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Section 3.4 Initial-Value Problems (IVPs)
Table 28 illustrates how particular solutions can be generated from the general solution. However, in real-world problems, these constants are not selected. Instead, they are determined from known or assumed information about the problem’s initial state. These known values are called
initial conditions , and when a differential equation is coupled with initial conditions, the resulting problem is an
initial-value problem (IVP).
For example, suppose you are modeling an object in free fall, and you know the following about the object:
The object falls under constant acceleration \(32\ \mathrm{ft/s^2}\text{.}\)
The object is dropped from \(100\ \mathrm{ft}\text{.}\)
The object is dropped from rest with an initial velocity of zero.
If \(h(t)\) is its height above the ground, then the information above can be translated into an IVP since (1) gives the differential equation:
\begin{equation*}
h''(t) = -32
\end{equation*}
and the remaining conditions give the initial conditions:
\begin{equation*}
h(0) = 100, \quad h'(0) = 0.
\end{equation*}
The general solution can be found by integrating \(h''(t)\) twice:
\begin{equation*}
h''(t) = -32 \quad\Rightarrow\quad
h'(t) = c_1 - 32t \quad\Rightarrow\quad
h(t) = c_2 + c_1t - 16t^2
\end{equation*}
Applying the initial conditions to \(h(t)\) and \(h'(t)\) allows us to solve for the constants \(c_1\) and \(c_2\text{:}\)
\(\displaystyle 100 = h(0) = c_2 + c_1(0) - 16(0)^2 \quad\Rightarrow\quad c_2 = 100\)
\(\displaystyle 0 = h'(0) = c_1 - 32(0) \quad\Rightarrow\quad c_1 = 0\)
Using these values, we get the particular solution:
\begin{equation*}
h(t) = 100 - 16t^2.
\end{equation*}
To summarize, solutions to differential equations yield general solutions, whereas solutions to initial-value problems yield particular solutions.
🌌 Example 31 .
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*}
Thus, the particular solution is:
\begin{equation*}
y = -e^{x^2} + 3.
\end{equation*}
The next example shows why multiple initial conditions are needed when a general solution contains more than one constant.
🌌 Example 32 .
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:
\begin{align*}
y(0) \amp = c_1 + c_2 = 4
\end{align*}
\begin{align*}
y'(0) \amp = -3c_1 + 4c_2 = -5
\end{align*}
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*}
The particular solution is:
\begin{equation*}
y = 3e^{-3t} + e^{4t}.
\end{equation*}
Checkpoint 33 . Check your Understanding.
(a) 📖❓ Initial Condition Meaning.
Assume we have a differential equation with dependent variable
\(y\) and independent variable
\(x\text{.}\)
What is an initial condition?
A known value of \(y\) or a known value of \(y\) ’s derivatives.
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; it’s a specific point referring to a known value of \(y\) or one of its derivatives.
The first step in solving a differential equation.
Incorrect. The initial condition is not the initial step for solving a differential equation.
(b) 📖❓ 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.
(c) 📖❓ 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 doesn’t determine whether you need a particular solution.
When initial conditions are provided.
Correct! A particular solution is obtained when the initial conditions are satisfied.
When the differential equation has a general solution.
Incorrect. The existence of a general solution doesn’t 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.
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