Warmup: Find solution to \( y' = 2y \) (1st order Linear, Seperable)
Using methods we know: \( y(x) = C e^{2x} \) is General Solution
Another way: Look for an exponential function as solution
if \( y = e^{rx} \) is a solution
want: \( y' = 2y \)
\[ r \cdot e^{rx} = 2 e^{rx} \implies (r - 2) e^{rx} = 0 \]
Since \( e^{rx} \neq 0 \), then \( (r - 2) = 0 \) when \( r = 2 \).
\( y = e^{2x} \) is solution
\( y'' \) is also involved in the equation
General form: \( f(x, y, y', y'') = 0 \)
eg:
\[ y'' = 0 \] \[ x y'' + 2y' + y^3 = \tan x \] \[ y'' \cdot (y')^2 = e^{-2x} \]
Linear in \( y, y', y'' \)
\[ A(x)y'' + B(x)y' + C(x)y = D(x) \]
is the General form for Linear \( 2^{\text{nd}} \) order ODE.
\( D(x) = 0 \rightsquigarrow \) Linear Homogeneous ODE
\( D(x) \neq 0 \rightsquigarrow \) Linear Non-Homogeneous ODE
Warning! Do not confuse with \( y' = \phi\left(\frac{y}{x}\right) \) (Homogeneous function).
\( = 1 - x \)
Linear, Non-Homogeneous
Non-linear
Linear, Non-Homogeneous
Linear, Homogeneous
Linear, Homogeneous
\[ A(x)y'' + B(x)y' + C(x)y = D(x) \]
with special case: \( A(x), B(x), C(x) \) constants, \( D(x) = 0 \)
\[ ay'' + by' + cy = 0, \quad a \neq 0 \]
Eg: \( y'' + 4y' + 3y = 0 \)
idea: Look for solutions of the form \( y = e^{rx} \)
\[ y' = re^{rx} \]
\[ y'' = r^2 e^{rx} \]
want \( y'' + 4y' + 3y = 0 \rightsquigarrow r^2 e^{rx} + 4re^{rx} + 3e^{rx} = 0 \)
\[ (r^2 + 4r + 3)\underset{>0}{e^{rx}} = 0 \]
\( e^{rx} \) is a solution only if
\[ \left. r^2 + 4r + 3 = 0 \right\} \text{ called characteristic equation.} \]
Find roots of characteristic equation: \(r^2 + 4r + 3 = 0\)
\[(r + 1)(r + 3) = 0\]
\(r = -1\), \(r = -3\) are two solutions
\(y_1(x) = e^{-x}\) and \(y_2(x) = e^{-3x}\) are two solutions to \(y'' + 4y' + 3y = 0\)
\[y' = -5e^{-x} - 9e^{-3x}\]
\[y'' = 5e^{-x} + 27e^{-3x}\]
\[y'' + 4y' + 3y = (5e^{-x} + 27e^{-3x}) + (-20e^{-x} - 36e^{-3x}) + (15e^{-x} + 9e^{-3x}) = 0\]
Yes, it is a solution
In fact:
\[ y(x) = c_1 y_1(x) + c_2 y_2(x) = c_1 e^{-x} + c_2 e^{-3x} \]
is a solution to
\[ y'' + 4y' + 3y = 0 \quad \text{for any } c_1, c_2 \in \mathbb{R}. \]
If \( y_1(x), y_2(x) \) are two “linearly independent” solutions to
\[ a y'' + b y' + c y = 0 \]
then
\[ y(x) = c_1 y_1(x) + c_2 y_2(x) \]
is the general solution.
Defn: \(y_1(x)\) & \(y_2(x)\) are linearly independent (L.I) if \(k_1 y_1(x) + k_2 y_2(x) = 0\) implies that \(k_1 = 0\) & \(k_2 = 0\)
(Same as saying \(y_1(x) \neq k y_2(x)\))
\[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_1' y_2 \]
\(W(y_1, y_2) \neq 0 \rightsquigarrow y_1, y_2\) are L.I
\(W(y_1, y_2) = 0 \rightsquigarrow y_1, y_2\) are Linearly Dependent
Eg:
\[ y_1 = e^{-x}, \quad y_2 = e^{-3x} \]
\[ y_1' = -e^{-x}, \quad y_2' = -3e^{-3x} \]
\[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} e^{-x} & e^{-3x} \\ -e^{-x} & -3e^{-3x} \end{vmatrix} \]
\[ = -3e^{-3x} \cdot e^{-x} + e^{-3x} \cdot e^{-x} \]
\[ = -2e^{-4x} \]
\[ \neq 0 \]
\( \Rightarrow \quad e^{-x}, \quad e^{-3x} \quad \text{are} \quad \text{L independent} \)
eg! \( y_1(x) = \sin x, \quad y_2(x) = 3\sin x \)
\( y_2 = 3y_1 \) is L. dependent
\( y_1' = \cos x, \quad y_2' = 3\cos x \)
\[ W(y_1, y_2) = \begin{vmatrix} \sin x & 3\sin x \\ \cos x & 3\cos x \end{vmatrix} = 3\sin x \cos x - 3\sin x \cos x \] \[ = 0, \]
eg: \( y_1 = e^x \), \( y_2 = x e^x \)
\( y_1' = e^x \), \( y_2' = x e^x + e^x \)
\[ W(y_1, y_2) = \begin{vmatrix} e^x & x e^x \\ e^x & x e^x + e^x \end{vmatrix} = e^x [x e^x + e^x] - e^x [x e^x] \] \[ = e^{2x} \neq 0 \]
\( e^x \), \( x e^x \) are L.I.
eg: Find General solution to \( y'' - 2y' + y = 0 \)
\( y = e^{rx} \) is solution
\[ \rightsquigarrow (r^2 - 2r + 1)e^{rx} = 0 \]
ch. equation: \( r^2 - 2r + 1 = 0 \)
\[ r = 1 \text{ is the only root} \]
\( y_1 = e^x \) is a solution to \( y'' - 2y' + y = 0 \)
But we need a second solution to write Gen. Sol.
\( \rightarrow \) Next Lecture!!