DE-BOOK What is a Solution?

Section What is a Solution?

📝: 🎧 Listen.

A central goal in differential equations is learning how to find their solutions. In fact, most problems you’ll encounter begin with a familiar prompt:

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“Find the solution to the following differential equation.”
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This "solve and move on" mindset can sometimes obscure a deeper, more important question:

“What does it actually mean to be a solution to a differential equation?”
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A solution to a differential equation is not just the outcome of a sequence of mathematical steps; it is a function that fulfills all the criteria of the equation. This section aims to provide a clearer understanding of what it means to be a solution to a differential equation, as well as methods for verifying and visualizing them.

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📝: 🎧 Listen.

In algebra, a solution is any value that “satisfies” the algebraic equation. In contrast, a solution to a differential equation is any function that “satisfies” the differential equation. So, understanding what it means to satisfy a differential equation is the first step toward understanding its solutions.

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Checkpoint 21. 📖❓ What Kind of Mathematical Object is a Solution?

In general, a solution to a differential equation is any that satisfies it.

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function
Yes, when you solve a differential equation, you are finding a function.
derivative
While derivatives are involved in differential equations, they do not describe solutions in general.
number
It is possible for a solution to be a number, but not in general.
exponential
Exponential functions are common solutions, but they do not represent all solutions in general.

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Satisfying an Algebraic Equation.

📝: 🎧 Listen.

In algebra, we say that a value satisfies an equation if, when we substitute that value in place of the variable, it makes both sides of the equation equal. This gives a true statement, like \(0=0\) or \(1=1\text{.}\)

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For example, consider the equation:

\begin{equation*} y^3 = 3y + 2. \end{equation*}

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Let’s check whether \(y = 2\) and \(y = 0\) are solutions by substituting them into the equation.

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First, try \(y = 2\text{.}\)

\begin{align*} (2)^3 \amp = 3(2) + 2 \\ 8 \amp = 6 + 2 \\ 8 \amp = 8 \end{align*}

That’s a true statement, so \(y = 2\) satisfies the equation.

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Now check \(y = 0\text{.}\)

\begin{align*} (0)^3 \amp = 3(0) + 2 \\ 0 \amp = 0 + 2 \\ 0 \amp \ne 2 \end{align*}

This is false, so \(y = 0\) does not satisfy the equation.

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To summarize: \(y = 2\) is a solution to the equation, while \(y = 0\) is not.

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Satisfying a Differential Equation.

📝: 🎧 Listen.

In differential equations, solutions are functions rather than single values. To verify a solution, we substitute the function into the dependent variable in the equation. If both sides simplify to the same expression, then the function is a solution.

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For instance, consider the differential equation:

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\begin{equation*} y' = 3y. \end{equation*}

To check whether \(y = x^3\) or \(y = e^{3x}\) is a solution, we substitute each into the equation and compare both sides.

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First, try \(y = x^3\text{.}\)

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\begin{align*} \left[x^3\right]' \amp = 3(x^3) \\ 3x^2 \amp \ne 3x^3 \end{align*}

Since these are not the same, \(y = x^3\) does not satisfy the equation.

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Now try \(y = e^{3x}\text{.}\)

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\begin{align*} \left[e^{3x}\right]' \amp = 3(e^{3x}) \\ 3e^{3x} \amp = 3e^{3x} \end{align*}

Since both sides match, this is a true statement and \(y = e^{3x}\) is a solution.

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Here, a “true statement” means both sides simplify to the same function. When we substitute \(y = x^3\text{,}\) we get different expressions. But when we substitute \(y = e^{3x}\text{,}\) both sides match. Technically, we could rewrite \(3e^{3x} = 3e^{3x}\) as \(0 = 0\) by subtracting one side from the other, but it’s enough to see that the functions match directly.

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Checkpoint 22. 📖❓ Quick Verification.

\(y = x^2 + 3\) is a solution to the differential equation

\begin{equation*} \frac{dy}{dx} - 3 = 2x\text{.} \end{equation*}

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True
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Incorrect. \(y = x^2 + 3\) is not a solution since

\begin{align*} \frac{dy}{dx} - 3 \amp = 2x \\ \frac{d}{dx}\left[x^2 + 3\right] - 3 \amp = 2x \\ 2x - 3 \amp = 2x \quad \leftarrow \text{false} \end{align*}
False
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Correct! \(y = x^2 + 3\) is not a solution since

\begin{align*} \frac{dy}{dx} - 3 \amp = 2x \\ \frac{d}{dx}\left[x^2 + 3\right] - 3 \amp = 2x \\ 2x - 3 \amp = 2x \quad \leftarrow \text{false} \end{align*}

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Verifying Solutions.

🌌 Example 23. A Warm-Up.

Verify if \(y = 2x^2\) is a solution to \(\quad xy' - 2x^2 = y\text{.}\)

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Solution.

We’ll move all terms to the left side to simplify the process:

\begin{equation*} xy' - 2x^2 - y = 0\text{.} \end{equation*}

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Now substitute in \(y = 2x^2\) to the left and show that it simplies to \(0\text{:}\)

\begin{align*} xy' - 2x^2 - y \amp = x\left( 2x^2 \right)' - 2x^2 - 2x^2 \\ \amp = 4x^2 - 2x^2 - 2x^2 \\ \amp = 0 \quad✅ \end{align*}

So \(y = 2x^2\) is a solution to \(xy' - 2x^2 = y\text{.}\)

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🌌 Example 24. A Second Order Equation.

Verify if \(P = \sin t\) is a solution to \(\quad 2P'' + P = \sin t\text{.}\)

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Solution.

Since \(P''\) appears, compute the derivatives first:

\begin{equation*} P = \sin t, \quad P' = \cos t, \quad P'' = -\sin t \end{equation*}

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Now rewrite the equation with all terms on one side and substitute:

\begin{align*} 2P'' + P - \sin t \amp = 2(-\sin t) + \sin t - \sin t \\ \amp = -2\sin t + \cancel{\sin t} - \cancel{\sin t} \\ \amp = -2\sin t \ne 0 \quad❌ \end{align*}

So \(P = \sin t\) is not a solution to \(2P'' + P = \sin t\text{.}\)

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You can also verify solutions that include constants, as the next examples show.

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🌌 Example 25. Verifying Solutions with a Common Form.

Show that the functions

\begin{equation*} y = e^{2x},\quad y = -5e^{2x},\quad y = \pi e^{2x},\quad y = 0 \end{equation*}

are all solutions to the differential equation

\begin{equation*} y' - 2y = 0\text{.} \end{equation*}

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Solution.

Rather than verify each function separately, notice that they all are of the form \(y = c e^{2x}\text{,}\) where \(c\) is a constant. Let’s verify the general case.

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Substitute \(y = c e^{2x}\) into the equation:

\begin{equation*} y' - 2y = (c e^{2x})' - 2(c e^{2x}) = 2c e^{2x} - 2c e^{2x} = 0 \quad✅ \end{equation*}

So \(y = c e^{2x}\) is a solution for any constant \(c\text{.}\) In particular, this includes the functions

\begin{equation*} y = e^{2x},\quad y = -5e^{2x},\quad y = \pi e^{2x},\quad y = 0 \end{equation*}

for the constants \(c = 1, -5, \pi, 0\) respectively.

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🌌 Example 26. Verifying a Solution with Two Constants.

Verify that \(y = c_1 e^{-x} + c_2 e^{x} + x^2 + 2\) is a solution to

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

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Solution.

Move all terms to one side:

\begin{equation*} y'' - y + x^2 = 0 \end{equation*}

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Now substitute \(y = c_1 e^{-x} + c_2 e^{x} + x^2 + 2\text{:}\)

\begin{align*} y'' - y + x^2 \amp = \left( c_1 e^{-x} + c_2 e^{x} + x^2 + 2 \right)'' - (c_1 e^{-x} + c_2 e^{x} + x^2 + 2) + x^2 \\ \amp = \left( -c_1 e^{-x} + c_2 e^{x} + 2 \right)' - c_1 e^{-x} - c_2 e^{x} - x^2 - 2 + x^2 \\ \amp = c_1 e^{-x} + c_2 e^{x} + 0 - c_1 e^{-x} - c_2 e^{x} - 2 + 0 \\ \amp = 0 \quad✅ \end{align*}

Thus, \(y = c_1 e^{-x} + c_2 e^{x} + x^2 + 2\) is a solution to \(y'' - y = -x^2\text{.}\)

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📤 Wrap-Up.

📝: 🎧 Listen.

🗝️ Key Takeaways...

In differential equations, a solution is a function that satisfies the equation. This means that when you plug it into the dependent variable, both sides simplify to the same expression.

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To verify a solution, take any necessary derivatives, substitute into the equation, and simplify. If both sides match, or reduce to something like \(0 = 0\text{,}\) then the function is a solution.

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