9.2 Method of Undetermined Coefficients and Variation of Parameters

Particular Solution Forms

Polynomial and Exponential Forms

When you're solving a nonhomogeneous linear ODE, the first step is figuring out what form your particular solution $y_p$ should take. The nonhomogeneous term (the right-hand side) tells you what to guess.

• Polynomial right-hand side: guess a polynomial of the same degree, but include all lower-degree terms too.
  • For $x^2 + 3x$, try $y_p = Ax^2 + Bx + C$. You need that constant $C$ even though the original has no constant term, because derivatives can shift degrees down.
• Exponential right-hand side $e^{ax}$: guess $Ae^{ax}$.
  • For $5e^{2x}$, try $y_p = Ae^{2x}$.
• Products of functions: guess the corresponding product form.
  • For $xe^x$, try $y_p = (Ax + B)e^x$. You need both $Ax$ and $B$ because differentiating the product will generate lower-degree polynomial terms.

Trigonometric and Special Cases

• Sine or cosine right-hand side: always include both sine and cosine in your guess, even if only one appears on the right-hand side. Derivatives of sine produce cosine and vice versa.
  • For $3\sin(4x)$, try $y_p = A\sin(4x) + B\cos(4x)$.
• Combination of types: use the superposition principle. Build a guess that covers each piece.
  • For $2x + 3e^x + \sin(x)$, try $y_p = Ax + B + Ce^x + D\sin(x) + E\cos(x)$.
• Overlap with the homogeneous solution (the modification rule): if any term in your guess already solves the homogeneous equation, multiply that term by $x^k$, where $k$ is the multiplicity of the corresponding root in the characteristic equation.
  • If $e^x$ is already a homogeneous solution (simple root), try $y_p = Axe^x$ instead of $Ae^x$.
  • If $e^x$ corresponds to a double root, you'd need $y_p = Ax^2 e^x$.

Undetermined Coefficients Method

Procedure and Implementation

This method works for constant-coefficient linear ODEs where the right-hand side is a combination of polynomials, exponentials, sines, and cosines. Here's the process:

1. Solve the homogeneous equation $y'' + py' + qy = 0$ to get $y_c$.
2. Write a guess for $y_p$ based on the form of the right-hand side (using the rules above).
3. Check for overlap between your guess and $y_c$. If there's overlap, multiply the overlapping terms by $x^k$.
4. Substitute $y_p$ and its derivatives into the original ODE.
5. Collect like terms and set coefficients equal on both sides.
6. Solve the resulting algebraic system for the undetermined constants.

Worked example: Solve $y'' + 4y = x^2$.

1. Homogeneous solution: characteristic equation $r^2 + 4 = 0$ gives $r = \pm 2i$, so $y_c = c_1\cos(2x) + c_2\sin(2x)$.
2. Right-hand side is a degree-2 polynomial, so guess $y_p = Ax^2 + Bx + C$. No overlap with $y_c$.
3. Compute derivatives: $y_p' = 2Ax + B$, $y_p'' = 2A$.
4. Substitute: $2A + 4(Ax^2 + Bx + C) = x^2$, which gives $4Ax^2 + 4Bx + (2A + 4C) = x^2$.
5. Equate coefficients: $4A = 1$, $4B = 0$, $2A + 4C = 0$.
6. Solve: $A = \frac{1}{4}$, $B = 0$, $C = -\frac{1}{8}$.

Particular solution: $y_p = \frac{1}{4}x^2 - \frac{1}{8}$.

Verification and Special Considerations

Always verify by plugging $y_p$ back into the original equation. For the example above: $y_p'' + 4y_p = 2(\frac{1}{4}) + 4(\frac{1}{4}x^2 - \frac{1}{8}) = \frac{1}{2} + x^2 - \frac{1}{2} = x^2$. It checks out.

The overlap rule is the most common source of errors. Consider $y'' - y = e^x$. The characteristic equation $r^2 - 1 = 0$ gives $r = \pm 1$, so $e^x$ is already part of $y_c$. Guessing $y_p = Ae^x$ would just give zero on the left side. Instead, multiply by $x$: try $y_p = Axe^x$, then solve for $A$.

Limitations: This method does not work when:

• The equation has variable coefficients (e.g., $y'' + xy = \ln(x)$)
• The right-hand side involves functions like $\tan(x)$, $\sec(x)$, $\ln(x)$, or $\frac{1}{x}$

For those cases, you need Variation of Parameters.

General Solution of Nonhomogeneous Equations

Combining Solutions

The general solution to a nonhomogeneous equation has two parts:

$y = y_c + y_p$

where $y_c$ is the complementary solution (general solution of the associated homogeneous equation) and $y_p$ is any particular solution of the nonhomogeneous equation.

The complementary solution carries all the arbitrary constants. For a second-order equation, that's two constants ($c_1$ and $c_2$); for an $n$th-order equation, it's $n$ constants.

Continuing the earlier example: $y = c_1\cos(2x) + c_2\sin(2x) + \frac{1}{4}x^2 - \frac{1}{8}$.

Verification and Applications

• Superposition for multiple terms: If the right-hand side has several pieces, like $y'' + y = \sin(x) + e^x$, you can find a particular solution for each piece separately, then add them. This often simplifies the algebra.
• Initial/boundary conditions: Apply these to the general solution (not just $y_c$). Plug in the conditions to determine $c_1$ and $c_2$.
  • For example, given $y(0) = 1$ and $y'(0) = 0$, substitute $x = 0$ into $y$ and $y'$, then solve the resulting system for $c_1$ and $c_2$.

Variation of Parameters Method

Method Overview and Setup

Variation of Parameters is a more general technique that works for any nonhomogeneous term, not just polynomials, exponentials, and trig functions. The tradeoff is that the integrals can get messy.

The idea: take the complementary solution $y_c = c_1 y_1 + c_2 y_2$ and replace the constants with unknown functions:

$y_p = u_1(x)\, y_1(x) + u_2(x)\, y_2(x)$

where $y_1$ and $y_2$ are linearly independent solutions of the homogeneous equation.

For example, to solve $y'' + y = \sec(x)$: the homogeneous solutions are $y_1 = \cos(x)$ and $y_2 = \sin(x)$.

Implementation and Calculations

For a second-order equation written in standard form $y'' + p(x)y' + q(x)y = g(x)$ (note: the coefficient of $y''$ must be 1), follow these steps:

1. Find $y_1$ and $y_2$ from the homogeneous equation.

2. Compute the Wronskian: $W = y_1 y_2' - y_2 y_1'$

3. Find $u_1'$ and $u_2'$ using the formulas: $u_1' = \frac{-y_2\, g(x)}{W}, \qquad u_2' = \frac{y_1\, g(x)}{W}$

4. Integrate to get $u_1$ and $u_2$. (No constants of integration needed here; those are absorbed into $y_c$.)

5. Write the particular solution: $y_p = u_1 y_1 + u_2 y_2$.

For higher-order equations, the pattern extends. A third-order equation uses $y_p = u_1 y_1 + u_2 y_2 + u_3 y_3$, and you solve a $3 \times 3$ system involving the Wronskian determinant.

Advantages and Limitations

Variation of Parameters handles right-hand sides that Undetermined Coefficients cannot touch:

• $y'' + y = \tan(x)$
• $y'' + y = \sec(x)$
• Equations with variable coefficients (as long as you can find the homogeneous solutions)

The main difficulty is that the integrals in step 4 can be hard or impossible to evaluate in closed form. For instance, $y'' + y = e^{x^2}$ leads to integrals with no elementary antiderivative.