Superposition Principle: Because the differential equation is linear and homogeneous, if \(
y_1(t) \) and \( y_2(t) \) are linearly independent solutions, any linear combination \( c_1 y_1 + c_2 y_2 \) is
also a solution and forms the general solution.
(in linear algebra terminology, \( y_1 \) and \( y_2 \) span the solution space)
\( c_1, c_2 \) come from two
initial conditions: \( y(t_0) = y_0 \), \( y'(t_0) = y_0' \)
Initial Value Problem (IVP): A second-order differential equation requires two initial
conditions to uniquely solve for the two unknown arbitrary constants \( c_1 \) and \( c_2 \) by setting up a \(
2 \times 2 \) linear system.
PAGE 2
Example: Second-Order Linear
ODE with Initial Values
Step 1: Characteristic Equation. For a linear homogeneous ODE
with constant coefficients, we test exponential solutions of the form \( y = e^{rt} \). Substituting this into
the ODE yields the characteristic polynomial.
Solutions: \( y = e^{rt} \)
Sub into diff. eq. \( \rightarrow r^2 + 8r - 9 = 0 \)
\[ y =
\underbrace{c_1 e^{-9t} + c_2 e^t}_{\substack{\text{competition between the two solutions} \\ \text{the weights }
c_1, c_2 \text{ determine the} \\ \text{long term behavior } (t \to \infty) \\ \text{the initial conditions determine } c_1, c_2}} \]
Asymptotic Behavior: Notice that as \( t \to \infty \), \(
e^{-9t} \to 0 \) while \( e^t \to \infty \). Hence, unless \( c_2 = 0 \), the term \( c_2 e^t \) dictates the
long-term growth or decay of the system.
let's look at how \( y(0) = 1, \; y'(0) = 0 \) affect the solution
need to differentiate \( y = c_1 e^{-9t} + c_2 e^t \) to use \( y'(0) = 0 \)
\[
y' = -9 c_1 e^{-9t} + c_2 e^t
\]
cannot lump into just \( C_1 \) since we care what \( c_1, c_2 \) are
Instructor Note: Unlike during integration where arbitrary constants can be grouped into an
overall constant, here \( c_1 \) and \( c_2 \) are specific coefficients attached to linearly independent
solutions \( e^{-9t} \) and \( e^t \). They must remain distinct to satisfy both initial conditions.
System of Equations: Combining equation ① from the initial position \( y(0) = 1 \implies c_1 +
c_2 = 1 \) with equation ② from the initial velocity \( -9 c_1 + c_2 = 0 \implies c_2 = 9 c_1 \). Substituting
yields \( 10 c_1 = 1 \implies c_1 = \frac{1}{10} \) and \( c_2 = \frac{9}{10} \).
solve ① and ② simultaneously, we get \( c_1 = \frac{1}{10},\; c_2 = \frac{9}{10} \)
Solution:\[ y = \frac{1}{10} e^{-9t} + \frac{9}{10} e^t \]
particular solution
Component Functions &
Asymptotic Behavior
Examining the individual behavior of the component exponential terms as \( t \to \infty \):
\( e^{-9t} \)
Figure: Coordinate graph of the decaying exponential function \( y = e^{-9t} \)
Figure: A coordinate plot showing exponential decay for \( y = e^{-9t} \). As \( t \to \infty \), the curve decays asymptotically toward the positive horizontal \( t \)-axis.
\( e^t \)
Figure: Coordinate graph of the growing exponential function \( y = e^t \)
Figure: A coordinate plot showing exponential growth for \( y = e^t \). As \( t \to \infty \), the curve increases rapidly toward positive infinity in the first quadrant.
Long-Term Limit Explanation: As \( t \to \infty \), the transient term \( \frac{1}{10} e^{-9t}
\to 0 \), while the growing term \( \frac{9}{10} e^t \to \infty \). Therefore, the overall trajectory is dominated
by \( e^t \), resulting in \( \lim_{t \to \infty} y(t) = +\infty \).
PAGE 4
Graph of \( y = \frac{1}{10}e^{-9t} + \frac{9}{10}e^t \)
Figure: coordinate graph showing curve \( y = \frac{1}{10}e^{-9t} + \frac{9}{10}e^t \) with horizontal tangent line at \( (0, 1) \)
Figure: Plot of the solution curve \( y = \frac{1}{10}e^{-9t} + \frac{9}{10}e^t \) versus time \( t \) on the interval \( [-0.50, 1.50] \). The function features an initial position at \( (0, 1) \) highlighted by a red point, where the derivative \( y'(0) = 0 \) produces a horizontal tangent line (dashed green line, \( m = 0 \)).
\( y(0) = 1 \)— initial position
\( y'(0) = 0 \)— initial slope
Initial Value Verification:
Evaluating \( y(t) = \frac{1}{10}e^{-9t} + \frac{9}{10}e^t \) at \( t = 0 \) yields \( y(0) = \frac{1}{10}(1) +
\frac{9}{10}(1) = 1 \).
Differentiating gives \( y'(t) = -\frac{9}{10}e^{-9t} + \frac{9}{10}e^t \). Evaluating at \( t = 0 \) yields \(
y'(0) = -\frac{9}{10} + \frac{9}{10} = 0 \), corresponding to the horizontal tangent line \( m = 0 \) at the
minimum point \( (0, 1) \).
PAGE 5
Effect of Changing Initial Slope on Long-Term
Behavior
Let's see how changing initial slope changes \(\lim_{t \to \infty} y\):
Given initial conditions:
\(y(0) = 1\) (same position)
\(y'(0) = -10\)
Note: The initial
position \(y(0) = 1\) remains unchanged from previous cases, but a significantly negative initial slope
\(y'(0) = -10\) is introduced to investigate its effect on the asymptotic behavior of the system.
Subtracting the
second equation from the first eliminates \(c_2\): \[1 - (-10) = c_1 - (-9c_1) \implies 11 = 10c_1 \implies c_1 =
\frac{11}{10}\] Substituting \(c_1\) back yields \(c_2 = 1 - \frac{11}{10} = -\frac{1}{10}\).
Particular solution:
\[y = \frac{11}{10} e^{-9t} - \frac{1}{10} e^t\]
Long-term limit:
\[\lim_{t \to \infty} y = -\infty\]
As \(t \to
\infty\), the decaying exponential \(e^{-9t} \to 0\), leaving the growing exponential \(-\frac{1}{10}e^t\).
Because its coefficient is negative, the entire expression tends to \(-\infty\). Thus, a sufficiently steep
negative initial velocity reverses the long-term divergence toward negative infinity.
PAGE 6
Graph of \( y = \frac{11}{10}e^{-9t} - \frac{1}{10}e^t
\)
The following graph displays the solution curve alongside its linear approximation (tangent line) at the initial
location \( (0, 1) \).
Figure: Graph of \( y = \frac{11}{10}e^{-9t} - \frac{1}{10}e^t \) with tangent line at \( (0, 1) \) having slope \( m = -10 \)
Figure: Graph of \( y = \frac{11}{10}e^{-9t} - \frac{1}{10}e^t \) plotted over the domain \( t \in [-0.4, 0.8] \) with vertical range \( y \in [-5, 10] \). The blue solid curve depicts the function decaying sharply from large positive values for \( t < 0 \) down towards \( y \approx 0 \) for \( t > 0.2 \). The green dashed line shows the tangent line \( y = -10t + 1 \) with slope \( m = -10 \) through the initial point \( (0, 1) \), which is marked by a red solid circle.
Observation on Behavior: The steep negative derivative \( y'(0) = -10 \) at \( t = 0 \) is
dominated by the rapid decay mode \( -9 \cdot \frac{11}{10}e^{-9t} = -9.9 \). For small positive values of \( t
\), the linear approximation \( y \approx 1 - 10t \) matches the curve closely before the fast exponential
component \( e^{-9t} \) dissipates.
PAGE 7
Asymptotic Behavior & Initial Conditions
Is it possible to fix \( y(0) = 1 \) and change \( y'(0) \) such
that \( \lim_{t \to \infty} y = 0 \)?
Goal: Determine whether an initial slope \( y'(0) = m \) exists
that suppresses the growing exponential mode \( e^t \), ensuring the solution decays to zero as \( t \to \infty
\).
Note: Both characteristic roots \(r_1 = 2\) and \(r_2 = 3\) are positive. Hence, as \(t \to
\infty\), both exponential modes \(e^{2t}\) and \(e^{3t}\) grow unbounded unless their coefficients are zero.
Solving the System: From the first equation, \(c_1 = 2 - c_2\). Substituting into the second
equation yields \(m = 2(2 - c_2) + 3c_2 = 4 + c_2\), which gives \(c_2 = m - 4\). Consequently, \(c_1 = 2 - (m -
4) = 6 - m\).
⋮
\[
c_1 = 6 - m, \qquad c_2 = m - 4
\]
PAGE 9
Long-Term Behavior and Equilibrium Analysis
General solution parameterized by \( m \):
\[ y = (6-m)e^{2t} + (m-4)e^{3t} \]
Testing Sample Values of \( m \)
Let's try \( m = 0 \):
\[ y = 6e^{2t} - 4e^{3t} \]
\[ \lim_{t \to \infty} y = -\infty \]
For \( m = 5 \):
\[ y = e^{2t} + e^{3t} \]
\[ \lim_{t \to \infty} y = \infty \]
Dominant Term Analysis: As \( t \to \infty \), the exponential term \( e^{3t} \) grows strictly
faster than \( e^{2t} \). The sign of its coefficient, \( (m - 4) \), dictates the asymptotic limit: if \( m > 4
\), \( y(t) \to +\infty \); if \( m < 4 \), \( y(t) \to -\infty \). At the transition \( m=4 \), the \( e^{3t} \)
term vanishes and \( y(t)=2e^{2t} \to +\infty \).
Feasibility of a Constant (Flat Line) Solution
Is it possible to choose \( m \) such that \( \lim_{t \to \infty} y = 2 \) (flat line)?
Figure: Coordinate sketch showing solution trajectories from \( y = 2 \) diverging to positive and negative infinity, contrasting with a hypothetical flat line.
Figure: A coordinate sketch illustrating possible behaviors of the solution \( y(t) \) on a set of axes with vertical axis \( y \) and horizontal axis \( t \). The curve starts at the initial condition \( y(0) = 2 \) on the vertical axis. One trajectory curves upwards towards positive infinity (labeled "to \( \infty \)"), while another curves downward towards negative infinity (labeled "to \( -\infty \)"). A horizontal dashed line extends from \( y = 2 \) labeled "possible?", questioning whether a constant solution \( y(t) = 2 \) can exist.
If the flat solution exists \( \to y = 2 \) for all \( t \).
Does it satisfy \( y'' - 5y' + 6y = 0 \)?
No, not possible.
Verification: If \( y(t) = 2 \) (a constant function) were a valid solution to the differential
equation, its derivatives would be \( y'(t) = 0 \) and \( y''(t) = 0 \). Substituting these into the homogeneous
equation:
\[ y'' - 5y' + 6y = 0 - 5(0) + 6(2) = 12 \neq 0 \]
Because \( 12 \neq 0 \), \( y = 2 \) is not an equilibrium or valid solution to the homogeneous equation.
PAGE 10
Solutions of \( y'' - 5y' + 6y = 0 \) with \( y(0) = 2
\) for Various \( y'(0) \)
Visualizing the sensitivity of the initial value problem to the initial derivative \( y'(0) \), demonstrating how
the boundary condition \( c_2 = 0 \) separates solutions tending to \( +\infty \) from those diverging to \(
-\infty \).
Figure: Coordinate plot displaying solution curves of \( y'' - 5y' + 6y = 0 \) with \( y(0) = 2 \) across varying initial slopes \( y'(0) \).
Figure: Graph of solution curves for the initial value problem \( y'' - 5y' + 6y = 0 \) with \( y(0) = 2 \), plotted over the time range \( t \in [0.0, 1.2] \) with \( y(t) \in [-10, 15] \). All trajectories originate at the marked initial state point \( (0, 2) \). Individual curves correspond to different values of the initial derivative \( y'(0) \): blue for \( y'(0) = 6 \), green for \( y'(0) = 5 \), purple for \( y'(0) = 4 \) (representing the boundary solution where \( c_2 = 0 \)), orange for \( y'(0) = 2 \), red for \( y'(0) = 0 \), and brown for \( y'(0) = -2 \). A horizontal dashed line at \( y = 0 \) marks the equilibrium axis.
Analytical Derivation &
Behavior
The characteristic equation associated with the second-order homogeneous linear differential equation \( y'' - 5y'
+ 6y = 0 \) is:
