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5.2 General Solutions of Linear Diff. Eqs

last time: \( y'' + p(x)y' + q(x)y = f(x) \)

Note: The coefficients \( p(x) \), \( q(x) \), and the function \( f(x) \) cannot contain \( y \) (else nonlinear).

The associated homogeneous eq is (where the right side is zero):

\[ y'' + p(x)y' + q(x)y = 0 \]

Solution to the homogeneous part:

\[ y = c_1 y_1 + c_2 y_2 \]

\( y_1 \) and \( y_2 \) are solutions so are their linear combos.

If not homogeneous (right side not zero) then solution is:

\[ y = \underbrace{c_1 y_1 + c_2 y_2}_{\text{homogeneous part}} + \underbrace{y_p}_{\text{part due to right side}} \]

We add them: The principle of superposition

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We will also try to understand the structure of higher-order linear eqs.

For example, 3rd-order: \( y''' = 0 \)

Solution is easy: \( y = c_1 x^2 + c_2 x + c_3 \)

  • linear combo of \( x^2, x, \text{ and } 1 \)
  • 3 functions spanning the space

General solution of a linear 3rd-order

\[ y = c_1 y_1 + c_2 y_2 + c_3 y_3 \]

That is true for \( n^{\text{th}} \)-order: \( y^{(n)} + p_1(x)y^{(n-1)} + p_2(x)y^{(n-2)} + \dots + p_n(x)y = f(x) \)

Solution:

\[ y = \underbrace{c_1 y_1 + c_2 y_2 + \dots + c_n y_n}_{\text{homogeneous part}} + \underbrace{y_p}_{\text{due to right side}} \]
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The n solutions are linearly independent

How to check? For example, for 3rd-order: \( y_1, y_2, y_3 \)

Linearly independent \( \rightarrow c_1 y_1 + c_2 y_2 + c_3 y_3 = 0 \iff c_1 = c_2 = c_3 = 0 \) for ALL \( x \) on an interval.

deriv:\[ c_1 y_1' + c_2 y_2' + c_3 y_3' = 0 \]
deriv:\[ c_1 y_1'' + c_2 y_2'' + c_3 y_3'' = 0 \]
\[ \begin{bmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

Unique solution of \( c_1 = c_2 = c_3 = 0 \) if and only if

\[ \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \end{vmatrix} \neq 0 \]

Wronskian of \( y_1, y_2, y_3 \)

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For example, are \( y_1 = 1, y_2 = \sin^2 x \) and \( y_3 = \cos^2 x \) linearly independent?

(This is a case we can tell by inspection:)

\[ \sin^2 x + \cos^2 x = 1 \]

So, \( c_1 \cdot 1 + c_2 \sin^2 x + c_3 \cos^2 x = 0 \)

has a solution with \( c_1, c_2, c_3 \) not all zero.

So they are not linearly independent.

The Wronskian can give us the answer whether we can tell by inspection

\[ W(1, \cos^2 x, \sin^2 x) = \begin{vmatrix} 1 & \cos^2 x & \sin^2 x \\ 0 & -2 \cos x \sin x & 2 \sin x \cos x \\ 0 & 2 \sin^2 x - 2 \cos^2 x & -(2 \sin^2 x - 2 \cos^2 x) \end{vmatrix} \]
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Wronskian and Linear Dependence

Expand along column 1:

\[ = \begin{vmatrix} -2 \cos x \sin x & 2 \sin x \cos x \\ 2 \sin^2 x - 2 \cos^2 x & -(2 \sin^2 x - 2 \cos^2 x) \end{vmatrix} \]
\[ = \dots = 0 \]

Wronskian is identically zero for ALL \( x \)

So, these three functions are linearly dependent on all intervals of \( x \).

If \( W \neq 0 \) all the time, then the functions are linearly independent on at least some intervals of \( x \).

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Generalizing the Wronskian

For 4 functions:

\[ W = \begin{vmatrix} y_1 & y_2 & y_3 & y_4 \\ y_1' & y_2' & y_3' & y_4' \\ y_1'' & y_2'' & y_3'' & y_4'' \\ y_1''' & y_2''' & y_3''' & y_4''' \end{vmatrix} \]

For \( n \) functions:

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

Square matrix

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Reduction of Order

Sometimes a high-order eq. can be reduced to a lower-order one if we know at least one solution (somehow).

Let's see a 2nd-order example:

\[ x^2 y'' + x y' - 9y = 0 \]
linear but NOT constant coefficient (this is an Euler's equation)

Suppose that, somehow, we know one solution

\[ y_1 = x^3 \]

Since this is 2nd-order, we know there is one more \( y_2 = ? \)

reduction of order method:
\[ y_2 = v(x) y_1 \]

To find \( v(x) \), sub \( y_2 = v(x) y_1 \) into the original diff. eq.

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\[ y_2 = v y_1 = v x^3 \] \[ y_2' = 3v x^2 + v' x^3 \] \[ y_2'' = 6v x + 6v' x^2 + v'' x^3 \]

sub into \( x^2 y'' + x y' - 9y = 0 \)

\[ x^2 (6v x + 6v' x^2 + v'' x^3) + x (3v x^2 + v' x^3) - 9(v x^3) = 0 \]

Simplify

\[ x^5 v'' + 7v' x^4 = 0 \] \[ \rightarrow v'' = -7v' x^{-1} \]
\[ \frac{d(v')}{dx} = -\frac{7}{x} (v') \]
\[ v'' = \frac{d(v')}{dx} \]

separable in \( v' \) and \( x \)

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\[ \frac{1}{v'} d(v') = -\frac{7}{x} dx \]

solve separable

\[ v' = C x^{-7} \]

Integration

\[ v = \int C x^{-7} dx = -\frac{C}{6} x^{-6} + D \]

Note: D is a constant

\[ v = -\frac{C}{6} x^{-6} + D \quad \text{and} \quad y_2 = v y_1 = v x^3 \]

Choosing Constants

Choose ANY convenient \( C, D \) except those that lead to \( v = 0 \) or \( v = \text{constant} \).

Here, we choose \( D = 0 \), \( C = -6 \)

\[ v = x^{-6} \]

Final Solution

So,

\[ y_2 = v y_1 = v x^3 = x^{-6} x^3 = \boxed{x^{-3}} \]