2.3 Method of Variation of Parameters

The universal method that works for ANY forcing function

What You'll Learn

By the end of this lesson, you will be able to:

Understand why variation of parameters is more general than undetermined coefficients
Compute the Wronskian determinant for a fundamental set of solutions
Apply the variation of parameters formula to find particular solutions
Solve any second-order linear ODE with any forcing function $g(t)$
Visualize how the method constructs the particular solution

Why Is This Method Powerful?

In the previous section, we learned undetermined coefficients — a quick method for finding particular solutions. But it only works when $g(t)$ has a special form: polynomials, exponentials, sines, cosines, or products of these.

What if $g(t) = \sec(t)$? Or $g(t) = e^t\sin(t)$? Or some arbitrary function? The method of variation of parameters is a completely general approach that works for any $g(t)$, no matter how exotic. It replaces the unknown constants in $y_c$ with unknown functions, solves for them, and yields the particular solution.

“

Joseph-Louis Lagrange developed the method of variation of parameters in the late 1700s. His insight was revolutionary: instead of guessing the form of a particular solution, allow the constants to vary — turn them into functions of the independent variable. This elegance transformed differential equations from specific tricks into a unified framework.

— Joseph-Louis Lagrange (1736–1813), visionary of variational methods

إِنَّ مَعَ الْعُسْرِ يُسْرًا

"Indeed, with hardship comes ease."

— Quran 94:5-6

Just as this powerful method eases the difficulty of solving complex differential equations, remember that mastery of difficult concepts comes through persistence. A method that works for any function is a gift — embrace its power.

Theory of Variation of Parameters

The Setup

Consider the second-order linear ODE in standard form:

a\,y'' + b\,y' + c\,y = g(t)

We have already found the complementary solution (homogeneous solution):

$$y_c(t) = C_1 y_1(t) + C_2 y_2(t)$$

where $y_1$ and $y_2$ are two linearly independent solutions of the homogeneous equation, and $C_1, C_2$ are arbitrary constants.

The idea of variation of parameters is: replace the constants with functions. Let:

y_p(t) = u_1(t)\,y_1(t) + u_2(t)\,y_2(t)

where $u_1(t)$ and $u_2(t)$ are unknown functions to be determined. Our goal is to find these functions such that $y_p$ satisfies the original non-homogeneous equation.

Deriving the Formulas

We start by substituting $y_p = u_1 y_1 + u_2 y_2$ into the ODE. To make progress, we impose one condition (a constraint) to simplify:

u_1'\,y_1 + u_2'\,y_2 = 0

Then:

y_p' = u_1\,y_1' + u_2\,y_2'

y_p'' = u_1'\,y_1' + u_1\,y_1'' + u_2'\,y_2' + u_2\,y_2''

Substituting into $a\,y_p'' + b\,y_p' + c\,y_p = g(t)$:

a(u_1'\,y_1' + u_1\,y_1'' + u_2'\,y_2' + u_2\,y_2'') + b(u_1\,y_1' + u_2\,y_2') + c(u_1\,y_1 + u_2\,y_2) = g(t)

Rearranging:

u_1(a\,y_1'' + b\,y_1' + c\,y_1) + u_2(a\,y_2'' + b\,y_2' + c\,y_2) + a(u_1'\,y_1' + u_2'\,y_2') = g(t)

Since $y_1$ and $y_2$ satisfy the homogeneous equation, the first two terms vanish:

a(u_1'\,y_1' + u_2'\,y_2') = g(t)

Combined with our constraint $u_1'\,y_1 + u_2'\,y_2 = 0$, we have a system of two equations:

u_1'\,y_1 + u_2'\,y_2 = 0

u_1'\,y_1' + u_2'\,y_2' = \frac{g(t)}{a}

Solving by Cramer's rule:

u_1' = -\frac{y_2 \cdot g(t)}{a \cdot W}, \quad u_2' = \frac{y_1 \cdot g(t)}{a \cdot W}

where $W = W(y_1, y_2) = y_1\,y_2' - y_2\,y_1'$ is the Wronskian determinant.

Theorem — Variation of Parameters Formula

Given the ODE $a\,y'' + b\,y' + c\,y = g(t)$ in standard form, with complementary solution $y_c = C_1 y_1 + C_2 y_2$, the particular solution is:

y_p(t) = u_1(t)\,y_1(t) + u_2(t)\,y_2(t)

where:

u_1(t) = -\int \frac{y_2(t) \cdot g(t)}{a \cdot W(t)} \, dt, \quad u_2(t) = \int \frac{y_1(t) \cdot g(t)}{a \cdot W(t)} \, dt

The general solution is:

y(t) = y_c(t) + y_p(t) = C_1\,y_1 + C_2\,y_2 + u_1\,y_1 + u_2\,y_2

The Wronskian Determinant

The Wronskian of two functions $y_1$ and $y_2$ is:

W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1\,y_2' - y_2\,y_1'

Geometric meaning: The Wronskian measures the "independence" of two solutions. If $W \neq 0$ at some point, then $y_1$ and $y_2$ are linearly independent at that point. If $y_1$ and $y_2$ form a fundamental set (which they do for the homogeneous equation), then $W \neq 0$ everywhere in the interval.

Key Insight: Why the Wronskian Matters

The Wronskian $W(y_1, y_2) = y_1\,y_2' - y_2\,y_1'$ appears in the denominator of the variation of parameters formula. If $W = 0$, the formula fails — because $y_1$ and $y_2$ would not be linearly independent. But for a fundamental set of solutions to the homogeneous equation, $W \neq 0$ is guaranteed. This is why the method always works (in principle) once you have the complementary solution.

Important: Standard Form

The coefficient of $y''$ must be exactly 1 for the formula to apply directly. If your equation is $2y'' + 6y' + 4y = 8$, first divide by 2 to get $y'' + 3y' + 2y = 4$. The $a$ in the formula is the coefficient in the standard form. If you forget to convert to standard form, you must remember to include the coefficient $a$ explicitly in the denominators.

Step-by-Step Method

Here is the streamlined procedure to apply variation of parameters:

Standard form. Write the equation as $y'' + py' + qy = g(t)$ (or $a\,y'' + b\,y' + c\,y = g(t)$ if $a \neq 1$). Identify $g(t)$ and the coefficient $a$.
Find the complementary solution. Solve the homogeneous equation $y'' + py' + qy = 0$ to get $y_c = C_1 y_1 + C_2 y_2$. Identify $y_1(t)$ and $y_2(t)$.
Compute the Wronskian. Calculate $W(y_1, y_2) = y_1\,y_2' - y_2\,y_1'$. (Simplify if possible.)
Compute $u_1$ and $u_2$. Use the formulas: $$u_1(t) = -\int \frac{y_2(t) \cdot g(t)}{a \cdot W(t)} \, dt, \quad u_2(t) = \int \frac{y_1(t) \cdot g(t)}{a \cdot W(t)} \, dt$$ Evaluate these integrals. (You may need integration by parts, trigonometric identities, etc.)
Construct the particular solution. Form $y_p = u_1(t)\,y_1(t) + u_2(t)\,y_2(t)$. Simplify if possible.
Write the general solution. The general solution is $y = y_c + y_p = C_1 y_1 + C_2 y_2 + y_p$. Apply initial conditions if given.

When to Use Variation of Parameters

Use this method when:

$g(t)$ is not a polynomial, exponential, sine, cosine, or simple product thereof (so undetermined coefficients won't work easily)
$g(t)$ is a quotient of trig functions like $\sec(t), \csc(t), \tan(t)$, etc.
You want the most general method that works for any $g(t)$

Undetermined coefficients is faster when $g(t)$ has the standard special forms and there's no resonance/overlap with $y_c$.

Worked Examples

Click each example to reveal the full solution. These are drawn from applied differential equations problems.

Step 1: Standard form. Already standard with $a = 1$, $g(t) = \tan(t)$.

Step 2: Complementary solution. For $y'' + y = 0$, the characteristic equation is $r^2 + 1 = 0$, giving $r = \pm i$. Thus:

y_c = C_1 \cos(t) + C_2 \sin(t), \quad y_1 = \cos(t), \quad y_2 = \sin(t)

Step 3: Wronskian.

W = \cos(t) \cdot \cos(t) - \sin(t) \cdot (-\sin(t)) = \cos^2(t) + \sin^2(t) = 1

Step 4: Compute $u_1$ and $u_2$.

u_1 = -\int \sin(t) \cdot \tan(t) \, dt = -\int \frac{\sin^2(t)}{\cos(t)} \, dt

Using $\sin^2(t) = 1 - \cos^2(t)$:

u_1 = -\int \frac{1-\cos^2(t)}{\cos(t)} \, dt = -\int \left(\sec(t) - \cos(t)\right) dt = \sin(t) - \ln|\sec(t) + \tan(t)|

u_2 = \int \cos(t) \cdot \tan(t) \, dt = \int \sin(t) \, dt = -\cos(t)

Step 5: Particular solution.

y_p = u_1 \cos(t) + u_2 \sin(t) = [\sin(t) - \ln|\sec(t)+\tan(t)|]\cos(t) + (-\cos(t))\sin(t)

y_p = \sin(t)\cos(t) - \cos(t)\ln|\sec(t)+\tan(t)| - \cos(t)\sin(t) = -\cos(t)\ln|\sec(t)+\tan(t)|

Step 6: General solution.

\boxed{y(t) = C_1\cos(t) + C_2\sin(t) - \cos(t)\ln|\sec(t)+\tan(t)|}

Note on $\sec(t)$ and $\tan(t)$

The logarithm $\ln|\sec(t) + \tan(t)|$ is the antiderivative of $\sec(t)$. This integral appears frequently in variation of parameters problems. Memorize it: $\int \sec(t)\,dt = \ln|\sec(t)+\tan(t)| + C$.

Step 1: Standard form. Already standard with $a = 1$, $g(t) = 9\sec^2(3t)$.

Step 2: Complementary solution. For $y'' + 9y = 0$, we have $r^2 + 9 = 0$, so $r = \pm 3i$. Thus:

y_c = C_1 \cos(3t) + C_2 \sin(3t), \quad y_1 = \cos(3t), \quad y_2 = \sin(3t)

Step 3: Wronskian.

W = \cos(3t) \cdot 3\cos(3t) - \sin(3t) \cdot (-3\sin(3t)) = 3[\cos^2(3t) + \sin^2(3t)] = 3

Step 4: Compute $u_1$ and $u_2$.

u_1 = -\int \frac{\sin(3t) \cdot 9\sec^2(3t)}{3} \, dt = -3\int \sin(3t)\sec^2(3t) \, dt = -3\int \frac{\sin(3t)}{\cos^2(3t)} \, dt

Substituting $w = \cos(3t)$, $dw = -3\sin(3t)\,dt$:

u_1 = -3 \cdot \left(-\frac{1}{3}\right)\int \frac{dw}{w^2} = \int w^{-2}\,dw = -\frac{1}{w} = -\frac{1}{\cos(3t)} = -\sec(3t)

u_2 = \int \frac{\cos(3t) \cdot 9\sec^2(3t)}{3} \, dt = 3\int \cos(3t)\sec^2(3t) \, dt = 3\int \sec(3t) \, dt

= 3 \cdot \frac{1}{3}\ln|\sec(3t)+\tan(3t)| = \ln|\sec(3t)+\tan(3t)|

Step 5: Particular solution.

y_p = -\sec(3t)\cos(3t) + \ln|\sec(3t)+\tan(3t)| \cdot \sin(3t) = -1 + \sin(3t)\ln|\sec(3t)+\tan(3t)|

Step 6: General solution.

\boxed{y(t) = C_1\cos(3t) + C_2\sin(3t) + \sin(3t)\ln|\sec(3t)+\tan(3t)| - 1}

Step 1: Standard form. Standard with $a = 1$, $g(t) = \sec^2(2t)$.

Step 2: Complementary solution. For $y'' + 4y = 0$, we have $r^2 + 4 = 0$, giving $r = \pm 2i$. Thus:

y_c = C_1\cos(2t) + C_2\sin(2t), \quad y_1 = \cos(2t), \quad y_2 = \sin(2t)

Step 3: Wronskian.

W = \cos(2t) \cdot 2\cos(2t) - \sin(2t) \cdot (-2\sin(2t)) = 2

Step 4 & 5: Compute $u_1, u_2$, and $y_p$.

u_1 = -\int \frac{\sin(2t)\sec^2(2t)}{2} \, dt = -\frac{1}{2}\int \tan(2t)\sec(2t) \, dt

Using $\int \tan(u)\sec(u)\,du = \sec(u) + C$:

u_1 = -\frac{1}{2} \cdot \frac{1}{2}\sec(2t) = -\frac{1}{4}\sec(2t)

u_2 = \int \frac{\cos(2t)\sec^2(2t)}{2} \, dt = \frac{1}{2}\int \sec(2t) \, dt = \frac{1}{2} \cdot \frac{1}{2}\ln|\sec(2t)+\tan(2t)| = \frac{1}{4}\ln|\sec(2t)+\tan(2t)|

y_p = -\frac{1}{4}\sec(2t)\cos(2t) + \frac{1}{4}\ln|\sec(2t)+\tan(2t)|\sin(2t) = -\frac{1}{4} + \frac{1}{4}\sin(2t)\ln|\sec(2t)+\tan(2t)|

Step 6: General solution.

\boxed{y(t) = C_1\cos(2t) + C_2\sin(2t) + \frac{1}{4}\sin(2t)\ln|\sec(2t)+\tan(2t)| - \frac{1}{4}}

Step 1: Standard form. Standard with $a = 1$, $g(t) = e^t \sin(t)$.

Step 2: Complementary solution. For $y'' - y = 0$, we have $r^2 - 1 = 0$, giving $r = \pm 1$. Thus:

y_c = C_1 e^{t} + C_2 e^{-t}, \quad y_1 = e^{t}, \quad y_2 = e^{-t}

Step 3: Wronskian.

W = e^{t} \cdot (-e^{-t}) - e^{-t} \cdot e^{t} = -1 - 1 = -2

Step 4: Compute $u_1$ and $u_2$.

u_1 = -\int \frac{e^{-t} \cdot e^t\sin(t)}{-2} \, dt = \frac{1}{2}\int \sin(t) \, dt = -\frac{1}{2}\cos(t)

u_2 = \int \frac{e^{t} \cdot e^t\sin(t)}{-2} \, dt = -\frac{1}{2}\int e^{2t}\sin(t) \, dt

For the integral $\int e^{2t}\sin(t)\,dt$, use integration by parts twice or the formula $\int e^{at}\sin(bt)\,dt = \frac{e^{at}(a\sin(bt)-b\cos(bt))}{a^2+b^2}$:

\int e^{2t}\sin(t)\,dt = \frac{e^{2t}(2\sin(t)-\cos(t))}{5}

Thus:

u_2 = -\frac{1}{2} \cdot \frac{e^{2t}(2\sin(t)-\cos(t))}{5} = -\frac{e^{2t}(2\sin(t)-\cos(t))}{10}

Step 5: Particular solution.

y_p = -\frac{1}{2}\cos(t) \cdot e^{t} + \left[-\frac{e^{2t}(2\sin(t)-\cos(t))}{10}\right] \cdot e^{-t}

= -\frac{1}{2}e^{t}\cos(t) - \frac{e^{t}(2\sin(t)-\cos(t))}{10} = e^{t}\left[-\frac{1}{2}\cos(t) - \frac{2\sin(t)-\cos(t)}{10}\right]

= e^{t}\left[\frac{-5\cos(t) - 2\sin(t) + \cos(t)}{10}\right] = \frac{e^{t}(-4\cos(t)-2\sin(t))}{10} = -\frac{e^{t}(2\cos(t)+\sin(t))}{5}

Step 6: General solution.

\boxed{y(t) = C_1 e^{t} + C_2 e^{-t} - \frac{e^{t}(2\cos(t)+\sin(t))}{5}}

Step 1: Standard form. Standard with $a = 1$, $g(t) = t^2 e^{-t}$.

Step 2: Complementary solution. For $y'' + 2y' + y = 0$, the characteristic equation is $r^2 + 2r + 1 = (r+1)^2 = 0$, giving $r = -1$ (repeated). Thus:

y_c = (C_1 + C_2 t)e^{-t}, \quad y_1 = e^{-t}, \quad y_2 = te^{-t}

Step 3: Wronskian.

W = e^{-t} \cdot (e^{-t} - te^{-t}) - te^{-t} \cdot (-e^{-t}) = e^{-2t} - te^{-2t} + te^{-2t} = e^{-2t}

Step 4: Compute $u_1$ and $u_2$.

u_1 = -\int \frac{te^{-t} \cdot t^2 e^{-t}}{e^{-2t}} \, dt = -\int t^3 \, dt = -\frac{t^4}{4}

u_2 = \int \frac{e^{-t} \cdot t^2 e^{-t}}{e^{-2t}} \, dt = \int t^2 \, dt = \frac{t^3}{3}

Step 5: Particular solution.

y_p = -\frac{t^4}{4} \cdot e^{-t} + \frac{t^3}{3} \cdot te^{-t} = e^{-t}\left(-\frac{t^4}{4} + \frac{t^4}{3}\right) = e^{-t} \cdot \frac{-3t^4 + 4t^4}{12} = \frac{t^4 e^{-t}}{12}

Step 6: General solution.

\boxed{y(t) = (C_1 + C_2 t)e^{-t} + \frac{t^4 e^{-t}}{12}}

Observation: Undetermined Coefficients Would Also Work

For this problem, since $g(t) = t^2 e^{-t}$ and we already have $e^{-t}$ and $te^{-t}$ in $y_c$, undetermined coefficients would require us to guess $y_p = (At^4 + Bt^3)e^{-t}$ (multiplying by $t^2$ due to resonance). Variation of parameters bypasses all the guesswork!

Step 1: Standard form. Standard with $a = 1$, $g(t) = te^t$.

Step 2: Complementary solution. For $y'' - 2y' + y = 0$, the characteristic equation is $r^2 - 2r + 1 = (r-1)^2 = 0$, giving $r = 1$ (repeated). Thus:

y_c = (C_1 + C_2 t)e^{t}, \quad y_1 = e^{t}, \quad y_2 = te^{t}

Step 3: Wronskian.

W = e^{t} \cdot (e^{t} + te^{t}) - te^{t} \cdot e^{t} = e^{2t} + te^{2t} - te^{2t} = e^{2t}

Step 4: Compute $u_1$ and $u_2$.

u_1 = -\int \frac{te^{t} \cdot te^t}{e^{2t}} \, dt = -\int t^2 \, dt = -\frac{t^3}{3}

u_2 = \int \frac{e^{t} \cdot te^t}{e^{2t}} \, dt = \int t \, dt = \frac{t^2}{2}

Step 5: Particular solution.

y_p = -\frac{t^3}{3} \cdot e^{t} + \frac{t^2}{2} \cdot te^{t} = e^{t}\left(-\frac{t^3}{3} + \frac{t^3}{2}\right) = \frac{t^3 e^{t}}{6}

Step 6: General solution.

\boxed{y(t) = (C_1 + C_2 t)e^{t} + \frac{t^3 e^{t}}{6}}

Step 1: Standard form. Standard with $a = 1$, $g(t) = \cosh(2t) = \frac{e^{2t}+e^{-2t}}{2}$.

Step 2: Complementary solution. For $y'' + 4y' + 4y = 0$, the characteristic equation is $r^2 + 4r + 4 = (r+2)^2 = 0$, giving $r = -2$ (repeated). Thus:

y_c = (C_1 + C_2 t)e^{-2t}, \quad y_1 = e^{-2t}, \quad y_2 = te^{-2t}

Step 3: Wronskian.

W = e^{-2t}(-2te^{-2t} + e^{-2t}) - te^{-2t}(-2e^{-2t}) = e^{-4t}(1 - 2t + 2t) = e^{-4t}

Step 4: Compute $u_1$ and $u_2$.

u_1 = -\int \frac{te^{-2t} \cdot \cosh(2t)}{e^{-4t}} \, dt = -\int t e^{2t}\cosh(2t) \, dt

Since $\cosh(2t) = \frac{e^{2t}+e^{-2t}}{2}$:

u_1 = -\frac{1}{2}\int t(e^{4t} + 1) \, dt = -\frac{1}{2}\left[\int te^{4t}\,dt + \int t\,dt\right]

Using integration by parts on $\int te^{4t}\,dt$:

\int te^{4t}\,dt = \frac{te^{4t}}{4} - \frac{e^{4t}}{16}

Thus:

u_1 = -\frac{1}{2}\left[\frac{te^{4t}}{4} - \frac{e^{4t}}{16} + \frac{t^2}{2}\right]

u_2 = \int \frac{e^{-2t} \cdot \cosh(2t)}{e^{-4t}} \, dt = \int e^{2t}\cosh(2t) \, dt = \frac{1}{2}\int e^{2t}(e^{2t} + e^{-2t}) \, dt

= \frac{1}{2}\left[\int e^{4t}\,dt + \int 1\,dt\right] = \frac{1}{2}\left[\frac{e^{4t}}{4} + t\right]

Step 5 & 6: Particular solution and general solution. After substituting into $y_p = u_1 y_1 + u_2 y_2$ and simplifying (which involves substantial algebra), the result is:

\boxed{y(t) = (C_1 + C_2 t)e^{-2t} + y_p(t)}

Note: The explicit form of $y_p$ is algebraically complex. In practice, one would compute and simplify using computer algebra. The key point is that the method yields a closed form.

Physical Context: This equation models a damped harmonic oscillator (e.g., a mass-spring system or RLC electrical circuit) driven by a non-sinusoidal forcing function $\sec(2t)$. The natural frequency is $\omega_0 = 2$, and the forcing at that frequency is unusual because $\sec(2t) = 1/\cos(2t)$ diverges periodically.

Step 1: Standard form. Standard with $a = 1$, $g(t) = \sec(2t)$.

Step 2: Complementary solution. For $y'' + 4y = 0$, we have $r^2 + 4 = 0$, giving $r = \pm 2i$. Thus:

y_c = C_1\cos(2t) + C_2\sin(2t), \quad y_1 = \cos(2t), \quad y_2 = \sin(2t)

Step 3: Wronskian.

W = \cos(2t) \cdot 2\cos(2t) - \sin(2t) \cdot (-2\sin(2t)) = 2

Step 4: Compute $u_1$ and $u_2$.

u_1 = -\int \frac{\sin(2t) \sec(2t)}{2} \, dt = -\frac{1}{2}\int \tan(2t) \, dt = \frac{1}{4}\ln|\cos(2t)|

u_2 = \int \frac{\cos(2t) \sec(2t)}{2} \, dt = \frac{1}{2}\int 1 \, dt = \frac{t}{2}

Step 5: Particular solution.

y_p = \frac{1}{4}\ln|\cos(2t)| \cdot \cos(2t) + \frac{t}{2} \cdot \sin(2t)

Step 6: General solution.

\boxed{y(t) = C_1\cos(2t) + C_2\sin(2t) + \frac{1}{4}\cos(2t)\ln|\cos(2t)| + \frac{t}{2}\sin(2t)}

Engineering Insight

The term $\dfrac{t}{2}\sin(2t)$ represents secular growth — a term that grows linearly with time. This arises because $\sec(2t)$ contains components at the resonant frequency. In a physical system, this would lead to unbounded oscillations (in the absence of damping). The logarithmic term $\dfrac{1}{4}\cos(2t)\ln|\cos(2t)|$ is periodic but behaves singularly when $\cos(2t) = 0$. This illustrates why $\sec(2t)$ is a problematic forcing function for this oscillator!

📝 Exam-Style Practice Problems

These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.

Practice 1: Variation of parameters for $y'' - 2y' + y = \dfrac{e^t}{t^2}$ ▶

Solution

Step 1: Find $y_c$. Characteristic equation: $(r-1)^2 = 0 \Rightarrow r = 1$ (repeated).

y_c = c_1 e^t + c_2 te^t, \quad y_1 = e^t, \quad y_2 = te^t

Step 2: Compute the Wronskian.

W = \begin{vmatrix} e^t & te^t \\ e^t & e^t + te^t \end{vmatrix} = e^t(e^t + te^t) - te^t \cdot e^t = e^{2t} + te^{2t} - te^{2t} = e^{2t}

Step 3: Compute $u_1'$ and $u_2'$. With $g(t) = \dfrac{e^t}{t^2}$:

u_1' = \frac{-y_2 g}{W} = \frac{-te^t \cdot \frac{e^t}{t^2}}{e^{2t}} = \frac{-\frac{e^{2t}}{t}}{e^{2t}} = -\frac{1}{t}

u_2' = \frac{y_1 g}{W} = \frac{e^t \cdot \frac{e^t}{t^2}}{e^{2t}} = \frac{\frac{e^{2t}}{t^2}}{e^{2t}} = \frac{1}{t^2}

Step 4: Integrate to find $u_1$ and $u_2$.

u_1 = \int -\frac{1}{t}\,dt = -\ln|t|

u_2 = \int \frac{1}{t^2}\,dt = -\frac{1}{t}

Step 5: Particular solution.

y_p = u_1 y_1 + u_2 y_2 = -\ln|t| \cdot e^t + \left(-\frac{1}{t}\right) \cdot te^t = -e^t\ln|t| - e^t

Step 6: General solution. (The $-e^t$ can be absorbed into $c_1 e^t$)

\boxed{y = c_1 e^t + c_2 te^t - e^t\ln|t|}

Practice 2: Variation of parameters for $y'' + 4y = \sec(2t)$ ▶

Solution

Step 1: Find $y_c$. Characteristic equation: $r^2 + 4 = 0 \Rightarrow r = \pm 2i$.

y_c = c_1\cos(2t) + c_2\sin(2t), \quad y_1 = \cos(2t), \quad y_2 = \sin(2t)

Step 2: Compute the Wronskian.

W = \begin{vmatrix} \cos(2t) & \sin(2t) \\ -2\sin(2t) & 2\cos(2t) \end{vmatrix} = 2\cos^2(2t) + 2\sin^2(2t) = 2

Step 3: Compute $u_1'$ and $u_2'$. With $g(t) = \sec(2t)$:

u_1' = \frac{-\sin(2t) \cdot \sec(2t)}{2} = \frac{-\tan(2t)}{2}

u_2' = \frac{\cos(2t) \cdot \sec(2t)}{2} = \frac{1}{2}

Step 4: Integrate.

u_1 = \int \frac{-\tan(2t)}{2}\,dt = \frac{1}{4}\ln|\cos(2t)|

u_2 = \int \frac{1}{2}\,dt = \frac{t}{2}

Step 5: Particular solution.

y_p = \frac{1}{4}\ln|\cos(2t)| \cdot \cos(2t) + \frac{t}{2} \cdot \sin(2t)

Step 6: General solution.

\boxed{y = c_1\cos(2t) + c_2\sin(2t) + \frac{1}{4}\cos(2t)\ln|\cos(2t)| + \frac{t}{2}\sin(2t)}

Interactive Variation of Parameters Calculator

Explore the variation of parameters method by selecting a preset equation or entering your own. Watch the complementary solution, Wronskian, and particular solution come to life!

Preset Examples

Coefficients: ay'' + by' + cy = g(t)

Forcing Function g(t)

Plot: Solutions yc, yp, and y = yc + yp

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Practice Problems

Test your mastery of variation of parameters! Try each problem, then check your answer.

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Problem 1 — Wronskian Computation

For $y'' + 4y = g(t)$ with $y_1 = \cos(2t)$ and $y_2 = \sin(2t)$, what is $W(y_1, y_2)$?

Problem 2 — Standard Form

Which equation is in standard form?

Problem 3 — When to Use Variation of Parameters

Which forcing function would make undetermined coefficients difficult or impossible?

Problem 4 — Complementary Solution

For $y'' + 6y' + 9y = g(t)$, what is the complementary solution $y_c$?

Problem 5 — Understanding Variation of Parameters

The fundamental idea of variation of parameters is to replace the constants $C_1, C_2$ in $y_c = C_1 y_1 + C_2 y_2$ with what?

Quick Reference Card

Variation of Parameters Formula

Given equation: $\quad ay'' + by' + cy = g(t)$

Complementary solution: $\quad y_c = C_1 y_1 + C_2 y_2$

Wronskian: $\quad W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1'$

Particular solution functions:

u_1(t) = -\int \frac{y_2(t) \cdot g(t)}{a \cdot W(t)} \, dt, \quad u_2(t) = \int \frac{y_1(t) \cdot g(t)}{a \cdot W(t)} \, dt

Particular solution: $\quad y_p = u_1 y_1 + u_2 y_2$

General solution: $\quad y = y_c + y_p$

2.3 Method of Variation of Parameters

Why Is This Method Powerful?

Theory of Variation of Parameters

The Setup

Deriving the Formulas

Step-by-Step Method

Worked Examples

📝 Exam-Style Practice Problems

Solution

Solution

Interactive Variation of Parameters Calculator

Results

Plot: Solutions yc, yp, and y = yc + yp

Practice Problems

Quick Reference Card