Variation of Parameters for Systems | Fundamental Matrix Method

Collecting the fundamental solutions as the columns of a fundamental matrix $𝐘 (t)$ lets the homogeneous general solution be written compactly as $𝐲 = 𝐘 (t) 𝐜$ . Replacing the constant vector $𝐜$ by a vector of functions $𝐮 (t)$ produces the single elegant formula \mathbf{u}' = \mathbf{Y}^{-1}\mathbf{g}, which solves every nonhomogeneous system whose homogeneous solutions are known.

Quick Reference

Item	Statement
Fundamental matrix	$𝐘 (t)$ whose columns are $𝐲^{(1)}, \dots, 𝐲^{(n)}$
Homogeneous general solution	$𝐲 (t) = 𝐘 (t) 𝐜$
Key property	\mathbf{Y}'(t) = \mathbf{A}(t)\mathbf{Y}(t)
Invertibility	$det 𝐘 (t)$ is the Wronskian, never zero
Ansatz	$𝐲_{p} = 𝐘 (t) 𝐮 (t)$
Resulting equation	\mathbf{u}'(t) = \mathbf{Y}^{-1}(t)\mathbf{g}(t)
Particular solution	$𝐲_{p} = 𝐘 (t) \int 𝐘^{- 1} (t) 𝐠 (t) d t$
$2 \times 2$ inverse	${[\begin{matrix} a & b \\ c & d \end{matrix}]}^{- 1} = \frac{1}{a d - b c} [\begin{matrix} d & - b \\ - c & a \end{matrix}]$
Constants	Omit the constant of integration; only one particular solution is required

The Fundamental Matrix

Let $𝐲^{(1)} (t), 𝐲^{(2)} (t), \dots, 𝐲^{(n)} (t)$ form a fundamental set of solutions of the homogeneous system \mathbf{y}' = \mathbf{A}(t)\mathbf{y}. Then the general solution

𝐲 (t) = c_{1} 𝐲^{(1)} (t) + c_{2} 𝐲^{(2)} (t) + \dots + c_{n} 𝐲^{(n)} (t)

can also be written as

\mathbf{y}(t) = \mathbf{Y}(t)\mathbf{c} , \tag{i}

where $𝐜$ is the $n \times 1$ column vector of constants

𝐜 = [\begin{matrix} c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{matrix}]

and $𝐘 (t)$ is the $n \times n$ matrix, called a fundamental matrix, whose columns are the fundamental solutions:

𝐘 (t) = [\begin{matrix} | & | & \dots & | \\ 𝐲^{(1)} (t) & 𝐲^{(2)} (t) & \dots & 𝐲^{(n)} (t) \\ | & | & \dots & | \end{matrix}] .

Writing $𝐘 𝐜$ instead of a sum is not merely cosmetic. Matrix-vector multiplication is the operation "multiply each column by the corresponding constant and add", so the two expressions are literally the same computation. The gain is that $𝐘$ can now be inverted, differentiated, and manipulated as a single object.

Two properties will be needed:

\mathbf{Y}'(t) = \mathbf{A}(t)\mathbf{Y}(t), because each column satisfies the homogeneous system.
$det (𝐘 (t))$ is the Wronskian of the fundamental solutions, and since they are linearly independent it is never zero. Hence $𝐘^{- 1} (t)$ always exists.

The Method

To find a particular solution of

\mathbf{y}' = \mathbf{A}(t)\mathbf{y} + \mathbf{g}(t) ,

we replace the constant vector $𝐜$ in (i) by a column vector of functions $𝐮 (t)$ , exactly as in the scalar version of variation of parameters:

The calculation below shows that

\mathbf{u}'(t) = \mathbf{Y}^{-1}(t)\mathbf{g}(t) \tag{iii}

and thus

\mathbf{u}(t) = \int\mathbf{Y}^{-1}(t)\mathbf{g}(t)\,dt . \tag{iv}

Derivation

Starting from $𝐲_{p} = 𝐘 (t) 𝐮 (t)$ and differentiating with the product rule,

\mathbf{y}_p' = \mathbf{Y}'(t)\mathbf{u}(t) + \mathbf{Y}(t)\mathbf{u}'(t) .

Note that $𝐘 (t)$ and \mathbf{Y}'(t) are $n \times n$ matrices while $𝐮 (t)$ and \mathbf{u}'(t) are $n \times 1$ , so the order of the products matters. The expression \mathbf{u}'(t)\mathbf{Y}(t) is not even defined.

Substituting into \mathbf{y}' = \mathbf{A}(t)\mathbf{y} + \mathbf{g}(t),

\underbrace{\mathbf{Y}'(t)\mathbf{u}(t) + \mathbf{Y}(t)\mathbf{u}'(t)}_{\mathbf{y}_p'} = \mathbf{A}(t)\underbrace{\mathbf{Y}(t)\mathbf{u}(t)}_{\mathbf{y}_p} + \mathbf{g}(t) .

Since $𝐘 (t)$ is a fundamental matrix of \mathbf{y}' = \mathbf{A}(t)\mathbf{y}, we have \mathbf{Y}'(t) = \mathbf{A}(t)\mathbf{Y}(t). Therefore the term $𝐀 (t) 𝐘 (t) 𝐮 (t)$ on the right is identical to \mathbf{Y}'(t)\mathbf{u}(t) on the left, and the two cancel:

\mathbf{Y}(t)\mathbf{u}'(t) = \mathbf{g}(t) .

Multiplying both sides on the left by $𝐘^{- 1} (t)$ ,

\underbrace{\mathbf{Y}^{-1}(t)\mathbf{Y}(t)}_{\mathbf{I}}\mathbf{u}'(t) = \mathbf{Y}^{-1}(t)\mathbf{g}(t)

𝐮 (t) = \int 𝐘^{- 1} (t) 𝐠 (t) d t .

Two remarks. First, $det (𝐘 (t))$ is the Wronskian of linearly independent solutions, hence nonzero, so the inverse always exists. Second, since we are looking for only one particular solution, the constant of integration is unnecessary: adding one contributes $𝐘 (t) 𝐤$ , a homogeneous solution absorbed into $𝐲_{h}$ .

Compare this with the scalar case of Section 4.4. There, Cramer's rule on a $2 \times 2$ system produced two separate formulas for u_1' and u_2'. Here the same content is packaged as a single matrix equation, and the two formulas reappear when $𝐘^{- 1}$ is written out. The matrix version is both shorter and valid at any size.

Worked Example

Find a particular solution for

\frac{d 𝐲}{d t} = [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}] 𝐲 + [\begin{matrix} t \\ e^{t} \end{matrix}] .

Solution

Here

𝐀 = [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}] and 𝐠 (t) = [\begin{matrix} t \\ e^{t} \end{matrix}] .

Step 1: the fundamental matrix. The eigenvalues of $𝐀$ are $λ_{1} = 1$ and $λ_{2} = 0$ , with associated eigenvectors

𝐯^{(1)} = [\begin{matrix} 1 \\ 2 \end{matrix}], 𝐯^{(2)} = [\begin{matrix} 1 \\ 3 \end{matrix}] .

Therefore two linearly independent solutions of the homogeneous system are

𝐲^{(1)} = [\begin{matrix} 1 \\ 2 \end{matrix}] e^{t}, 𝐲^{(2)} = [\begin{matrix} 1 \\ 3 \end{matrix}] \underset{1}{\underset{⏟}{e^{0 \cdot t}}},

and the general solution of the homogeneous system is

𝐲_{h} = \underset{𝐘 (t)}{\underset{⏟}{[\begin{matrix} e^{t} & 1 \\ 2 e^{t} & 3 \end{matrix}]}} \underset{𝐜}{\underset{⏟}{[\begin{matrix} c_{1} \\ c_{2} \end{matrix}]}} .

Step 2: invert $𝐘$ .

det (𝐘 (t)) = 3 e^{t} - 2 e^{t} = e^{t}

so, using the $2 \times 2$ inverse formula,

\begin{aligned} \mathbf{Y}^{-1}(t) &= \frac{1}{e^{t}}\begin{bmatrix} 3 & -1 \\ -2e^{t} & e^{t} \end{bmatrix} \\ &= \begin{bmatrix} 3e^{-t} & -e^{-t} \\ -2 & 1 \end{bmatrix} \end{aligned}

Step 3: compute \mathbf{u}'.

\begin{aligned} \mathbf{u}' &= \mathbf{Y}^{-1}(t)\mathbf{g}(t) \\ &= \begin{bmatrix} 3e^{-t} & -e^{-t} \\ -2 & 1 \end{bmatrix}\begin{bmatrix} t \\ e^{t} \end{bmatrix} \\ &= \begin{bmatrix} 3te^{-t} - 1 \\ -2t + e^{t} \end{bmatrix} \end{aligned}

where the first entry used $- e^{- t} \cdot e^{t} = - 1$ .

Step 4: integrate.

𝐮 = [\begin{matrix} \int (3 t e^{- t} - 1) d t \\ \int (- 2 t + e^{t}) d t \end{matrix}] = [\begin{matrix} - 3 e^{- t} (t + 1) - t \\ - t^{2} + e^{t} \end{matrix}]

For the first integral, integrate $t e^{- t}$ by parts with $u = t$ and $d v = e^{- t} d t$ , so $v = - e^{- t}$ :

\int t e^{- t} d t = - t e^{- t} - \int (- e^{- t}) d t = - t e^{- t} - e^{- t} = - e^{- t} (t + 1) .

No constant of integration is added, since it is unnecessary.

Step 5: assemble.

\begin{aligned} \mathbf{y}_p &= \mathbf{Y}\mathbf{u} \\ &= \begin{bmatrix} e^{t} & 1 \\ 2e^{t} & 3 \end{bmatrix}\begin{bmatrix} -3e^{-t}(t+1) - t \\ -t^{2} + e^{t} \end{bmatrix} \\ &= \begin{bmatrix} -3t - 3 - te^{t} - t^{2} + e^{t} \\ -6t - 6 - 2te^{t} - 3t^{2} + 3e^{t} \end{bmatrix} . \end{aligned}

For the first component, $e^{t} [- 3 e^{- t} (t + 1) - t] = - 3 (t + 1) - t e^{t}$ , and adding the second entry of $𝐮$ gives $- 3 t - 3 - t e^{t} - t^{2} + e^{t}$ .

The general solution is $𝐲 = 𝐘 𝐜 + 𝐲_{p}$ .

Comparison. Section 5.3 solved the same system by undetermined coefficients and obtained

𝐲_{p}^{UC} = [\begin{matrix} - 3 t - t^{2} - t e^{t} \\ 3 - 6 t - 3 t^{2} + (1 - 2 t) e^{t} \end{matrix}] .

The difference between the two particular solutions is

[\begin{matrix} - 3 + e^{t} \\ - 9 + 2 e^{t} \end{matrix}] = - 3 [\begin{matrix} 1 \\ 3 \end{matrix}] + e^{t} [\begin{matrix} 1 \\ 2 \end{matrix}],

a combination of the two homogeneous solutions, exactly as the structure theorem predicts. Different methods produce different particular solutions and the same general solution.

Exercises

Use variation of parameters to find a particular solution of

\mathbf{y}' = \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix}\mathbf{y} + \begin{bmatrix} 1 \\ t \end{bmatrix} .

Solution

The matrix is diagonal, so $𝐘 (t) = [\begin{matrix} e^{2 t} & 0 \\ 0 & e^{- t} \end{matrix}]$ and $𝐘^{- 1} (t) = [\begin{matrix} e^{- 2 t} & 0 \\ 0 & e^{t} \end{matrix}]$ .

\mathbf{u}' = \begin{bmatrix} e^{-2t} \\ te^{t} \end{bmatrix} \quad \Longrightarrow \quad \mathbf{u} = \begin{bmatrix} -\tfrac{1}{2}e^{-2t} \\ (t-1)e^{t} \end{bmatrix}

using integration by parts for the second entry. Therefore

𝐲_{p} = [\begin{matrix} e^{2 t} & 0 \\ 0 & e^{- t} \end{matrix}] [\begin{matrix} - \frac{1}{2} e^{- 2 t} \\ (t - 1) e^{t} \end{matrix}] = [\begin{matrix} - \frac{1}{2} \\ t - 1 \end{matrix}] .

Check: y_1' = 0 and $2 y_{1} + 1 = - 1 + 1 = 0$ . ✓ And y_2' = 1 while $- y_{2} + t = - (t - 1) + t = 1$ . ✓

Find a particular solution of \mathbf{y}' = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\mathbf{y} + \begin{bmatrix} 0 \\ e^{t} \end{bmatrix}, given the fundamental matrix $𝐘 (t) = [\begin{matrix} e^{t} & t e^{t} \\ 0 & e^{t} \end{matrix}]$ .

Solution

Here $det 𝐘 = e^{2 t}$ , so

𝐘^{- 1} = \frac{1}{e^{2 t}} [\begin{matrix} e^{t} & - t e^{t} \\ 0 & e^{t} \end{matrix}] = [\begin{matrix} e^{- t} & - t e^{- t} \\ 0 & e^{- t} \end{matrix}] .

Then

\mathbf{u}' = \begin{bmatrix} e^{-t} & -te^{-t} \\ 0 & e^{-t} \end{bmatrix}\begin{bmatrix} 0 \\ e^{t} \end{bmatrix} = \begin{bmatrix} -t \\ 1 \end{bmatrix} \quad \Longrightarrow \quad \mathbf{u} = \begin{bmatrix} -t^{2}/2 \\ t \end{bmatrix} .

Therefore

𝐲_{p} = [\begin{matrix} e^{t} & t e^{t} \\ 0 & e^{t} \end{matrix}] [\begin{matrix} - t^{2} / 2 \\ t \end{matrix}] = [\begin{matrix} - \frac{t^{2}}{2} e^{t} + t^{2} e^{t} \\ t e^{t} \end{matrix}] = [\begin{matrix} \frac{t^{2}}{2} e^{t} \\ t e^{t} \end{matrix}] .

Note that the coefficient matrix here is defective, which is why $𝐘$ contains the term $t e^{t}$ .

Explain why $𝐘^{- 1}$ must multiply $𝐠$ on the left rather than the right.

Solution

From \mathbf{Y}\mathbf{u}' = \mathbf{g}, isolating \mathbf{u}' requires cancelling $𝐘$ from the left, which means multiplying both sides on the left by $𝐘^{- 1}$ : \mathbf{Y}^{-1}\mathbf{Y}\mathbf{u}' = \mathbf{Y}^{-1}\mathbf{g}.

Multiplying on the right is not even defined. $𝐠$ is $n \times 1$ and $𝐘^{- 1}$ is $n \times n$ , so the product $𝐠 𝐘^{- 1}$ requires the inner dimensions $1$ and $n$ to match, which they do not. Matrix multiplication is not commutative, and dimension checking catches most errors of this kind immediately.

Frequently Asked Questions

What is a fundamental matrix?

The $n \times n$ matrix whose columns are $n$ linearly independent solutions of the homogeneous system. It satisfies \mathbf{Y}' = \mathbf{A}\mathbf{Y}, and multiplying it by a constant vector $𝐜$ reproduces the general homogeneous solution.

Is the fundamental matrix unique?

No. Any set of $n$ independent solutions works, and reordering columns or rescaling them gives another valid fundamental matrix. Different choices produce different intermediate $𝐮$ , but the same general solution. Some texts prefer the normalized choice with $𝐘 (t_{0}) = 𝐈$ , which makes initial conditions trivial to apply.

Why is

𝐘 (t)

always invertible?

Because its determinant is the Wronskian of a set of linearly independent solutions, which by the theory of Section 3.3 is never zero on the interval. This is precisely what makes the method work at every value of $t$ .

Do I need the constant of integration?

No. Adding a constant vector $𝐤$ to $𝐮$ contributes $𝐘 (t) 𝐤$ to $𝐲_{p}$ , which is a homogeneous solution and gets absorbed into $𝐲_{h}$ . Since any particular solution suffices, take the simplest antiderivative componentwise.

Which is better for systems, undetermined coefficients or variation of parameters?

For $2 \times 2$ systems, variation of parameters is often faster, because inverting a $2 \times 2$ matrix takes one line while working out the correct modified trial form can take several. Variation of parameters is also the only option when $𝐀$ varies with $t$ or when $𝐠$ contains functions outside the allowed class.

Why do I get a different particular solution from each method?

Because "a particular solution" is not unique: any two differ by a homogeneous solution. Both answers give the same general solution once $𝐲_{h}$ is added, with the arbitrary constants shifted. The comparison at the end of the worked example shows exactly this.

Nonhomogeneous Systems: Variation of Parameters

Quick Reference

The Fundamental Matrix

The Method

Worked Example

Exercises

Frequently Asked Questions

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