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Section 3.4 Initial-Value Problems (IVPs)

Table 32 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:
  1. The object falls under constant acceleration \(32\ \mathrm{ft/s^2}\text{.}\)
  2. The object is dropped from \(100\ \mathrm{ft}\text{.}\)
  3. 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 \(c_2\text{:}\)
  1. \(\displaystyle 100 = h(0) = c_2 + c_1(0) - 16(0)^2 \quad\Rightarrow\quad c_2 = 100\)
  2. \(\displaystyle 0 = h'(0) = c_1 - 32(0) \quad\Rightarrow\quad c_1 = 0\)
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 35.

Find the particular solution to the initial-value problem:
\begin{equation} \frac{dy}{dx} = 2xy - 6x, \quad y(0) = 2\tag{3.1} \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 36.

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 37. 🤔💭 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.
Before moving on, it’s worth pausing on two questions we’ve been quietly taking for granted: does an initial value problem always have a solution, and if it does, is that solution the only one? For most of the equations in this book the answer to both is yes, and there is a classical result that says exactly when:

✳️ Existence & Uniqueness (Informal).

Consider the first-order initial value problem
\begin{equation*} \frac{dy}{dx} = f(x, y), \qquad y(x_0) = y_0. \end{equation*}
If \(f\) is continuous and changes smoothly with respect to \(y\) near the starting point \((x_0, y_0)\)
 2 
The precise condition is that \(f\) and its partial derivative \(\frac{\partial f}{\partial y}\) (the derivative treating \(x\) as constant) are both continuous on a rectangle containing \((x_0, y_0)\text{.}\) This result is known as the Picard-Lindelöf theorem.
, then the initial value problem has exactly one solution on some interval around \(x_0\text{.}\)
Both conclusions matter. Existence says the problem is not asking for the impossible — some solution curve really does pass through the initial point. Uniqueness says the initial condition pins down the entire future (and past) of the solution: two solutions that agree at one point must agree everywhere they are both defined. Uniqueness is also why an order-\(n\) equation needs \(n\) initial conditions: its general solution carries \(n\) arbitrary constants, and each condition determines one of them, just as we saw with \(y(0)\) and \(y'(0)\) above.
When the smoothness hypothesis fails, uniqueness really can break. Here is the classic cautionary example.

🌌 Example 38. An IVP with More Than One Solution.

Show that the initial value problem
\begin{equation*} \frac{dy}{dx} = 3y^{2/3}, \qquad y(0) = 0 \end{equation*}
has at least two different solutions.
Solution.
The constant function \(y_1(x) = 0\) works: both sides are \(0\text{,}\) and \(y_1(0) = 0\text{.}\) But so does \(y_2(x) = x^3\text{:}\) substituting gives
\begin{equation*} y_2' = 3x^2 \qquad \text{and} \qquad 3y_2^{2/3} = 3\left(x^3\right)^{2/3} = 3x^2, \end{equation*}
and \(y_2(0) = 0\text{.}\) Two genuinely different solutions pass through the same initial point (in fact, infinitely many do — a solution can follow \(y = 0\) for a while and then peel off onto a shifted cubic).
The culprit is the hypothesis: \(f(x,y) = 3y^{2/3}\) is continuous, but it does not change smoothly with respect to \(y\) at \(y = 0\) — its \(y\)-derivative \(2y^{-1/3}\) blows up there. Right where the guarantee fails, uniqueness fails with it.
The takeaway: when you solve an IVP built from reasonably smooth functions — which describes nearly every model in this book — you can trust that the particular solution you found is the solution. But the guarantee has hypotheses, and the example above shows they are not just legal fine print.
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