Problems 15 through 24 deal with ODEs and techniques discussed in Section \(4.8 .\) Consider the equation $$ \frac{d^{2} x}{d t^{2}}-2 \frac{d x}{d t}-3 x=0 $$ (a) Show that \(x_{1}(t)=e^{3 n}\) and \(x_{2}(t)=e^{-t}\) are two solutions. (b) Show that \(x(t)=c_{1} x_{1}(t)+c_{2} x_{2}(t)\) is also a solution.

Second-order differential equations are fundamental in mathematics and physical sciences. They involve the second derivative of an unknown function. The standard form of a second-order linear homogeneous differential equation with constant coefficients looks like this:
\[ a \frac{d^2x}{dt^2} + b \frac{dx}{dt} + c x = 0 \].
Here, \( a \), \( b \), and \( c \) are constants.
The equation we are dealing with is:
\[ \frac{d^2 x}{d t^{2}}-2 \frac{d x}{d t}-3 x=0 \].
Our goal is to find a function \( x(t) \) that satisfies this equation. Such functions are called 'solutions' to the differential equations.
Understanding these equations is crucial because they provide models for many physical systems, from mechanical vibrations to electrical circuits. Once we identify a solution, it tells us how the system behaves over time.

To verify if a given function is a solution to a differential equation, we follow these steps:

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1. Compute the necessary derivatives of the function.
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2. Substitute these derivatives back into the original equation.
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3. Simplify to confirm that the equation holds true.

In this exercise, we have two candidate solutions: \( x_{1}(t) = e^{3t} \) and \( x_{2}(t) = e^{-t} \).
For \( x_{1}(t) = e^{3t} \), its first and second derivatives are \( 3e^{3t} \) and \( 9e^{3t} \), respectively. Substituting these into the equation:
\[ \frac{d^{2} x_{1}}{d t^2} - 2 \frac{d x_{1}}{d t} - 3 x_{1} = 9e^{3t} - 2(3e^{3t}) - 3(e^{3t}) = 0, \]
which confirms that \( e^{3t} \) is a solution.
Next, for \( x_{2}(t) = e^{-t} \), the derivatives are \( -e^{-t} \) and \( e^{-t} \). Substituting these:
\[ \frac{d^{2} x_{2}}{d t^2} - 2 \frac{d x_{2}}{d t} - 3 x_{2} = e^{-t} + 2e^{-t} - 3e^{-t} = 0, \]
which confirms that \( e^{-t} \) is also a solution.
Verifying solutions is a critical skill as it ensures that the proposed functions correctly solve the differential equation.

A powerful property of linear homogeneous differential equations is that any linear combination of solutions is also a solution.
Given two solutions \( x_{1}(t) \) and \( x_{2}(t) \), their linear combination is expressed as:
\[ x(t) = c_{1}x_{1}(t) + c_{2}x_{2}(t), \]
where \( c_{1} \) and \( c_{2} \) are constants.
For our exercise, we combine \( x_{1}(t) = e^{3t} \) and \( x_{2}(t) = e^{-t} \) to form:
\[ x(t) = c_{1}e^{3t} + c_{2}e^{-t}. \]
To verify this as a solution, we compute its derivatives:
\[ \frac{dx}{dt} = 3c_{1}e^{3t} - c_{2}e^{-t}, \]
\[ \frac{d^2 x}{dt^2} = 9c_{1}e^{3t} + c_{2}e^{-t}. \]
Substitute these into the original differential equation:
\[ 9c_{1}e^{3t} + c_{2}e^{-t} - 2(3c_{1}e^{3t} - c_{2}e^{-t}) - 3(c_{1}e^{3t} + c_{2}e^{-t}) = 0. \]
Simplifying:
\[ c_{1}(9e^{3t} - 6e^{3t} - 3e^{3t}) + c_{2}(e^{-t} + 2e^{-t} - 3e^{-t}) = 0, \]
which simplifies to:
\[ 0 = 0, \]
confirming that \[ x(t) = c_{1}e^{3t} + c_{2}e^{-t} \] is indeed a solution.
This concept elegantly shows how combining multiple known solutions can form a new solution, helping us understand the behavior of complex systems.