. This idea also applies to more general equations where the derivative involves both \(x\) and \(y\text{:}\)

\begin{equation*} \frac{d}{dx}\left[g(x,y)\right] = f(x) \end{equation*}

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Here, \(f(x)\) must depend only on \(x\text{,}\) and \(g(x,y)\) may involve both variables. This structure allows us to apply our first solution technique: direct integration.

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✍🏻 Method 1. Direct Integration.

To solve a differential equation of the form

\begin{equation} \frac{d}{dx}\left[g(x,y)\right] = f(x)\text{,}\tag{8} \end{equation}

follow these steps:

Integrate Both Sides 🔗: Apply integration with respect to \(x\) to eliminate the derivative.
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Solve for \(y\) 🔗: After integration, isolate \(y\) to get the general solution.
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Checkpoint 42.

(a) 📖❓ 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*}

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True
Correct!
False
Incorrect. This equation is in the form (8).

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(b) 📖❓ 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*}

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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 \(x\text{.}\)

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Consider the following examples to see how this approach applies to some more interesting problems.

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🌌 Example 43. Integrate both sides to Recover \(y\).

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*}

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Solution.

First, observe that this problem is in the form

\begin{equation*} \frac{d}{dx}\left[g(x,y)\right] = f(x) \end{equation*}

where

\begin{equation*} g(x,y) = 5x \cdot y \quad \text{ and } \quad f(x) = \frac{1}{x^2}\text{.} \end{equation*}

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So, we can solve for \(y\) using direct integration:

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Step 1: Integrate both Sides

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Integrating both sides with respect to \(x\text{,}\) we get

\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*}

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Step 2: Isolate the Solution

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Now that the dependent variable \(y\) is free from the derivative, we can isolate it to find the general solution, as follows

\begin{align*} 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) \quad \leftarrow \ \text{general solution} \end{align*}

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Since we are given the condition \(y\left(1\right) = -4\text{,}\) we need to find \(c\text{.}\)

\begin{align*} -4 \amp = \frac{1}{5}\left(-1 + c\right)\qquad \leftarrow \small\text{ multiply both sides by } 5 \\ -20 \amp = -1 + c \quad \implies \quad c = -19 \end{align*}

Therefore, the particular solution (for \(c=-19\)) is

\begin{equation*} y = \frac{1}{5x}\left(-\frac{1}{x} - 19\right)\text{.} \end{equation*}

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Checkpoint 44. 📖❓ First Step.

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 \(y\text{?}\)

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Release \(y\) by integrating both sides with respect to \(x\text{.}\)
Correct! Integrating both sides is the first step in solving for \(y\text{.}\)
Release \(x\) and \(y\) by integrating both sides with respect to \(y\text{.}\)
Incorrect. Integrating both sides with respect to \(y\) would not eliminate the derivative since the derivative is with respect to \(x\text{.}\)
Compute the derivative of \(5x \cdot y\) using the product rule.
Incorrect. This would actually make the equation more complicated.
Isolate \(x\text{.}\)
Incorrect. This would not help solve for \(y\text{.}\)

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🌌 Example 45. Isolate the Derivative before Integrating.

Compute the general solution of the differential equation

\begin{equation*} \frac{1}{\cos x}\frac{d}{dx}\Big[y\sin(2x)\Big] - 1 = \sec x \end{equation*}

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Solution.

First, we get the derivative by itself on the left-hand side

📝: ☝️ Recall the Identity.

\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*}

Now, we integrate by sides with respect to \(x\text{,}\) releasing \(y\) 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*}

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Direct integration is one of the simplest techniques for solving differential equations, provided the structure allows it. If you can isolate a derivative of \(y\) or an expression involving \(y\) and the right-hand side depends only on \(x\text{,}\) you’re in business. Just integrate both sides and solve for \(y\text{.}\)

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This approach highlights how closely tied differential equations are to the core ideas of calculus.

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📤 Wrap-Up.

🗝️ Key Takeaways...

When you can rearrange a differential equation so that a derivative is isolated on one side and equals a function of the independent variable, like

\begin{equation*} \frac{dy}{dx} = f(x) \quad\text{or}\quad \frac{d}{dx}\left[x^2\cos(y)\right] = f(x) \end{equation*}

the next step is to integrate both sides of the equation.

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Doing so removes the derivative and allows you to isolate the unknown function \(y\text{.}\)
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This process is called direct integration. It’s the simplest method for solving differential equations, and it lays the foundation for many techniques you’ll learn later.
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Check Your Understanding.

Checkpoint 46. 📖❓ Compute the General Solution.

Given the differential equation

\begin{equation*} y'= e^{2t} - 4t \end{equation*}

Find the general solution.

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Press Activate to submit your answer.

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\(y(t) =\)

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Don’t forget the constant of integration. Do not use scripts on the constant (e.g., \(c_2\)).

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You have attempted of activities on this page.

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