Variation Of Parameters Particular Solution: Build Forced Solutions

On this page: The Principle | Conditions | Misconceptions | EE Questions | Retrieval Practice | Worked Example | Solve a Problem | Related Principles | FAQ

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The Principle

Statement

Variation of parameters turns a known homogeneous fundamental pair into a particular solution for a forced second-order linear equation. Instead of guessing the shape of $y_p$, it lets the coefficients of $y_1$ and $y_2$ vary with $x$ and solves for the integrals those coefficients must contain.

The particular-solution formula is:

$y_{p}=-y_{1}\int \frac{y_{2}g}{W}\,dx+y_{2}\int \frac{y_{1}g}{W}\,dx$

This principle supplies the particular piece used by Linear Nonhomogeneous Solution Structure. After one $y_p$ is found, the general solution has the form $y=y_h+y_p$ under the stated conditions.

Mathematical Form

$y_{p}=-y_{1}\int \frac{y_{2}g}{W}\,dx+y_{2}\int \frac{y_{1}g}{W}\,dx$

Where:

• $y_1$, $y_2$ = a known fundamental solution pair for the associated homogeneous equation
• $g(x)$ = the forcing term in normalized standard form
• $W=y_1y_2^{\prime}-y_1^{\prime}y_2$ = the Wronskian of the pair
• $y_p$ = one particular solution of the nonhomogeneous equation

Why the Wronskian appears

Variation of parameters replaces the constant weights $C_1$ and $C_2$ from Fundamental Solution Set General Form with functions. The Wronskian is the determinant that makes those two changing weights solvable from the forcing term. If $W=0$, the pair is not a usable fundamental pair for this formula.

With $y_p=u_1y_1+u_2y_2$ and the auxiliary condition $u_1^{\prime}y_1+u_2^{\prime}y_2=0$, the forcing equation gives

$u_1^{\prime}=-\frac{y_2g}{W},\qquad u_2^{\prime}=\frac{y_1g}{W}$

Integrating those two coefficient derivatives gives the signs in the formula.

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Conditions of Applicability

Condition: $normalized y^{\prime\prime}+P(x)y^{\prime}+Q(x)y=g(x)$; pair known; $W\neq 0$

Practical modeling notes

• “Normalized” means the coefficient of $y^{\prime\prime}$ has already been divided to $1$, so the right side is the resulting $g(x)$.
• The pair $y_1,y_2$ must solve the associated homogeneous equation $y^{\prime\prime}+P(x)y^{\prime}+Q(x)y=0$ on the same working interval.
• The Wronskian must be nonzero on the interval where the formula is used. A zero in the denominator is a domain warning, not a minor algebra issue.
• The integrals are indefinite in the formula because changing their constants only adds homogeneous terms, which can be absorbed into $y_h$.

When It Doesn’t Apply

This formula is not the first move in every forced equation:

• Unnormalized equations: if the equation is $a(x)y^{\prime\prime}+b(x)y^{\prime}+c(x)y=r(x)$, normalize first so $g(x)=r(x)/a(x)$.
• Unknown homogeneous pair: variation of parameters starts after a fundamental pair is known.
• Dependent pair: if $W=0$, the pair cannot support the formula.
• Simpler standard forcing: for constant-coefficient equations with standard nonresonant forcing, Undetermined Coefficients Trial Family may be faster.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Misconceptions

Misconception 1: “Variation of parameters finds the homogeneous solutions too”

The truth: the formula assumes $y_1$ and $y_2$ are already known.

Why this matters: if the associated homogeneous equation is still unsolved, variation of parameters has no pair to vary.

Misconception 2: “The right side can be copied before normalizing”

The truth: $g(x)$ is the forcing after the second-order equation has leading coefficient $1$.

Why this matters: using the unnormalized right side changes both integrals and usually gives the wrong particular solution.

Misconception 3: “I must keep arbitrary constants in both integrals”

The truth: constants in the two antiderivatives change $y_p$ by $-C_1y_1+C_2y_2$, which is already part of the homogeneous solution.

Why this matters: set those constants to zero when selecting one $y_p$; keep the arbitrary constants only in $y_h$.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• What jobs do $y_1$, $y_2$, $g(x)$, $W$, and $y_p$ each perform in the formula?
• Why does the formula contain two integrals instead of a single guessed coefficient?

For the Principle

• Before using the formula, what must be checked about the differential equation, the fundamental pair, and the Wronskian?
• Why can constants from the indefinite integrals be ignored when the goal is one particular solution?

Between Principles

• How does this principle supply the $y_p$ piece needed by Linear Nonhomogeneous Solution Structure?

Generate an Example

• Describe one forcing term where variation of parameters would be more natural than trying to guess an undetermined-coefficients trial.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the principle in words: _____For a normalized second-order linear nonhomogeneous equation, a known fundamental pair gives a particular solution through two Wronskian-weighted integrals.

Write the canonical equation: _____$y_{p}=-y_{1}\int \frac{y_{2}g}{W}\,dx+y_{2}\int \frac{y_{1}g}{W}\,dx$

State the canonical condition: _____$normalized y^{\prime\prime}+P(x)y^{\prime}+Q(x)y=g(x);\, \text{pair known};\, W\neq 0$

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Worked Example

Use this worked example to practice Self-Explanation.

Problem

Use variation of parameters to find one particular solution of

$y^{\prime\prime}-y=\frac{1}{x}$

on an interval where $x>0$. For the associated homogeneous equation, use the fundamental pair $y_1=e^x$, $y_2=e^{-x}$. Choose a fixed base point $x_0>0$ for the definite integrals.

Step 1: Verbal Decoding

Target: $y_p$
Given: $g(x)$, $y_1$, $y_2$
Constraints: normalized second-order linear nonhomogeneous equation; known fundamental pair; working interval avoids $x=0$; Wronskian nonzero

Step 2: Visual Decoding

Sketch a two-column setup: one column for $-y_1\int (y_2g/W)\,dx$ and one column for $y_2\int (y_1g/W)\,dx$. Mark $x>0$ so the forcing has no singularity on the interval. (Both columns contribute one piece of $y_p$.)

Step 3: Mathematical Modeling

1. $y_p(x)=-e^x\int_{x_0}^{x} \frac{e^{-t}(1/t)}{W}\,dt+e^{-x}\int_{x_0}^{x} \frac{e^t(1/t)}{W}\,dt$

Step 4: Mathematical Procedures

1. $W=e^x(-e^{-x})-e^xe^{-x}$
2. $W=-2$
3. $y_p(x)=-e^x\int_{x_0}^{x} \frac{e^{-t}}{-2t}\,dt+e^{-x}\int_{x_0}^{x} \frac{e^t}{-2t}\,dt$
4. $y_p(x)=\frac{e^x}{2}\int_{x_0}^{x} \frac{e^{-t}}{t}\,dt-\frac{e^{-x}}{2}\int_{x_0}^{x} \frac{e^t}{t}\,dt$
5. Let $A(x)=\dfrac{e^x}{2}\displaystyle\int_{x_0}^{x}\dfrac{e^{-t}}{t}\,dt$.
6. Let $B(x)=\dfrac{e^{-x}}{2}\displaystyle\int_{x_0}^{x}\dfrac{e^t}{t}\,dt$.
7. $\underline{y_p(x)=A(x)-B(x),\qquad x_0>0}$

Step 5: Reflection

• Domain check: the interval $x>0$ avoids the singular forcing term $1/x$.
• Connection to concept: the answer is allowed to keep non-elementary antiderivatives because the formula constructs $y_p$ by integration.
• Independence check: $W=-2\neq 0$, so the pair is usable in the formula.

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Before moving on: self-explain the model

Try explaining Step 3 out loud or in writing: why $e^x$ and $e^{-x}$ are the pair being varied, why $g(x)=1/x$ after normalization, and why the Wronskian belongs in both denominators.

Mathematical model with explanation

Principle: Variation Of Parameters Particular Solution - $y_{p}=-y_{1}\int \frac{y_{2}g}{W}\,dx+y_{2}\int \frac{y_{1}g}{W}\,dx$.

Conditions: the equation is normalized, $e^x$ and $e^{-x}$ are a known fundamental pair for $y^{\prime\prime}-y=0$, and $W=-2\neq 0$ on the interval.

Relevance: the forcing $1/x$ is not a standard undetermined-coefficients forcing, but variation of parameters can still build one particular solution.

Description: The homogeneous pair supplies the two basis functions. The forcing term determines how their coefficients vary, and the Wronskian keeps the two coefficient equations independent.

Goal: substitute the pair, forcing term, and Wronskian into the formula to construct a valid $y_p$.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem

Use variation of parameters to find one particular solution of

$y^{\prime\prime}+y=\tan x$

on an interval where $\cos x\neq 0$. For the associated homogeneous equation, use the fundamental pair $y_1=\cos x$, $y_2=\sin x$.

Hint (if needed): the Wronskian of $\cos x$ and $\sin x$ is $1$.

Show Solution

Step 1: Verbal Decoding

Target: $y_p$
Given: $g(x)$, $y_1$, $y_2$
Constraints: normalized second-order linear nonhomogeneous equation; known fundamental pair; working interval avoids zeros of cosine; Wronskian nonzero

Step 2: Visual Decoding

Sketch a number line and mark consecutive zeros of $\cos x$, then choose one open interval between them. Set up two integral slots matching the two formula terms. (The forcing $\tan x$ is legal only inside such an interval.)

Step 3: Mathematical Modeling

1. $y_p=-\cos x\int \frac{\sin x\tan x}{W}\,dx+\sin x\int \frac{\cos x\tan x}{W}\,dx$

Step 4: Mathematical Procedures

1. $W=\cos x\cos x-(-\sin x)\sin x$
2. $W=1$
3. $y_p=-\cos x\int \sin x\tan x\,dx+\sin x\int \cos x\tan x\,dx$
4. $\int \cos x\tan x\,dx=\int \sin x\,dx$
5. $\int \cos x\tan x\,dx=-\cos x$
6. $\int \sin x\tan x\,dx=\int \frac{\sin^2 x}{\cos x}\,dx$
7. $\int \sin x\tan x\,dx=\int(\sec x-\cos x)\,dx$
8. $\int \sin x\tan x\,dx=\ln|\sec x+\tan x|-\sin x$
9. $y_p=-\cos x\left(\ln|\sec x+\tan x|-\sin x\right)-\sin x\cos x$
10. $\underline{y_p=-\cos x\ln|\sec x+\tan x|}$

Step 5: Reflection

• Verification: substituting $y_p=-\cos x\ln|\sec x+\tan x|$ gives $y_p^{\prime\prime}+y_p=\tan x$ on the chosen interval.
• Domain check: the formula and forcing require $\cos x\neq 0$.
• Simplification check: the terms $+\cos x\sin x$ and $-\sin x\cos x$ cancel algebraically; any integration constants would add only a homogeneous combination of $\cos x$ and $\sin x$.

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Related Principles

Principle	Relationship to Variation Of Parameters Particular Solution
Fundamental Solution Set General Form	Supplies the known pair $y_1,y_2$ that variation of parameters needs.
Linear Nonhomogeneous Solution Structure	Explains where the constructed particular solution fits in the full solution.
Undetermined Coefficients Trial Family	Gives a faster particular-solution route for standard nonresonant forcing in constant-coefficient equations.

See Differential Equations Subdomain for the higher-order linear sequence, and Principle Structures for organizing formulas, conditions, and problem types.

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FAQ

What is Variation Of Parameters Particular Solution?

Variation Of Parameters Particular Solution is the formula that builds one particular solution of a normalized second-order linear nonhomogeneous equation from a known fundamental pair. It uses $y_1$, $y_2$, the forcing $g(x)$, and the Wronskian $W$ in two integrals.

When does variation of parameters apply?

It applies when the equation is normalized as $y^{\prime\prime}+P(x)y^{\prime}+Q(x)y=g(x)$, the pair is known, and $W\neq 0$. Normalize first, use a pair from the associated homogeneous equation, and work on an interval where the Wronskian does not vanish.

Why do I have to normalize the equation first?

The formula is written for leading coefficient $1$. If the original equation is $a(x)y^{\prime\prime}+b(x)y^{\prime}+c(x)y=r(x)$, then the formula uses $g(x)=r(x)/a(x)$ after dividing through by $a(x)$.

How is variation of parameters different from undetermined coefficients?

Undetermined coefficients guesses a trial family from standard forcing terms in constant-coefficient equations. Variation of parameters uses a known homogeneous fundamental pair and can handle many forcings that do not fit a simple trial family.

Do the integrals need elementary antiderivatives?

Not always. The formula can define a particular solution through integrals even when the antiderivatives are not elementary. For many learning problems, setting up the correct integrals is already the main modeling step.

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Related Guides

• Principle Structures - Organize the formula, conditions, and related solution structures.
• Fundamental Solution Set General Form - Review the known pair required before using this formula.
• Linear Nonhomogeneous Solution Structure - Connect a particular solution to the full forced-equation family.
• Problem Solving - Practice choosing the right structure before calculating.

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How This Fits in Unisium

Within the differential equations subdomain, Unisium treats Variation Of Parameters Particular Solution as the general-purpose route from a known homogeneous pair to one forced-equation solution. The Unisium Study System pairs this guide with elaborative encoding, retrieval practice, and structured problem solving so you learn the stable decision: normalize, verify the pair, compute the Wronskian, then build $y_p$ from the integrals.

Ready to practice differential equations with structure? Check access and join the Unisium waitlist or explore the full framework in Masterful Learning.

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