\[ a y'' + b y' + c y = 0 \]
\( a, b, c \) constants
Solution Intuition
We can guess the likely solutions:
-
→ \( y \) is related to its 1st and 2nd derivs.
-
Candidate families: exponential, sine, cosine
Why these functions? The derivative of \( e^{rt} \) is \( r e^{rt} \), and the derivatives of \(
\sin(kt) \) and \( \cos(kt) \) cycle back to sines and cosines. Linear combinations of a function and its
derivatives can only sum to zero if the derivatives reproduce the same functional form.
Let's look at exponential first: \( y = e^{rt} \) for some \( r \).
Example
\[ y'' + 2y' - 3y = 0 \]
Want solution that looks like \( y = e^{rt} \).
Sub \( y = e^{rt} \), \( y' = r e^{rt} \), \( y'' = r^2 e^{rt} \) into diff. eq.:
\[
\begin{aligned}
r^2 e^{rt} + 2 r e^{rt} - 3 e^{rt} &= 0 \\[6pt]
e^{rt} (r^2 + 2r - 3) &= 0
\end{aligned}
\]
Since \( e^{rt} \neq 0 \),
So,
\[ r^2 + 2r - 3 = 0 \]
this is called the Characteristic Equation
Next Steps: Factoring the characteristic equation gives \( (r + 3)(r - 1) = 0 \), yielding roots
\( r_1 = 1 \) and \( r_2 = -3 \). The two linearly independent solutions are \( y_1 = e^{t} \) and \( y_2 =
e^{-3t} \), giving the general solution \( y(t) = c_1 e^{t} + c_2 e^{-3t} \).
