Comprehensive Guide to Ordinary Differential Equations: Complementary Functions and Particular Integrals

General Theory of Linear Differential Equations

• Standard Form of Linear Differential Equations:

  • An $n$-th order linear ordinary differential equation with constant coefficients has the general form:     $f(D) y = q(x)$     where $D = \frac{d}{dx}$ is the differential operator, $f(D)$ is a polynomial in $D$, $y$ is the dependent variable, and $q(x)$ is the non-homogeneous term (right-hand side function of $x$).

• Structure of the General Solution:

  • The general solution of an $n$-th order linear differential equation consists of two distinct components:     $y = \text{CF} + \text{PI}$     where $\text{CF}$ represents the Complementary Function and $\text{PI}$ represents the Particular Integral.

• Role of the Complementary Function (CF):

  • The Complementary Function is obtained entirely from the left-hand side (LHS) operator $f(D)$ by setting the equation to its homogeneous form ($q(x) = 0$).

  • It solves the homogeneous linear differential equation $f(D) y = 0$ and incorporates $n$ arbitrary constants for an $n$-th order differential equation.

• Role of the Particular Integral (PI):

  • The Particular Integral is derived using the right-hand side (RHS) non-homogeneous term $q(x)$.

  • The specific method to determine $\text{PI}$ depends on the mathematical form of $q(x)$, such as exponential functions $e^{ax}$, polynomials $x^m$, trigonometric functions $\cos(ax)$ or $\sin(ax)$, logarithmic functions $\ln(ax)$, or trigonometric functions like $\tan(ax)$.

Complementary Functions for Homogeneous Equations

• Procedure for Finding the Complementary Function:

  • Step 1: Form the homogeneous equation by setting the right-hand side equal to zero ($q(x) = 0$).

  • Step 2: Form the Auxiliary Equation (AE) by substituting $D = m$ into $f(D)$, resulting in the polynomial equation $f(m) = 0$.

  • Step 3: Solve the polynomial auxiliary equation for its roots $m_1, m_2, \dots, m_n$. Standard calculators can be used to solve quadratic or cubic equations directly unless the roots are complex.

• Classification of Auxiliary Equation Roots and CF Rules:

  • Case 1: Real and Distinct Roots:

  • If the roots $m_1, m_2$ are real numbers and $m_1 \neq m_2$:       $\text{CF} = A e^{m_1 x} + B e^{m_2 x}$

  • The order of assignment of constants to roots is arbitrary; $A e^{m_1 x} + B e^{m_2 x}$ is identical to $A e^{m_2 x} + B e^{m_1 x}$.

  • Case 2: Real and Equal (Repeated) Roots:

  • If two roots are real and equal ($m_1 = m_2 = m$):       $\text{CF} = (A x + B) e^{m x}$

  • If three roots are real and equal ($m_1 = m_2 = m_3 = m$):       $\text{CF} = (A x^2 + B x + C) e^{m x}$

  • Case 3: Complex Conjugate Roots:

  • If the roots are complex conjugates of the form $m = \alpha \pm i \beta$, where $\alpha$ is the real part and $\beta$ is the imaginary coefficient:       $\text{CF} = e^{\alpha x} (A \cos(\beta x) + B \sin(\beta x))$

  • If the real part $\alpha = 0$ (purely imaginary roots $m = \pm i \beta$), then $e^{0 x} = 1$, simplifying the expression to:       $\text{CF} = A \cos(\beta x) + B \sin(\beta x)$

  • For quadratic equations where complex roots cannot be determined by factoring by inspection, use the quadratic formula:       $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$       for an equation written in standard quadratic form $a m^2 + b m + c = 0$.

• Worked Examples for Complementary Functions:

  • Example 1: Second-Order Equation with Distinct Real Roots:

  • Differential Equation: $(D^2 - 5D + 6)y = 0$

  • Auxiliary Equation: $m^2 - 5m + 6 = 0$

  • Factoring: $(m - 3)(m - 2) = 0 \implies m_1 = 2, m_2 = 3$

  • Complementary Function: $\text{CF} = A e^{2x} + B e^{3x}$ (or $\text{CF} = A e^{3x} + B e^{2x}$)

  • Example 2: Second-Order Equation with Repeated Real Roots:

  • Differential Equation: $(D^2 + 2D + 1)y = 0$

  • Auxiliary Equation: $m^2 + 2m + 1 = 0$

  • Factoring: $(m + 1)^2 = 0 \implies m_1 = -1, m_2 = -1$

  • Complementary Function: $\text{CF} = (A x + B) e^{-x}$

  • Example 3: Third-Order Equation with Three Repeated Real Roots:

  • Differential Equation: $(D^3 + 3D^2 + 3D + 1)y = 0$

  • Auxiliary Equation: $m^3 + 3m^2 + 3m + 1 = 0$

  • Factoring: $(m + 1)^3 = 0 \implies (m + 1)(m + 1)(m + 1) = 0 \implies m = -1, -1, -1$

  • Caveat: Do not drop repeated factors or state $m = -1$ as a single root; all three repeated roots must be accounted for in the degree of the polynomial pre-factor.

  • Complementary Function: $\text{CF} = (A x^2 + B x + C) e^{-x}$

  • Example 4: Second-Order Equation with Complex Roots:

  • Differential Equation: $(D^2 + 2D + 5)y = 0$

  • Auxiliary Equation: $m^2 + 2m + 5 = 0$

  • Applying Quadratic Formula ($a = 1, b = 2, c = 5$):       $m = \frac{-2 \pm \sqrt{2^2 - 4(1)(5)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$

  • Identification: Real part $\alpha = -1$, imaginary coefficient $\beta = 2$

  • Complementary Function: $\text{CF} = e^{-x} (A \cos(2x) + B \sin(2x))$

  • Example 5: Third-Order Equation with Mixed Real and Complex Roots:

  • Differential Equation: $(D^3 - 1)y = 0$

  • Auxiliary Equation: $m^3 - 1 = 0$

  • Warning: Do not simplify improperly as $m^3 = 1 \implies m = 1, 1, 1$.

  • Factoring Identity: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

  • Factored Form: $(m - 1)(m^2 + m + 1) = 0$

  • Root 1: $m - 1 = 0 \implies m_1 = 1$

  • Roots 2 & 3: For $m^2 + m + 1 = 0$, apply quadratic formula ($a = 1, b = 1, c = 1$):       $m = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm \sqrt{-3}}{2} = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$

  • Complementary Function: Combining real root $m = 1$ and complex conjugate pair $\alpha = -\frac{1}{2}, \beta = \frac{\sqrt{3}}{2}$:       $\text{CF} = e^{-\frac{1}{2} x} \left( A \cos\left(\frac{\sqrt{3}}{2} x\right) + B \sin\left(\frac{\sqrt{3}}{2} x\right) \right) + C e^x$

  • Example 6: Fourth-Order Equation with Mixed Real and Imaginary Roots:

  • Differential Equation: $(D^4 - 1)y = 0$

  • Auxiliary Equation: $m^4 - 1 = 0$

  • Warning: Do not simplify as $m^4 = 1 \implies m = 1, 1, 1, 1$.

  • Factoring Strategy: Use $a^2 - b^2 = (a - b)(a + b)$, writing $(m^2)^2 - 1^2 = 0$:       $(m^2 - 1)(m^2 + 1) = 0$

  • Real Roots: $m^2 - 1 = 0 \implies m = \pm 1$

  • Imaginary Roots: $m^2 + 1 = 0 \implies m^2 = -1 \implies m = \pm i = 0 \pm 1i$

  • Roots Summary: Four distinct roots $m = 1, -1, i, -i$

  • Complementary Function:       $\text{CF} = A e^x + B e^{-x} + C \cos(x) + D \sin(x)$       (written equivalently as $\text{CF} = A \cos(x) + B \sin(x) + C e^x + D e^{-x}$)

Particular Integrals for Exponential Functions

• Methodology for Type 1 ($q(x) = e^{ax}$):

  • Standard Particular Integral Form:     $\text{PI} = \frac{1}{f(D)} e^{ax}$

  • Basic Substitution Rule: Replace the differential operator $D$ with the constant $a$ ($D = a$).

  • Case 1: Non-Zero Denominator ($f(a) \neq 0$):

  • If substituting $D = a$ results in a non-zero denominator:       $\text{PI} = \frac{1}{f(a)} e^{ax}$

  • Case 2: Zero Denominator / Rule Failure ($f(a) = 0$):

  • Substituting $D = a$ yielding $f(a) = 0$ causes division by zero, indicating rule failure.

  • Action: Pre-multiply the expression by $x$ and differentiate the denominator polynomial $f(D)$ with respect to $D$:       $\text{PI} = x \frac{1}{f'(D)} e^{ax}$

  • Evaluate $f'(a)$. If $f'(a) \neq 0$, the Particular Integral is:       $\text{PI} = x \frac{1}{f'(a)} e^{ax}$

  • Iterative Failure Rule: If $f'(a) = 0$ remains zero, repeat the procedure by multiplying by $x$ again (yielding $x^2$) and differentiating the denominator again ($f''(D)$):       $\text{PI} = x^2 \frac{1}{f''(D)} e^{ax}$

  • Continue iterating until a non-zero value is obtained in the denominator upon evaluating at D = a$.\n\n* **Worked Examples for Type 1 Exponential Functions**:\n * **Problem 1: Non-Zero Denominator Case**:\n * Differential Equation: (D^2 + 3D + 2)y = e^x\n * Complementary Function: Homogeneous equation (D^2 + 3D + 2)y = 0 \implies m^2 + 3m + 2 = 0 \implies (m + 1)(m + 2) = 0 \implies m = -1, -2.\n      \text{CF} = A e^{-x} + B e^{-2x}\n * Particular Integral:\n      \text{PI} = \frac{1}{D^2 + 3D + 2} e^x\n      Here a = 1$. Substitute$D = 1 into denominator:\n      \text{Denominator} = 1^2 + 3(1) + 2 = 6 eq 0\n      \text{PI} = \frac{1}{6} e^x\n * General Solution:\n      y = \text{CF} + \text{PI} = A e^{-x} + B e^{-2x} + \frac{1}{6} e^x\n * **Problem 2: Single Rule Failure Case**:\n * Differential Equation: (D^2 - 3D + 2)y = e^x\n * Complementary Function: Homogeneous equation (D^2 - 3D + 2)y = 0 \implies m^2 - 3m + 2 = 0 \implies (m - 1)(m - 2) = 0 \implies m = 1, 2.\n      \text{CF} = A e^x + B e^{2x}\n * Particular Integral:\n      \text{PI} = \frac{1}{D^2 - 3D + 2} e^x\n      Here a = 1$. Substitute$D = 1 into denominator:\n      \text{Denominator} = 1^2 - 3(1) + 2 = 0\n      Rule fails due to zero denominator.\n * Apply Failure Rule: Multiply by x$and differentiate denominator w.r.t$D$($\frac{d}{dD}(D^2 - 3D + 2) = 2D - 3):\n      \text{PI} = x \frac{1}{2D - 3} e^x\n      Substitute D = 1$into$2D - 3:\n      \text{Denominator} = 2(1) - 3 = -1 eq 0\n      \text{PI} = x \frac{1}{-1} e^x = -x e^x\n * General Solution:\n      y = \text{CF} + \text{PI} = A e^x + B e^{2x} - x e^x\n * **Problem 3: Double Rule Failure Case**:\n * Differential Equation: (D^2 + 2D + 1)y = e^{-x}\n * Complementary Function: Auxiliary equation m^2 + 2m + 1 = 0 \implies (m + 1)^2 = 0 \implies m = -1, -1.\n      \text{CF} = (A x + B) e^{-x}\n * Particular Integral:\n      \text{PI} = \frac{1}{D^2 + 2D + 1} e^{-x}\n      Here a = -1$. Substitute$D = -1 into denominator:\n      \text{Denominator} = (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0\n      Rule fails.\n * First Failure Step: Multiply by x$and differentiate denominator w.r.t$D$($\frac{d}{dD}(D^2 + 2D + 1) = 2D + 2):\n      \text{PI} = x \frac{1}{2D + 2} e^{-x}\n      Substitute D = -1$into$2D + 2:\n      \text{Denominator} = 2(-1) + 2 = 0\n      Rule fails again.\n * Second Failure Step: Multiply by x$again ($x \times x = x^2$) and differentiate denominator w.r.t$D$($\frac{d}{dD}(2D + 2) = 2):\n      \text{PI} = x^2 \frac{1}{2} e^{-x} = \frac{x^2}{2} e^{-x}\n * General Solution:\n      y = \text{CF} + \text{PI} = (A x + B) e^{-x} + \frac{x^2}{2} e^{-x}\n\n# Particular Integrals for Trigonometric Functions\n\n* **Methodology for Type 2 (q(x) = \sin(ax)$or$\cos(ax)**):\n * Standard Particular Integral Form:\n    \text{PI} = \frac{1}{f(D^2)} \sin(ax) \quad \text{or} \quad \text{PI} = \frac{1}{f(D^2)} \cos(ax)\n * Substitution Rule: Replace D^2$with$-a^2.\n * Critical Warning on Order of Operations: The expression -a^2$means$-(a^2)$, NOT$(-a)^2$. For example, if$a = 2$,$D^2$is replaced by$-2^2 = -4$, not$(-2)^2 = +4$.

  • Handling Odd Powers of $D$: For odd powers of $D$ like $D^3$, rewrite the term as $D \cdot D^2$ to allow substitution of $D^2 = -a^2$.

  • Case 1: Non-Zero Denominator ($f(-a^2) \neq 0$):

  • If evaluating $f(-a^2)$ yields a non-zero denominator:       \text{PI} = \frac{1}{f(-a^2)} \sin(ax) \quad \text{or} \t\text{PI} = \frac{1}{f(-a^2)} \cos(ax)

  • Case 2: Zero Denominator ($f(-a^2) = 0$):

  • If substitution yields zero, pre-multiply by $x$ and differentiate the denominator with respect to $D$:       $\text{PI} = x \frac{1}{f'(D)} \sin(ax)$

  • If the derivative leaves an isolated $D$ in the denominator, multiply numerator and denominator by $D$ to re-introduce $D^2$ in the denominator, substitute $D^2 = -a^2$, and differentiate the numerator function.

  • Conjugate Method for Linear Operator Terms in Denominator:

  • When substituting $D^2 = -a^2$ leaves a linear expression in $D$ (e.g., $\alpha D + \beta$), multiply numerator and denominator by its algebraic conjugate $\alpha D - \beta$:       $\frac{1}{\alpha D + \beta} = \frac{\alpha D - \beta}{(\alpha D + \beta)(\alpha D - \beta)} = \frac{\alpha D - \beta}{\alpha^2 D^2 - \beta^2}$

  • Practical Recommendations for Sign Error Prevention:

    • If the denominator contains negative terms like $-2D - 3$, factor out $-1$ to write $-(2D + 3)$, then multiply numerator and denominator by conjugate $(2D - 3)$.

    • If the denominator is $-4D + 5$, factor out $-1$ to write $-(4D - 5)$, then multiply numerator and denominator by conjugate $(4D + 5)$.

• Worked Examples for Type 2 Trigonometric Functions:

  • Problem 1: Standard Non-Zero Denominator Case:

  • Differential Equation: $(D^2 + 4)y = \cos(x)$

  • Complementary Function: Auxiliary equation $m^2 + 4 = 0 \implies m = \pm 2i = 0 \pm 2i$.       $\text{CF} = A \cos(2x) + B \sin(2x)$

  • Particular Integral:       $\text{PI} = \frac{1}{D^2 + 4} \cos(x)$       Here $a = 1$. Replace $D^2$ with $-1^2 = -1$:       $\text{Denominator} = -1 + 4 = 3 \neq 0$       $\text{PI} = \frac{1}{3} \cos(x)$

  • General Solution:       $y = \text{CF} + \text{PI} = A \cos(2x) + B \sin(2x) + \frac{1}{3} \cos(x)$

  • Problem 2: Zero Denominator Case with Operational Operator Multiplication:

  • Differential Equation: $(D^2 + 4)y = \cos(2x)$

  • Complementary Function: Auxiliary equation $m^2 + 4 = 0 \implies m = \pm 2i$.       $\text{CF} = A \cos(2x) + B \sin(2x)$

  • Particular Integral:       $\text{PI} = \frac{1}{D^2 + 4} \cos(2x)$       Here $a = 2$. Replace $D^2$ with $-2^2 = -4$:       $\text{Denominator} = -4 + 4 = 0$       Rule fails.

  • Multiply by $x$ and differentiate denominator w.r.t $D$ ($\frac{d}{dD}(D^2 + 4) = 2D$):       $\text{PI} = x \frac{1}{2D} \cos(2x)$

  • Multiply numerator and denominator by $D$ to form $D^2$ in denominator:       $\text{PI} = x \frac{D}{2D^2} \cos(2x)$       Substitute $D^2 = -4$ into denominator:       $\text{Denominator} = 2(-4) = -8$       $\text{PI} = x \frac{1}{-8} D(\cos(2x))$       Evaluate $D(\cos(2x)) = \frac{d}{dx}(\cos(2x)) = -2 \sin(2x)$, giving:       $\text{PI} = -\frac{x}{8} (-2 \sin(2x)) = \frac{x \sin(2x)}{4}$

  • General Solution:       $y = \text{CF} + \text{PI} = A \cos(2x) + B \sin(2x) + \frac{x \sin(2x)}{4}$

  • Problem 3: Conjugate Method for First-Degree Operator Denominator:

  • Differential Equation: $(D^2 + 5D + 6)y = \sin(3x)$

  • Complementary Function: Auxiliary equation $m^2 + 5m + 6 = 0 \implies (m + 2)(m + 3) = 0 \implies m = -2, -3$.       $\text{CF} = A e^{-2x} + B e^{-3x}$

  • Particular Integral:       $\text{PI} = \frac{1}{D^2 + 5D + 6} \sin(3x)$       Here $a = 3$. Substitute $D^2 = -3^2 = -9$ into denominator:       $\text{Denominator} = -9 + 5D + 6 = 5D - 3$

  • Multiply numerator and denominator by conjugate $(5D + 3)$:       $\text{PI} = \frac{5D + 3}{(5D - 3)(5D + 3)} \sin(3x) = \frac{5D + 3}{25 D^2 - 9} \sin(3x)$       Substitute $D^2 = -9$ into denominator:       $\text{Denominator} = 25(-9) - 9 = -225 - 9 = -234 \neq 0$

  • Apply operator numerator to $\sin(3x)$:       $(5D + 3)\sin(3x) = 5 D(\sin(3x)) + 3 \sin(3x) = 5(3 \cos(3x)) + 3 \sin(3x) = 15 \cos(3x) + 3 \sin(3x)$       $\text{PI} = \frac{1}{-234} (15 \cos(3x) + 3 \sin(3x))$

  • General Solution:       $y = \text{CF} + \text{PI} = A e^{-2x} + B e^{-3x} - \frac{1}{234} (15 \cos(3x) + 3 \sin(3x))$

Particular Integrals for Polynomial Functions

• Methodology for Type 3 ($q(x) = x^m$):

  • Standard Particular Integral Form:     $\text{PI} = \frac{1}{f(D)} x^m$

  • Procedure:

  • Step 1: Factor out the lowest degree term or constant term from $f(D)$ to transform the operator expression into the form $[1 \pm g(D)]$ in the denominator.

  • Step 2: Move the operator expression to the numerator as a negative binomial power, yielding $[1 \pm g(D)]^{-1}$ or $[1 \pm g(D)]^{-2}$.

  • Step 3: Determine derivative truncation limit: Compute derivatives of $x^m$. Since $D^{m+1}(x^m) = 0$, all series terms containing powers of $D$ greater than $m$ evaluate to zero and are truncated.

  • Step 4: Expand using standard Binomial expansion series and apply term-by-term to x^m$.\n\n* **Standard Binomial Expansion Formulas (Valid for \lvert x \rvert < 1)**:\n * First Binomial Identity:\n    (1 - x)^{-1} = 1 + x + x^2 + x^3 + \dots\n * Second Binomial Identity:\n    (1 + x)^{-1} = 1 - x + x^2 - x^3 + \dots\n    (Odd powers are negative, even powers are positive)\n * Third Binomial Identity:\n    (1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \dots\n * Fourth Binomial Identity:\n    (1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \dots\n    (Odd powers are negative, integer coefficients follow sequence 1, 2, 3, 4, \dots)\n\n* **Worked Examples for Type 3 Polynomial Functions**:\n * **Problem 1: Quadratic Polynomial RHS Case**:\n * Differential Equation: (D^2 - 4D + 4)y = x^2\n * Complementary Function: Auxiliary equation m^2 - 4m + 4 = 0 \implies (m - 2)^2 = 0 \implies m = 2, 2$.       $\text{CF} = (A x + B) e^{2x}$

  • Particular Integral:       $\text{PI} = \frac{1}{D^2 - 4D + 4} x^2 = \frac{1}{(D - 2)^2} x^2 = \frac{1}{(2 - D)^2} x^2$       Factor out $2^2 = 4$ from denominator:       $(2 - D)^2 = 2^2 \left(1 - \frac{D}{2}\right)^2 = 4 \left(1 - \frac{D}{2}\right)^2$       Bring operator to numerator:       $\text{PI} = \frac{1}{4} \left(1 - \frac{D}{2}\right)^{-2} x^2$

  • Truncation Analysis for $x^2$:       $D(x^2) = 2x, \quad D^2(x^2) = 2, \quad D^3(x^2) = 0$

  • Expanding Series $(1 - X)^{-2} = 1 + 2X + 3X^2 + \dots$ with $X = \frac{D}{2}$:       $\left(1 - \frac{D}{2}\right)^{-2} = 1 + 2\left(\frac{D}{2}\right) + 3\left(\frac{D}{2}\right)^2 = 1 + D + \frac{3}{4} D^2$

  • Operating on $x^2$:       $\text{PI} = \frac{1}{4} \left[ 1(x^2) + D(x^2) + \frac{3}{4} D^2(x^2) \right] = \frac{1}{4} \left[ x^2 + 2x + \frac{3}{4}(2) \right] = \frac{1}{4} \left[ x^2 + 2x + \frac{3}{2} \right]$       Simplifying inside brackets using common denominator:       $\text{PI} = \frac{1}{4} \left( \frac{2x^2 + 4x + 3}{2} \right) = \frac{1}{8} (2x^2 + 4x + 3)$

  • General Solution:       $y = \text{CF} + \text{PI} = (A x + B) e^{2x} + \frac{1}{8} (2x^2 + 4x + 3)$

  • Problem 2: Linear Polynomial RHS Case:

  • Differential Equation: $(D^2 + 5D + 6)y = x + 1$

  • Complementary Function: Auxiliary equation m^2 + 5m + 6 = 0 \implies (m + 2)(m + 3) = 0 \implies m = -2, -3$.\n      \text{CF} = A e^{-2x} + B e^{-3x}\n * Particular Integral:\n      \text{PI} = \frac{1}{D^2 + 5D + 6} (x + 1) = \frac{1}{6 + 5D + D^2} (x + 1)\n      Factor out constant 6 from denominator:\n      \text{PI} = \frac{1}{6 \left[ 1 + \frac{5D + D^2}{6} \right]} (x + 1) = \frac{1}{6} \left[ 1 + \frac{5D + D^2}{6} \right]^{-1} (x + 1)\n * Truncation Analysis for (x + 1):\n      D(x + 1) = 1, \quad D^2(x + 1) = 0\n * Binomial Expansion (1 + X)^{-1} = 1 - X + X^2 - \dots$with$X = \frac{5D + D^2}{6}:\n      Because D^2(x + 1) = 0$, second-order terms$X^2$containing$D^2$vanish. Retain only$1 - X:\n      \text{PI} = \frac{1}{6} \left[ 1 - \frac{5D + D^2}{6} \right] (x + 1) = \frac{1}{6} \left[ (x + 1) - \frac{5 D(x + 1) + D^2(x + 1)}{6} \right]\n      \text{PI} = \frac{1}{6} \left[ x + 1 - \frac{5(1) + 0}{6} \right] = \frac{1}{6} \left[ x + 1 - \frac{5}{6} \right]\n      Simplify constant 1 - \frac{5}{6} = \frac{1}{6}:\n      \text{PI} = \frac{1}{6} \left( x + \frac{1}{6} \right)\n * General Solution:\n      y = \text{CF} + \text{PI} = A e^{-2x} + B e^{-3x} + \frac{1}{6} \left( x + \frac{1}{6} \right)$$