## Section 2.1 What is a Solution?

Regardless what kind of equation you are working with, a solution is a value or function that “satisfies” the equation. The term

For example, suppose I want to check if \(y=2\) and \(y = 0\) are solutions to the equation

*satisfies*simply means that when you plug the value into the equation, it simplifies to a statement that is undeniably true.^{ 1 }

This “undeniably true” statement is sometimes called an identity.

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

To do this, we verify that both \(2\) and \(0\)
Since \(y=2\) yields an true statement we say it satisfies the equation and is a solution. In contrast, \(y=0\) does not give a true statement, so it does not satisfy the equation and is not a solution.

*satisfy*the equation by separately plugging \(2\) and \(0\) in for each \(y\text{,}\) simplify and see if we end up with an undeniably true statement, like so
\begin{align*}
(2)^3 =\amp\ 3(2) + 2 \\
8 =\amp\ 6 + 2 \\
8 =\amp\ 8 \quad \leftarrow \text{true}
\end{align*}

\begin{align*}
(0)^3 =\amp\ 3(0) + 2 \\
0 =\amp\ 0 + 2 \\
0 =\amp\ 2 \quad \leftarrow \text{false}
\end{align*}

The same idea applies to differential equations, except in that

*solutions to differential equations are functions instead of numbers*. To see this, let’s verify if \(y = 3x\) and \(y = e^{3x}\) are solutions to the differential equation \(\displaystyle y^\prime = 3y \text{.}\)Separately plugging \(3x\) and \(e^{3x}\) into the equation yields
Since \(y=3x\) results in a false statement, it does not satisfy the equation and is not a solution. However, \(y=e^{3x}\) does satisfy the equation and is a solution.

\begin{align*}
y^\prime =\amp\ 3y \\
\left[3x\right]^\prime =\amp\ 3(3x) \\
3 =\amp\ 9x \quad \leftarrow \text{false}
\end{align*}

\begin{align*}
y^\prime =\amp\ 3y \\
\left[e^{3x}\right]^\prime =\amp\ 3e^{3x} \\
e^{3x} \cdot \left[3x\right]^\prime =\amp\ 3e^{3x} \\
e^{3x} \cdot 3 =\amp\ 3e^{3x} \\
3e^{3x} =\amp\ 3e^{3x} \quad \leftarrow \text{true}
\end{align*}

To summarize, verifying a solution involves substituting the function into the differential equation and ensuring that the equality is satisfied.

### Reading Questions Check your Understanding

####
1. *A solution to a differential equation is a function that the equation*.

*A solution to a differential equation is a function that the equation*.

####
2. *What does it mean for a function to satisfy a differential equation?*

*What does it mean for a function to satisfy a differential equation?*

- If you plug the function into the equation, you get a true statement.
- Yes, a function that satisfies a differential equation yields a true statement when plugged into the equation.
- If you plug the function into the equation, you get the solution.
- Incorrect. The function is being checked to see if it is a solution, you do not get the solution by plugging it in.
- If you take the derivative of the function, you get a true statement.
- Incorrect. Carefully read the section again.
- If you integrate the function, you get a true statement.
- Incorrect. Carefully read the section again.

*What does it mean for a function to satisfy a differential equation?*

####
3. *The function, \(y = x^3\text{,}\) satisfies the differential equation \(y' = 3y\)*.

*The function, \(y = x^3\text{,}\) satisfies the differential equation \(y' = 3y\)*.

True.

- \(y = x^3\) is not a solution since\begin{align*} y' =\amp\ 3y \\ \left[x^3\right]^{\prime} =\amp\ 3(x^3) \\ 3x^2 =\amp\ 3x^3 \quad \leftarrow \text{false} \end{align*}
False.

- \(y = x^3\) is not a solution since\begin{align*} y' =\amp\ 3y \\ \left[x^3\right]^{\prime} =\amp\ 3(x^3) \\ 3x^2 =\amp\ 3x^3 \quad \leftarrow \text{false} \end{align*}

####
4. *Which variable in equation \(\ds u'' + t^2 u = 0 \) represents the solution?*

*Which variable in equation \(\ds u'' + t^2 u = 0 \) represents the solution?*

####
5. *In general, a “solution” to a differential equation is a *.

*In general, a “solution” to a differential equation is a*.

- constant
- It is possible for a solution to be a constant, but not in general.
- function
- Yes, when you solve a differential equation, you get a function.
- number
- It is possible for a solution to be a number, but not in general.
- limit
- Sorry, no.

*In general, a “solution” to a differential equation is a*

####
6. *Which variable in \(\ds \frac{dP}{ds} + \frac{P}{s^2} = 17s \) does the solution depend on?*

*Which variable in \(\ds \frac{dP}{ds} + \frac{P}{s^2} = 17s \) does the solution depend on?*

- dependent variable, \(s\)
- Incorrect. The solution depends on \(s\text{,}\) but \(s\) is not a dependent variable.
- independent variable, \(s\)
- Yes! the solution, \(P\text{,}\) depends on the independent variable \(s\text{.}\)
- dependent variable, \(P\)
- Incorrect. \(P\) is the solution, so it does not depend on \(P\text{.}\)
- independent variable, \(P\)
- Incorrect. The variable \(P\) is not the independent variable.

*Which variable in \(\ds \frac{dP}{ds} + \frac{P}{s^2} = 17s \) does the solution depend on?*

####
7. *What is the primary goal of solving a differential equation?*

*What is the primary goal of solving a differential equation?*

- To find an unknown function that satisfies the equation.
- Correct! The goal of solving a differential equation is to find the function that meets the equation’s conditions.
- To find the derivative of a function.
- Incorrect. While derivatives are involved, the goal is to find the function, not just its derivative.
- To identify the constants in an equation.
- Incorrect. Identifying constants might be part of the process, but it is not the primary goal.
- To determine the independent variable.
- Incorrect. The independent variable is usually known; we solve for the dependent variable.

*What is the primary goal of solving a differential equation?*

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