All are linear combos of \( y_1 \) and \( y_2 \) (fundamental solutions) so all satisfy DE.
Check Wronskian: \( W \neq 0 \), then they are fundamental solutions.
char. eq. \( r^2 + ar + b = 0 \implies r = r_1, r_2 \)
if roots are complex: \( r = \lambda \pm i\mu \)
both complex or both real
We begin with the differential equation:
The characteristic equation is:
Using the quadratic formula to solve for \(r\):
Thus, the roots are:
The first fundamental solution is:
complex value \(y_1\)
The second fundamental solution is:
Using the property \(e^{-it} = e^{i(-t)}\), we simplify to:
The general solution is a linear combination of the fundamental solutions:
To find the real solution, we group the terms:
Let \(k_1 = c_1 + c_2\) and \(k_2 = i(c_1 - c_2)\).
\(k_1\) and \(k_2\) are real and depend on Initial Conditions (IC's).
\(c_1\) and \(c_2\) are complex conjugate pairs.
In general, if the roots of the characteristic equation are complex conjugates:
Then the general solution is:
Note: \( C_1, C_2 \) are real, and depend on Initial Conditions (ICs).
Consider the differential equation:
The characteristic equation is:
The general solution is:
Solve the following second-order differential equation with initial conditions:
Using \( y(0) = 3 \):
\[ 3 = C_1 \]
Using \( y'(0) = 1 \) and substituting \( C_1 = 3 \):
\[ 1 = 3\left(-\frac{1}{2}\right) + C_2(1) \implies C_2 = \frac{5}{2} \]
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Particular Solution:
\[ \lim_{t \to \infty} y = 0 \quad \text{because of } e^{-\frac{1}{2}t} \]
\( \sin, \cos \to \) oscillations
combine \( \to \) decaying oscillations
In general, \( r = \lambda \pm i\mu \)
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The graph below illustrates the particular solution \( y = 3e^{-\frac{1}{2}t} \cos t + \frac{5}{2} e^{-\frac{1}{2}t} \sin t \). The solid blue line represents the oscillating function, while the dashed green lines represent the exponential envelopes \( e^{-\frac{1}{2}t} \) and \( -e^{-\frac{1}{2}t} \).
