(I) \(n^{th}\)-order Linear Differential Equations

An \(n^{th}\)-order differential equation has the general form: \(G(x, y, y', \dots, y^{(n)}) = 0\).

The standard form for an \(n^{th}\)-order linear differential equation is:

$$y^{(n)} + p_1(x)y^{(n-1)} + \dots + p_n(x)y = f(x) \quad (1)$$

The associated homogeneous linear differential equation is:

$$y^{(n)} + p_1(x)y^{(n-1)} + \dots + p_n(x)y = 0 \quad (1h)$$

where the coefficient functions \(p_i(x)\) and the forcing function \(f(x)\) are continuous on some open interval \(I\).

Theorem 2: Existence and Uniqueness for Linear Equations

Suppose that \(p_1, \dots, p_n\) and \(f\) are continuous on an open interval \(I\) containing a point \(a\). Then the Initial Value Problem:

$$y^{(n)} + p_1(x)y^{(n-1)} + \dots + p_n(x)y = f(x)$$

$$y(a)=b_0, y'(a)=b_1, \dots, y^{(n-1)}(a)=b_{n-1}$$

has a unique solution on \(I\).

Theorem 1: Principle of Superposition for Homogeneous Equations

Let \(y_1, \dots, y_n\) be \(n\) solutions of the homogeneous equation (1h) on an interval \(I\). Then the linear combination:

$$y = c_1y_1 + \dots + c_ny_n$$

is also a solution on \(I\). These theorems are generalizations of Theorems 1 and 2 from Section 3.1.

(II) Linearly Independent Solutions and General Solutions

Definition: A set of \(n\) functions \(f_1, f_2, \dots, f_n\) is said to be linearly dependent on an interval \(I\) if there exists a linear dependence relation: \(c_1f_1 + \dots + c_nf_n = 0\) for all \(x\) on \(I\), where the constants \(c_1, \dots, c_n\) are not all zero. Otherwise, they are linearly independent.

Example 1: Determine Linear Independence

(1) \(f_1 = \sin(2x), f_2 = \sin(x)\cos(x), f_3 = e^x\):

Since \(\sin(2x) = 2\sin(x)\cos(x)\), these functions are linearly dependent. The relation is \(\sin(2x) - 2\sin(x)\cos(x) + 0 \cdot e^x = 0\).

(2) \(f_1 = x+1, f_2 = x^2, f_3 = x^3-x\):

Setting \(c_1(x+1) + c_2x^2 + c_3(x^3-x) = 0\) and rearranging powers of \(x\):

\(c_3x^3 + c_2x^2 + (c_1-c_3)x + c_1 = 0 \implies c_3=0, c_2=0, c_1=0\).

Therefore, \(f_1, f_2,\) and \(f_3\) are linearly independent.

Suppose \(f_1, \dots, f_n\) are each \(n-1\) times differentiable. Their Wronskian is:

$$W = \begin{vmatrix} f_1 & f_2 & \dots & f_n \\ f_1' & f_2' & \dots & f_n' \\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(n-1)} & f_2^{(n-1)} & \dots & f_n^{(n-1)} \end{vmatrix}$$

Theorem 3: Wronskian of Solutions

Suppose that \(y_1, \dots, y_n\) are \(n\) solutions of the homogeneous equation on interval \(I\) where the coefficient functions are continuous. There are two possibilities:

(a) If \(y_1, \dots, y_n\) are linearly dependent on \(I\), then \(W \equiv 0\) on \(I\).

(b) If \(y_1, \dots, y_n\) are linearly independent on \(I\), then \(W \ne 0\) at each point on \(I\).

Theorem 4: General Solutions of Homogeneous Linear Equations

Let \(y_1, \dots, y_n\) be \(n\) linearly independent solutions of the homogeneous equation on \(I\). Then the linear combination \(y = c_1y_1 + \dots + c_ny_n\) is a general solution. The set \(\{y_1, y_2, \dots, y_n\}\) is called a fundamental set of solutions or a basis of the solution set.

Example 2

Use the Wronskian to prove that \(y_1=e^x, y_2=e^{2x}, y_3=e^{3x}\) are solutions of the third-order equation \(y''' - 6y'' + 11y' - 6y = 0\) and are linearly independent. Then find a particular solution satisfying \(y(0)=0, y'(0)=0, y''(0)=3\).

Calculating the Wronskian:

$$W = \begin{vmatrix} e^x & e^{2x} & e^{3x} \\ e^x & 2e^{2x} & 3e^{3x} \\ e^x & 4e^{2x} & 9e^{3x} \end{vmatrix} = \begin{vmatrix} e^x & e^{2x} & e^{3x} \\ 0 & e^{2x} & 2e^{3x} \\ 0 & 2e^{2x} & 6e^{3x} \end{vmatrix} = 2e^{6x} \ne 0$$

So, \(y_1, y_2,\) and \(y_3\) are linearly independent. The general solution is:

\(y = c_1e^x + c_2e^{2x} + c_3e^{3x}\)

\(y' = c_1e^x + 2c_2e^{2x} + 3c_3e^{3x}\)

\(y'' = c_1e^x + 4c_2e^{2x} + 9c_3e^{3x}\)

Applying initial conditions at \(x=0\):

\(c_1 + c_2 + c_3 = 0\)

\(c_1 + 2c_2 + 3c_3 = 0 \implies c_2 + 2c_3 = 0\)

\(c_1 + 4c_2 + 9c_3 = 3 \implies 3c_2 + 8c_3 = 3\)

Solving the system yields \(c_1 = \frac{3}{2}, c_2 = -3, c_3 = \frac{3}{2}\). The particular solution is:

$$y(x) = \frac{3}{2}e^x - 3e^{2x} + \frac{3}{2}e^{3x}$$

(III) Non-homogeneous Equations

Theorem 5: Solutions of Non-homogeneous Equations

Let \(y_p\) be a particular solution of the non-homogeneous equation (1) on interval \(I\). Let \(y_1, \dots, y_n\) be \(n\) linearly independent solutions of the associated homogeneous equation (1h). Then the general solution of equation (1) is:

$$y = y_c + y_p \quad \text{where } y_c = c_1y_1 + \dots + c_ny_n$$

(IV) Method of Reduction of Order

Suppose \(y_1(x)\) is one known solution of \(y'' + p(x)y' + q(x)y = 0\). The method of reduction of order consists of substituting \(y_2(x) = v(x)y_1(x)\) into the equation to determine \(v(x)\) such that \(y_2(x)\) is a second linearly independent solution.

Derivation:

\(y = v(x)y_1(x)\)

\(y' = v'(x)y_1(x) + v(x)y_1'(x)\)

\(y'' = v''(x)y_1(x) + 2v'(x)y_1'(x) + v(x)y_1''(x)\)

Substituting into \(y'' + p(x)y' + q(x)y = 0\):

\(v''(x)y_1(x) + 2v'(x)y_1'(x) + v(x)y_1''(x) + p(x)[v'(x)y_1(x) + v(x)y_1'(x)] + q(x)v(x)y_1(x) = 0\)

Since \(y_1\) is a solution (\(y_1'' + p y_1' + q y_1 = 0\)), the terms containing \(v(x)\) cancel out, leaving:

$$[y_1(x)]v'' + [2y_1'(x) + p(x)y_1(x)]v' = 0$$
Example 3

Use the method of reduction of order to solve \(x^2y'' + xy' - 9y = 0\) for \(x > 0\), given \(y_1(x) = x^3\).

Let \(y_2 = vx^3\). Then:

\(y_2' = 3x^2v + v'x^3\)

\(y_2'' = 6xv + 6x^2v' + v''x^3\)

Substitute into the differential equation:

\(x^2(6xv + 6x^2v' + v''x^3) + x(3x^2v + v'x^3) - 9vx^3 = 0\)

\(x^5v'' + 7x^4v' = 0 \implies v'' + \frac{7}{x}v' = 0\)

Let \(w = v'\), then \(w' = v''\). The equation becomes the first-order linear equation:

\(w' + \frac{7}{x}w = 0 \implies \frac{w'}{w} = -\frac{7}{x} \implies \ln(w) = -7\ln(x) + C_1\)

\(w = x^{-7} \implies v' = x^{-7}\).

Integrating gives \(v(x) = \int x^{-7} dx = -\frac{1}{6}x^{-6}\). Picking \(v(x) = x^{-6}\):

\(y_2(x) = y_1(x)v(x) = x^3 \cdot x^{-6} = x^{-3} = \frac{1}{x^3}\).

The general solution is:

$$y = c_1x^3 + \frac{c_2}{x^3}$$