Lesson 14 (09/25/20): General Solutions to Linear Homogeneous Equations (3.2)

Lesson 14 (09/25/20): General Solutions to Linear Homogeneous Equations (3.2)

Review example: Solve \( y'' + y' - 12y = 0 \), \( y(0) = 3 \), \( y'(0) = 2 \)

Characteristic equation:

\[ r^2 + r - 12 = 0 \] \[ (r - 3)(r + 4) = 0 \] \[ r = 3, \quad r = -4 \]

\( y_1(x) = e^{3x} \), \( y_2(x) = e^{-4x} \) are two L.I. solutions

General solution: \( y(x) = c_1 e^{3x} + c_2 e^{-4x} \)

\[ y(0) = 3 = c_1 + c_2 \] \[ y'(x) = 3c_1 e^{3x} - 4c_2 e^{-4x} \] \[ y'(0) = 2 = 3c_1 - 4c_2 \] \[ c_1 + c_2 = 3 \quad \text{①} \] \[ 3c_1 - 4c_2 = 2 \quad \text{②} \] \[ 4\text{①} + \text{②} \] \[ 7c_1 = 14 \implies c_1 = 2 \]

plug in ①

\[ 1 + c_2 = 3 \implies c_2 = 1 \] \[ y(x) = 2e^{3x} + e^{-4x} \]

eg: Solve \( y'' - 2y' + y = 0 \)

Ch. equation: \[ r^2 - 2r + 1 = 0 \] \[ (r - 1)^2 = 0 \] \[ r = 1 \text{ is the only root} \]

\( y_1 = e^x \) is a solution to \( y'' - 2y' + y = 0 \).

Try! \( y_2 = x e^x \rightsquigarrow y_2' = x e^x + e^x \)

\[ y_2'' = x e^x + 2e^x \] \[ y_2'' - 2y_2' + y_2 = (x e^x + 2e^x) - 2(x e^x + e^x) + x e^x = 0 \]

\( y_2 = x e^x \) is a solution.

\( y_1, y_2 \) are L.I. (Last Lecture)

\[ y = C_1 e^x + C_2 x e^x \text{ is General solution.} \]

Second order linear homogeneous, constant coeff.

\[ ay'' + by' + cy = 0 \]

ch. equation: \[ ar^2 + br + c = 0 \]

roots: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

1. \( b^2 - 4ac > 0 \)

   \[ r_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}, \quad r_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \quad \text{are two roots} \]

   \( y_1 = e^{r_1 x}, \quad y_2 = e^{r_2 x} \quad \text{two solutions} \rightsquigarrow y = C_1 e^{r_1 x} + C_2 e^{r_2 x} \text{ is general sol.} \)

2. \( b^2 - 4ac = 0 \)

   \( r = -b/2a \) is the only root

   \( y_1 = e^{rx}, \quad y_2 = xe^{rx} \) are two solutions \( \rightsquigarrow y = C_1 e^{rx} + C_2 x e^{rx} \) is gen. solution

3. \( b^2 - 4ac < 0 \rightsquigarrow \) Next week

Method of Reduction of Order

Linear homogeneous second order equation

\[ a(x) y'' + b(x) y' + c(x) = 0 \]

If \( y_1(x) \) is a solution

Look for second solution of the form \( v(x) y_1(x) \)

\(\rightarrow\) Let \( y_2(x) = v(x) y_1(x) \)

want \[ a(x) y_2'' + b(x) y_2' + c(x) y_2 = 0 \]

Find \( v(x) \) by solving this and using \( y_1(x) \) is a solution

\[ a(x) y_1'' + b(x) y_1' + c(x) y_1 = 0 \]

\[ y'' - 2y' + y = 0 \]

We know: \( y_1(x) = e^x \) is a solution

Find \( v(x) \) such that \( y_2(x) = v(x) y_1(x) = v(x) e^x \) is sol.

\[ y_2' = v(x) e^x + v'(x) e^x \] \[ y_2'' = v(x) e^x + v'(x) e^x + v'(x) e^x + v''(x) e^x \]

Want \( y_2'' - 2y_2' + y_2 = 0 \)

\[ \underbrace{\left[ v(x) e^x + 2v'(x) e^x + v''(x) e^x \right]}_{y_2''} - 2 \underbrace{\left[ v(x) e^x + v'(x) e^x \right]}_{y_2'} + \underbrace{\left[ v(x) e^x \right]}_{y_2} = 0 \] \[ \rightsquigarrow v''(x) e^x = 0 \implies \text{Solve } v''(x) = 0 \rightsquigarrow v(x) = c_1 + c_2 x \]

Choose \( c_1, c_2 \) such that \( v(x) y_1 \) is L.I to \( y_1 \implies \underline{\text{Choose }} v(x) = x \).

\[ v(x) = c_1 + c_2 x \]

\( v(x) y_1(x) = (c_1 + c_2 x) e^x \) is a solution for any value of \( c_1, c_2 \)

Look for \( c_1, c_2 \) such that \( (c_1 + c_2 x) e^x \) is L.I to \( e^x \).

Example: \( y_1(x) = x \) is a solution to \( x^2 y'' - 2x y' + 2y = 0 \)

Find \( y_2(x) \) such that \( y = c_1 y_1 + c_2 y_2 \) is general solution

Verify: \( y_1' = 1 \), \( y_1'' = 0 \)

\[ x^2 y_1'' - 2x y_1' + 2y_1 = x^2(0) - 2x(1) + 2(x) = 0 \quad \checkmark \]

Look for:

\( y_2 = v(x) y_1 \implies y_2 = v(x) \cdot x \),   \( y_2' = v'(x) \cdot x + v(x) \)

\[ y_2'' = v''(x) \cdot x + v'(x) + v'(x) \] \[ x^2 y_2'' - 2x y_2' + 2y_2 = x^2 [v''(x) \cdot x + 2v'(x)] - 2x [v'(x) \cdot x + v(x)] + 2v(x) \cdot x = 0 \] \[ v''(x) \cdot x^3 = 0 \implies v''(x) = 0 \] \[ v(x) = C_1 + C_2 x \]

Find \( C_1, C_2 \) such that \( y_2 = (C_1 + C_2 x)x \) is L.I to \( y_1 = x \)

\( \Rightarrow y_2(x) = x^2 \) is a solution, L.I to \( y_1 = x \)

Summary: (Homogeneous, Linear, \(2^{\text{nd}}\) order)

\[ a(x)y'' + b(x)y' + c(x)y = 0 \]

Principle of superposition:

if \(y_1, y_2\) are linearly independent solutions then \(y(x) = c_1 y_1 + c_2 y_2\) is general solution

Special cases:

\(a(x), b(x), c(x)\) are constants

look for \(e^{rx}\) as solution

\(r\) is root to \(ar^2 + br + c = 0\).

Linear independence:

\(y_1, y_2\) are L.I if

\[ c_1 y_1 + c_2 y_2 = 0 \text{ has } c_1 = 0, c_2 = 0 \text{ as the only solution.} \]

— \(W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} \neq 0\) then \(y_1, y_2\) are L.I

Generalize to \(n^{\text{th}}\) order Linear, Homogeneous ODE

\[ a_n(x) y^{(n)}(x) + a_{n-1}(x) y^{(n-1)}(x) + \cdots + a_3(x) y'''(x) + a_2(x) y''(x) + a_1(x) y'(x) + a_0(x) = 0 \]

Linearly independent functions

\(y_1, y_2, \dots, y_n\) are \(n\) L.I functions if

\[ c_1 y_1 + c_2 y_2 + \cdots + c_n y_n = 0 \]

has \(c_1 = 0, c_2 = 0, \dots, c_n = 0\) is the only solution

Principle of superposition:

If \(y_1, y_2, \dots, y_n\) are \(n\) L.I solutions then \(y = c_1 y_1 + c_2 y_2 + \cdots + c_n y_n\) is the General solution

Wronskian

\[ W(y_1, y_2, y_3, \dots, y_n) = \begin{vmatrix} y_1 & y_2 & y_3 & \cdots & y_n \\ y_1' & y_2' & y_3' & \cdots & y_n' \\ y_1'' & y_2'' & y_3'' & \cdots & y_n'' \\ \vdots & \vdots & \vdots & & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \cdots & \cdots & y_n^{(n-1)} \end{vmatrix} \]

if \[ W(y_1, \dots, y_n) \begin{cases} \neq 0 & y_1, \dots, y_n \text{ are L.I} \\ = 0 & y_1, \dots, y_n \text{ are linearly dependent} \end{cases} \]

eg: Show that \( 1, x, x^2 \) are L.I functions

\[ W(1, x, x^2) = \begin{vmatrix} 1 & x & x^2 \\ 0 & 1 & 2x \\ 0 & 0 & 2 \end{vmatrix} = 2 \neq 0 \]

\( \Rightarrow 1, x, x^2 \) are L.I

Example: Show that \(2\), \(\cos^2 x\), \(3\sin^2 x\) are linearly dependent

Find \(c_1, c_2, c_3 \neq 0\) s.t.

\[ 2c_1 + c_2 \cos^2 x + 3c_3 \sin^2 x = 0 \] \[ \left. \begin{aligned} c_1 &= -3 \\ c_2 &= 6 \\ c_3 &= 2 \end{aligned} \right\} \quad -6 + 6\cos^2 x + 6\sin^2 x \] \[ = -6 + 6 [\cos^2 x + \sin^2 x] \] \[ = 0 \]

Check: \(W(2, \cos^2 x, 3\sin^2 x) = 0\)