- Template
- General Solution
- Specific
- Particular Solution
- Collection
- Family of Solutions
Match the word that best describes each type of solution.
Section Solution Types & Initial Conditions
π: π§ Listen.
To make sense of all these solutions, we classify solutions into three key categories: general solution, particular solution, and family of solutions. Understanding the distinctions between these types is an important part of solving differential equations.
An Analogy.
To help clarify the differences between these types of solutions, letβs use an analogy.
Imagine a family of solutions as a collection of all possible dates in a year. The general solution is like the format for writing dates, such as
mm/dd/yyyy. Each particular solution is like a specific date, such as 10/04/2039. The family of solutions would be the entire list of all dates in this format.
Of course, solutions to differential equations arenβt dates, but the analogy highlights the idea: the general solution gives the structure, particular solutions are specific instances, and the family includes them all.
Checkpoint 27. πβ Describe the Different Solution Types.
Solution Types.
To make sense of all the possible solutions, we classify solutions into three key categories:
- General solution:
- This includes one or more arbitrary constants, like \(c\) in \(y = c e^{2x}\text{.}\) You can think of it as a template from which specific solutions can be generated.
- Particular solution:
- A particular solution results from assigning a specific value to each arbitrary constant in the general solution. For example, if \(c = 5\text{,}\) then \(y = 5 e^{2x}\text{.}\)
- Family of solutions:
- A family of solutions is the complete collection of particular solutions generated by varying the constants in the general solution.
Checkpoint 28. πβ Describe a Family of Solutions.
What is a family of solutions?
- A collection of all possible solutions to a differential equation.
- Correct! The family of solutions includes every possible particular solution.
- The general solution to a differential equation.
- Incorrect. The general solution represents a form of the family of solutions, not the entire set.
- A single specific solution to a differential equation.
- Incorrect. This describes a particular solution, not the family of solutions.
- A solution without any constants.
- Incorrect. A solution without constants is typically a particular solution, not the entire family.
Consider the equation \(y' - 2y = 0\text{.}\) The general solution is \(y = c e^{2x}\text{,}\) and some particular solutions are
\begin{equation*}
y = 5e^{2x}, \quad y = 3.1e^{2x}, \quad y = -2e^{2x},
\end{equation*}
corresponding to \(c = 5\text{,}\) \(3.1\text{,}\) and \(-2\text{.}\) If we could list every possible value of \(c\text{,}\) weβd have the full family of solutions.
Checkpoint 29. πβ General or Particular.
Solving a differential equation usually starts with the general solution, which includes one or more arbitrary constants. Remember, each constant can take on any value, which leads to infinitely many possible particular solutions.
β οΈ 30. Not All Solutions with Constants Are General Solutions β οΈ.
Keep in mind, solutions with arbitrary constants are not general solutions by default. For example, both of the functions
\begin{equation*}
y = \frac{1}{2}x^2 + c_1 x \quad \text{and} \quad y = \frac{1}{2}x^2 + c_1 x + c_2
\end{equation*}
are solutions to the equation \(y'' = 1\text{,}\) but only the second is general solution. The first is a special case of the second, obtained when \(c_2 = 0\text{,}\) so it can be the general solution.
Initial Conditions & Initial-Value Problems (IVPs).
In real-world problems, we often narrow down the set of possible solutions by specifying initial conditions, known values of the solution or its derivatives at a specific point. A differential equation paired with initial conditions is called an initial-value problem (IVP). If enough information is provided, the IVP gives a unique particular solution that satisfies both the equation and the conditions.
Checkpoint 31.
(a) πβ Initial Condition Meaning.
Assume we have a differential equation with dependent variable \(y\) and independent variable \(x\text{.}\)
What is an initial condition?
- Correct! An initial condition specifies the value of the solution or its derivative at a particular point.
- The starting value of \(x\text{.}\)
- Incorrect. The initial condition is related to the solution or its derivatives at a specific point.
- Any point in the \(xy\)-plane.
- Incorrect. An initial condition is not just any point; itβs a specific point referring to a known value of \(y\) or one of itβs derivatives
- The first step in solving a differential equation.
- Incorrect. The initial condition is not the initial step for solving a differential equation.
(b) πβ Is it an IVP.
The differential equation
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x
\end{equation*}
is an example of an initial-value problem.
- True
- Incorrect. An initial-value problem must include initial conditions, which are missing here.
- False
- Correct! This is just a differential equation, without initial conditions, it is not an initial-value problem.
For example, suppose an object is in free fall under constant acceleration \(32\ \mathrm{ft/s^2}\text{.}\) If \(h(t)\) is its height above the ground, then:
\begin{equation*}
h''(t) = -32.
\end{equation*}
The general solution is:
\begin{equation*}
h(t) = c_1 + c_2t - 16t^2.
\end{equation*}
This solution models any object falling under gravity. To describe a specific situation, we apply two initial conditions:
-
The object is dropped from \(100\ \mathrm{ft}\text{.}\)
-
It is dropped from rest with an initial velocity of zero.
In terms of \(h(t)\text{,}\) these conditions become:
\begin{equation*}
h(0) = 100, \quad h'(0) = 0.
\end{equation*}
Now we substitute the first condition into \(h(t)\) and the second into \(h'(t) = c_2 - 32t\text{:}\)
The resulting particular solution is:
\begin{equation*}
h(t) = 100 - 16t^2,
\end{equation*}
which models an object dropped specifically from \(100\ \mathrm{ft}\) with no initial velocity.
From General Solutions to Particular Solutions.
To find a particular solution, we start with a general solution and apply given initial conditions to determine the values of any constants. The following examples illustrate this process.
π Example 32. Solving for a Constant.
Find the particular solution to the initial value problem:
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x, \quad y(0) = 2
\end{equation*}
given that the general solution is:
\begin{equation*}
y = ce^{x^2} + 3.
\end{equation*}
Solution.
The initial condition \(y(0) = 2\) tells us that \(y\) must equal 2 when \(x = 0\text{.}\) Substituting into the general solution:
\begin{align*}
y = ce^{x^2} + 3 \quad \overset{y\ =\ 2,\ x\ =\ 0}{\Rightarrow} \quad 2 \amp = ce^{0^2} + 3 \\
2 \amp = c + 3 \\
c \amp = -1
\end{align*}
Checkpoint 33. πβ When is your Answer a Particular Solution.
In which case would you need to find a particular solution rather than just a general solution?
- When the differential equation is linear.
- Incorrect. Linearity doesnβt determine whether you need a particular solution.
- When initial conditions are provided.
- Correct! A particular solution is found when you need to satisfy initial conditions.
- When the differential equation is has a general solution.
- Incorrect. The existence of a general solution doesnβt determine whether you need a particular solution.
- When the function depends on more than one variable.
- Incorrect. Multivariable functions are not the reason for finding a particular solution.
When a general solution contains more than one constant, we need multiple initial conditions to determine them, as the next example shows.
π Example 34. Two Initial Conditions.
Verify that the function
\begin{equation*}
y = c_1 e^{-3t} + c_2 e^{4t}
\end{equation*}
is a solution to
\begin{equation*}
y'' - y' - 12y = 0,
\end{equation*}
and then find the particular solution satisfying:
\begin{equation*}
y(0) = 4, \quad y'(0) = -5.
\end{equation*}
Solution 1. Verifying the General Solution
Compute the derivatives:
\begin{align*}
y \amp = c_1 e^{-3t} + c_2 e^{4t} \\
y' \amp = -3c_1 e^{-3t} + 4c_2 e^{4t} \\
y'' \amp = 9c_1 e^{-3t} + 16c_2 e^{4t}
\end{align*}
Substitute into the equation:
\begin{align*}
y'' - y' - 12y \amp = \left(9c_1 e^{-3t} + 16c_2 e^{4t}\right)\\
\amp\ - \left(-3c_1 e^{-3t} + 4c_2 e^{4t}\right) - 12\left(c_1 e^{-3t} + c_2 e^{4t}\right)\\
\amp = 0
\end{align*}
So \(y(t) = c_1 e^{-3t} + c_2 e^{4t}\) is indeed a solution.
Solution 2. Finding the Particular Solution
Apply the initial conditions:
Solve the system:
\begin{align*}
c_1 + c_2 \amp = 4 \\
-3c_1 + 4c_2 \amp = -5
\end{align*}
Substituting and solving yields:
\begin{equation*}
c_1 = 3, \quad c_2 = 1.
\end{equation*}
π€ Wrap-Up.
π: π§ Listen.
ποΈ Key Takeaways...
-
A differential equation can have many solutions, often infinitely many, because the constants in the solution can take on any value.
-
A general solution includes one or more arbitrary constants and captures all possible solutions in a single form.
-
A particular solution comes from assigning specific values to those constants.
-
A family of solutions is the complete set of particular solutions generated by varying the constants in the general solution.
-
Initial conditions give known values for the solution or its derivatives.
-
When a differential equation is paired with initial conditions, we call it an initial value problem (IVP).
-
When we apply these conditions, we can determine the constants in the general solution, resulting in a unique particular solution.
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