DIFFYQS Nonhomogeneous systems

Perhaps it is better understood as a definite integral. In this case it will be easy to also solve for the initial conditions. Consider the equation with initial conditions

{\vec{x}}^{'} (t) + P \vec{x} (t) = \vec{f} (t), \vec{x} (0) = \vec{b} .

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The solution can then be written as

\begin{matrix} (3.6) & \vec{x} (t) = e^{- t P} \int_{0}^{t} e^{s P} \vec{f} (s) d s + e^{- t P} \vec{b} . \end{matrix}

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Again, the integration means that each component of the vector

e^{s P} \vec{f} (s)

is integrated separately. It is not hard to see that (3.6) really does satisfy the initial condition

\vec{x} (0) = \vec{b} .

\vec{x} (0) = e^{- 0 P} \int_{0}^{0} e^{s P} \vec{f} (s) d s + e^{- 0 P} \vec{b} = I \vec{b} = \vec{b} .

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Example 3.9.1.

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Suppose that we have the system

\begin{aligned} x_{1}^{'} + 5 x_{1} - 3 x_{2} & = e^{t}, \\ x_{2}^{'} + 3 x_{1} - x_{2} & = 0, \end{aligned}

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with initial conditions

x_{1} (0) = 1, x_{2} (0) = 0 .

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Let us write the system as

{\vec{x}}^{'} + [\begin{matrix} 5 & - 3 \\ 3 & - 1 \end{matrix}] \vec{x} = [\begin{matrix} e^{t} \\ 0 \end{matrix}], \vec{x} (0) = [\begin{matrix} 1 \\ 0 \end{matrix}] .

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The matrix

P = [\begin{matrix} 5 & - 3 \\ 3 & - 1 \end{matrix}]

has a doubled eigenvalue 2 with defect 1, and we leave it as an exercise to double check we computed

e^{t P}

correctly. Once we have

e^{t P},

we find

e^{- t P},

simply by negating

t .

e^{t P} = [\begin{matrix} (1 + 3 t) e^{2 t} & - 3 t e^{2 t} \\ 3 t e^{2 t} & (1 - 3 t) e^{2 t} \end{matrix}], e^{- t P} = [\begin{matrix} (1 - 3 t) e^{- 2 t} & 3 t e^{- 2 t} \\ - 3 t e^{- 2 t} & (1 + 3 t) e^{- 2 t} \end{matrix}] .

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Instead of computing the whole formula at once, let us do it in stages. First

\begin{aligned} \int_{0}^{t} e^{s P} \vec{f} (s) d s & = \int_{0}^{t} [\begin{array}{c} (1 + 3 s) e^{2 s} & - 3 s e^{2 s} \\ 3 s e^{2 s} & (1 - 3 s) e^{2 s} \end{array}] [\begin{array}{c} e^{s} \\ 0 \end{array}] d s \\ = \int_{0}^{t} [\begin{array}{c} (1 + 3 s) e^{3 s} \\ 3 s e^{3 s} \end{array}] d s \\ = [\begin{array}{c} \int_{0}^{t} (1 + 3 s) e^{3 s} d s \\ \int_{0}^{t} 3 s e^{3 s} d s \end{array}] \\ = [\begin{array}{c} t e^{3 t} \\ \frac{(3 t - 1) e^{3 t} + 1}{3} \end{array}] (used integration by parts). \end{aligned}

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Then

\begin{aligned} \vec{x} (t) & = e^{- t P} \int_{0}^{t} e^{s P} \vec{f} (s) d s + e^{- t P} \vec{b} \\ = [\begin{array}{c} (1 - 3 t) e^{- 2 t} & 3 t e^{- 2 t} \\ - 3 t e^{- 2 t} & (1 + 3 t) e^{- 2 t} \end{array}] [\begin{array}{c} t e^{3 t} \\ \frac{(3 t - 1) e^{3 t} + 1}{3} \end{array}] + [\begin{array}{c} (1 - 3 t) e^{- 2 t} & 3 t e^{- 2 t} \\ - 3 t e^{- 2 t} & (1 + 3 t) e^{- 2 t} \end{array}] [\begin{array}{c} 1 \\ 0 \end{array}] \\ = [\begin{array}{c} t e^{- 2 t} \\ - \frac{e^{t}}{3} + (\frac{1}{3} + t) e^{- 2 t} \end{array}] + [\begin{array}{c} (1 - 3 t) e^{- 2 t} \\ - 3 t e^{- 2 t} \end{array}] \\ = [\begin{array}{c} (1 - 2 t) e^{- 2 t} \\ - \frac{e^{t}}{3} + (\frac{1}{3} - 2 t) e^{- 2 t} \end{array}] . \end{aligned}

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Phew!

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Let us check that this really works.

x_{1}^{'} + 5 x_{1} - 3 x_{2} = (4 t e^{- 2 t} - 4 e^{- 2 t}) + 5 (1 - 2 t) e^{- 2 t} + e^{t} - (1 - 6 t) e^{- 2 t} = e^{t} .

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Similarly (exercise)

x_{2}^{'} + 3 x_{1} - x_{2} = 0 .

The initial conditions are also satisfied (exercise).

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For systems, the integrating factor method only works if

P

does not depend on

t,

that is,

P

is constant. The problem is that in general

\frac{d}{d t} [e^{\int P (t) d t}] \neq P (t) e^{\int P (t) d t},

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because matrix multiplication is not commutative.

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Subsubsection 3.9.1.2 Eigenvector decomposition

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For the next method, note that eigenvectors of a matrix give the directions in which the matrix acts like a scalar. If we solve the system along these directions, the computations are simpler as we treat the matrix as a scalar. We then put those solutions together to get the general solution for the system.

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Take the equation

\begin{matrix} (3.7) & {\vec{x}}^{'} (t) = A \vec{x} (t) + \vec{f} (t) . \end{matrix}

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Assume

A

has

n

linearly independent eigenvectors

{\vec{v}}_{1}, {\vec{v}}_{2}, \dots, {\vec{v}}_{n} .

Write

\begin{matrix} (3.8) & \vec{x} (t) = {\vec{v}}_{1} ξ_{1} (t) + {\vec{v}}_{2} ξ_{2} (t) + \dots + {\vec{v}}_{n} ξ_{n} (t) . \end{matrix}

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That is, we wish to write our solution as a linear combination of eigenvectors of

A .

If we solve for the scalar functions

ξ_{1}

through

ξ_{n},

we have our solution

\vec{x} .

Let us decompose

\vec{f}

in terms of the eigenvectors as well. We wish to write

\begin{matrix} (3.9) & \vec{f} (t) = {\vec{v}}_{1} g_{1} (t) + {\vec{v}}_{2} g_{2} (t) + \dots + {\vec{v}}_{n} g_{n} (t) . \end{matrix}

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That is, we wish to find

g_{1}

through

g_{n}

that satisfy (3.9). Since all the eigenvectors are independent, the matrix

E = [{\vec{v}}_{1} {\vec{v}}_{2} \dots {\vec{v}}_{n}]

is invertible. Write the equation (3.9) as

\vec{f} = E \vec{g},

where the components of

\vec{g}

are the functions

g_{1}

through

g_{n} .

Then

\vec{g} = E^{- 1} \vec{f} .

Hence it is always possible to find

\vec{g}

when there are

n

linearly independent eigenvectors.

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We plug (3.8) into (3.7), and note that

A {\vec{v}}_{k} = λ_{k} {\vec{v}}_{k} :

\begin{aligned} \overset{{\vec{x}}^{'}}{\overset{⏞}{{\vec{v}}_{1} ξ_{1}^{'} + {\vec{v}}_{2} ξ_{2}^{'} + \dots + {\vec{v}}_{n} ξ_{n}^{'}}} & = \overset{A \vec{x}}{\overset{⏞}{A ({\vec{v}}_{1} ξ_{1} + {\vec{v}}_{2} ξ_{2} + \dots + {\vec{v}}_{n} ξ_{n})}} + \overset{\vec{f}}{\overset{⏞}{{\vec{v}}_{1} g_{1} + {\vec{v}}_{2} g_{2} + \dots + {\vec{v}}_{n} g_{n}}} \\ = A {\vec{v}}_{1} ξ_{1} + A {\vec{v}}_{2} ξ_{2} + \dots + A {\vec{v}}_{n} ξ_{n} + {\vec{v}}_{1} g_{1} + {\vec{v}}_{2} g_{2} + \dots + {\vec{v}}_{n} g_{n} \\ = {\vec{v}}_{1} λ_{1} ξ_{1} + {\vec{v}}_{2} λ_{2} ξ_{2} + \dots + {\vec{v}}_{n} λ_{n} ξ_{n} + {\vec{v}}_{1} g_{1} + {\vec{v}}_{2} g_{2} + \dots + {\vec{v}}_{n} g_{n} \\ = {\vec{v}}_{1} (λ_{1} ξ_{1} + g_{1}) + {\vec{v}}_{2} (λ_{2} ξ_{2} + g_{2}) + \dots + {\vec{v}}_{n} (λ_{n} ξ_{n} + g_{n}) . \end{aligned}

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If we identify the coefficients of the vectors

{\vec{v}}_{1}

through

{\vec{v}}_{n},

we get the equations

\begin{aligned} ξ_{1}^{'} & = λ_{1} ξ_{1} + g_{1}, \\ ξ_{2}^{'} & = λ_{2} ξ_{2} + g_{2}, \\ ⋮ \\ ξ_{n}^{'} & = λ_{n} ξ_{n} + g_{n} . \end{aligned}

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Each one of these equations is independent of the others. They are all linear first order equations and can easily be solved by the standard integrating factor method for single equations. That is, for the

k^{th}

equation we write

ξ_{k}^{'} (t) - λ_{k} ξ_{k} (t) = g_{k} (t) .

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We use the integrating factor

e^{- λ_{k} t}

to find that

\frac{d}{d t} [ξ_{k} (t) e^{- λ_{k} t}] = e^{- λ_{k} t} g_{k} (t) .

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We integrate and solve for

ξ_{k}

to get

ξ_{k} (t) = e^{λ_{k} t} \int e^{- λ_{k} t} g_{k} (t) d t + C_{k} e^{λ_{k} t} .

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If we are looking for just any particular solution, we can set

C_{k}

to be zero. If we leave these constants in, we get the general solution. Write

\vec{x} (t) = {\vec{v}}_{1} ξ_{1} (t) + {\vec{v}}_{2} ξ_{2} (t) + \dots + {\vec{v}}_{n} ξ_{n} (t),

and we are done.

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As always, it is perhaps better to write these integrals as definite integrals. Suppose that we have an initial condition

\vec{x} (0) = \vec{b} .

Take

\vec{a} = E^{- 1} \vec{b}

to find

\vec{b} = {\vec{v}}_{1} a_{1} + {\vec{v}}_{2} a_{2} + \dots + {\vec{v}}_{n} a_{n},

just like before. Then if we write

ξ_{k} (t) = e^{λ_{k} t} \int_{0}^{t} e^{- λ_{k} s} g_{k} (s) d s + a_{k} e^{λ_{k} t},

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we get the particular solution

\vec{x} (t) = {\vec{v}}_{1} ξ_{1} (t) + {\vec{v}}_{2} ξ_{2} (t) + \dots + {\vec{v}}_{n} ξ_{n} (t)

satisfying

\vec{x} (0) = \vec{b},

because

ξ_{k} (0) = a_{k} .

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We remark that the technique we just outlined is the eigenvalue method applied to nonhomogeneous systems. If a system is homogeneous, that is, if

\vec{f} = \vec{0},

then the equations we get are

ξ_{k}^{'} = λ_{k} ξ_{k},

and so

ξ_{k} = C_{k} e^{λ_{k} t}

are the solutions and that is precisely what we got in Section 3.4.

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Example 3.9.2.

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Let

A = [\begin{matrix} 1 & 3 \\ 3 & 1 \end{matrix}] .

Solve

{\vec{x}}^{'} = A \vec{x} + \vec{f}

where

\vec{f} (t) = [\begin{matrix} 2 e^{t} \\ 2 t \end{matrix}]

for

\vec{x} (0) = [\begin{matrix} 3 / 16 \\ - 5 / 16 \end{matrix}] .

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The eigenvalues of

A

are

- 2

and 4 and corresponding eigenvectors are

[\begin{matrix} 1 \\ - 1 \end{matrix}]

and

[\begin{matrix} 1 \\ 1 \end{matrix}]

respectively. This calculation is left as an exercise. We write down the matrix

E

of the eigenvectors and compute its inverse (using the inverse formula for

2 \times 2

matrices)

E = [\begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix}], E^{- 1} = \frac{1}{2} [\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}] .

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We are looking for a solution of the form

\vec{x} = [\begin{matrix} 1 \\ - 1 \end{matrix}] ξ_{1} + [\begin{matrix} 1 \\ 1 \end{matrix}] ξ_{2} .

We first need to write

\vec{f}

in terms of the eigenvectors. That is we wish to write

\vec{f} = [\begin{matrix} 2 e^{t} \\ 2 t \end{matrix}] = [\begin{matrix} 1 \\ - 1 \end{matrix}] g_{1} + [\begin{matrix} 1 \\ 1 \end{matrix}] g_{2} .

Thus

[\begin{matrix} g_{1} \\ g_{2} \end{matrix}] = E^{- 1} [\begin{matrix} 2 e^{t} \\ 2 t \end{matrix}] = \frac{1}{2} [\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}] [\begin{matrix} 2 e^{t} \\ 2 t \end{matrix}] = [\begin{matrix} e^{t} - t \\ e^{t} + t \end{matrix}] .

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g_{1} = e^{t} - t

and

g_{2} = e^{t} + t .

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We further need to write

\vec{x} (0)

in terms of the eigenvectors. That is, we wish to write

\vec{x} (0) = [\begin{matrix} 3 / 16 \\ - 5 / 16 \end{matrix}] = [\begin{matrix} 1 \\ - 1 \end{matrix}] a_{1} + [\begin{matrix} 1 \\ 1 \end{matrix}] a_{2} .

Hence

[\begin{matrix} a_{1} \\ a_{2} \end{matrix}] = E^{- 1} [\begin{matrix} ^{3} /_{16} \\ ^{- 5} /_{16} \end{matrix}] = [\begin{matrix} ^{1} /_{4} \\ ^{- 1} /_{16} \end{matrix}] .

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a_{1} =^{1} /_{4}

and

a_{2} =^{- 1} /_{16} .

We plug our

\vec{x}

into the equation and get

\begin{aligned} \overset{{\vec{x}}^{'}}{\overset{⏞}{[\begin{array}{c} 1 \\ - 1 \end{array}] ξ_{1}^{'} + [\begin{array}{c} 1 \\ 1 \end{array}] ξ_{2}^{'}}} & = \overset{A \vec{x}}{\overset{⏞}{A [\begin{array}{c} 1 \\ - 1 \end{array}] ξ_{1} + A [\begin{array}{c} 1 \\ 1 \end{array}] ξ_{2}}} + \overset{\vec{f}}{\overset{⏞}{[\begin{array}{c} 1 \\ - 1 \end{array}] g_{1} + [\begin{array}{c} 1 \\ 1 \end{array}] g_{2}}} \\ = [\begin{array}{c} 1 \\ - 1 \end{array}] (- 2 ξ_{1}) + [\begin{array}{c} 1 \\ 1 \end{array}] 4 ξ_{2} + [\begin{array}{c} 1 \\ - 1 \end{array}] (e^{t} - t) + [\begin{array}{c} 1 \\ 1 \end{array}] (e^{t} + t) . \end{aligned}

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We get the two equations

\begin{aligned} ξ_{1}^{'} = - 2 ξ_{1} + e^{t} - t, & where ξ_{1} (0) = a_{1} = \frac{1}{4}, \\ ξ_{2}^{'} = 4 ξ_{2} + e^{t} + t, & where ξ_{2} (0) = a_{2} = \frac{- 1}{16} . \end{aligned}

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We solve with integrating factor. Computation of the integral is left as an exercise to the student. You will need integration by parts.

ξ_{1} = e^{- 2 t} \int e^{2 t} (e^{t} - t) d t + C_{1} e^{- 2 t} = \frac{e^{t}}{3} - \frac{t}{2} + \frac{1}{4} + C_{1} e^{- 2 t} .

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C_{1}

is the constant of integration. As

ξ_{1} (0) =^{1} /_{4},

then

^{1} /_{4} =^{1} /_{3} +^{1} /_{4} + C_{1}

and hence

C_{1} =^{- 1} /_{3} .

Similarly

ξ_{2} = e^{4 t} \int e^{- 4 t} (e^{t} + t) d t + C_{2} e^{4 t} = - \frac{e^{t}}{3} - \frac{t}{4} - \frac{1}{16} + C_{2} e^{4 t} .

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ξ_{2} (0) =^{- 1} /_{16}

we have

^{- 1} /_{16} =^{- 1} /_{3} -^{1} /_{16} + C_{2}

and hence

C_{2} =^{1} /_{3} .

The solution is

\vec{x} (t) = [\begin{matrix} 1 \\ - 1 \end{matrix}] \underset{ξ_{1}}{\underset{⏟}{(\frac{e^{t} - e^{- 2 t}}{3} + \frac{1 - 2 t}{4})}} + [\begin{matrix} 1 \\ 1 \end{matrix}] \underset{ξ_{2}}{\underset{⏟}{(\frac{e^{4 t} - e^{t}}{3} - \frac{4 t + 1}{16})}} = [\begin{matrix} \frac{e^{4 t} - e^{- 2 t}}{3} + \frac{3 - 12 t}{16} \\ \frac{e^{- 2 t} + e^{4 t} - 2 e^{t}}{3} + \frac{4 t - 5}{16} \end{matrix}] .

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That is,

x_{1} = \frac{e^{4 t} - e^{- 2 t}}{3} + \frac{3 - 12 t}{16}

and

x_{2} = \frac{e^{- 2 t} + e^{4 t} - 2 e^{t}}{3} + \frac{4 t - 5}{16} .

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Exercise 3.9.1.

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Check that

x_{1}

and

x_{2}

solve the problem. Check both that they satisfy the differential equation and that they satisfy the initial conditions.

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Subsubsection 3.9.1.3 Undetermined coefficients

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The method of undetermined coefficients also works for systems. The only difference is that we use unknown vectors rather than just numbers. Same caveats apply to undetermined coefficients for systems as for single equations. This method does not always work. Furthermore, if the right-hand side is complicated, we have to solve for lots of variables. Each element of an unknown vector is an unknown number. In system of 3 equations with say 4 unknown vectors (this would not be uncommon), we already have 12 unknown numbers to solve for. The method can turn into a lot of tedious work if done by hand. As the method is essentially the same as for single equations, let us just do an example.

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Example 3.9.3.

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Let

A = [\begin{matrix} - 1 & 0 \\ - 2 & 1 \end{matrix}] .

Find a particular solution of

{\vec{x}}^{'} = A \vec{x} + \vec{f}

where

\vec{f} (t) = [\begin{matrix} e^{t} \\ t \end{matrix}] .

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Note that we can solve this system in an easier way (can you see how?), but for the purposes of the example, let us use the eigenvalue method plus undetermined coefficients. The eigenvalues of

A

are

- 1

and 1 and corresponding eigenvectors are

[\begin{matrix} 1 \\ 1 \end{matrix}]

and

[\begin{matrix} 0 \\ 1 \end{matrix}]

respectively. Hence our complementary solution is

{\vec{x}}_{c} = α_{1} [\begin{matrix} 1 \\ 1 \end{matrix}] e^{- t} + α_{2} [\begin{matrix} 0 \\ 1 \end{matrix}] e^{t},

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for some arbitrary constants

α_{1}

and

α_{2} .

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We would want to guess a particular solution of

\vec{x} = \vec{a} e^{t} + \vec{b} t + \vec{c} .

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However, something of the form

\vec{a} e^{t}

appears in the complementary solution. Because we do not yet know if the vector

\vec{a}

is a multiple of

[\begin{matrix} 0 \\ 1 \end{matrix}],

we do not know if a conflict arises. It is possible that there is no conflict, but to be safe we should also try

\vec{b} t e^{t} .

Here we find the crux of the difference between a single equation and systems. We try both terms

\vec{a} e^{t}

and

\vec{b} t e^{t}

in the solution, not just the term

\vec{b} t e^{t} .

Therefore, we try

\vec{x} = \vec{a} e^{t} + \vec{b} t e^{t} + \vec{c} t + \vec{d} .

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Thus we have 8 unknowns. We write

\vec{a} = [\begin{matrix} a_{1} \\ a_{2} \end{matrix}],

\vec{b} = [\begin{matrix} b_{1} \\ b_{2} \end{matrix}],

\vec{c} = [\begin{matrix} c_{1} \\ c_{2} \end{matrix}],

and

\vec{d} = [\begin{matrix} d_{1} \\ d_{2} \end{matrix}] .

We plug

\vec{x}

into the equation. First let us compute

{\vec{x}}^{'} .

{\vec{x}}^{'} = (\vec{a} + \vec{b}) e^{t} + \vec{b} t e^{t} + \vec{c} = [\begin{matrix} a_{1} + b_{1} \\ a_{2} + b_{2} \end{matrix}] e^{t} + [\begin{matrix} b_{1} \\ b_{2} \end{matrix}] t e^{t} + [\begin{matrix} c_{1} \\ c_{2} \end{matrix}] .

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Now

{\vec{x}}^{'}

must equal

A \vec{x} + \vec{f},

which is

\begin{aligned} A \vec{x} + \vec{f} & = A \vec{a} e^{t} + A \vec{b} t e^{t} + A \vec{c} t + A \vec{d} + \vec{f} \\ = [\begin{array}{c} - a_{1} \\ - 2 a_{1} + a_{2} \end{array}] e^{t} + [\begin{array}{c} - b_{1} \\ - 2 b_{1} + b_{2} \end{array}] t e^{t} + [\begin{array}{c} - c_{1} \\ - 2 c_{1} + c_{2} \end{array}] t + [\begin{array}{c} - d_{1} \\ - 2 d_{1} + d_{2} \end{array}] + [\begin{array}{c} 1 \\ 0 \end{array}] e^{t} + [\begin{array}{c} 0 \\ 1 \end{array}] t \\ = [\begin{array}{c} - a_{1} + 1 \\ - 2 a_{1} + a_{2} \end{array}] e^{t} + [\begin{array}{c} - b_{1} \\ - 2 b_{1} + b_{2} \end{array}] t e^{t} + [\begin{array}{c} - c_{1} \\ - 2 c_{1} + c_{2} + 1 \end{array}] t + [\begin{array}{c} - d_{1} \\ - 2 d_{1} + d_{2} \end{array}] . \end{aligned}

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We identify the coefficients of

e^{t},

t e^{t},

t

and any constant vectors in

{\vec{x}}^{'}

and in

A \vec{x} + \vec{f}

to find the equations:

\begin{aligned} a_{1} + b_{1} & = - a_{1} + 1, & 0 & = - c_{1}, \\ a_{2} + b_{2} & = - 2 a_{1} + a_{2}, & 0 & = - 2 c_{1} + c_{2} + 1, \\ b_{1} & = - b_{1}, & c_{1} & = - d_{1}, \\ b_{2} & = - 2 b_{1} + b_{2}, & c_{2} & = - 2 d_{1} + d_{2} . \end{aligned}

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We could write the

8 \times 9

augmented matrix and start row reduction, but it is easier to just solve the equations in an ad hoc manner. Immediately we see that

b_{1} = 0,

c_{1} = 0,

d_{1} = 0 .

Plugging these back in, we get that

c_{2} = - 1

and

d_{2} = - 1 .

The remaining equations that tell us something are

\begin{aligned} a_{1} & = - a_{1} + 1, \\ a_{2} + b_{2} & = - 2 a_{1} + a_{2} . \end{aligned}

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a_{1} =^{1} /_{2}

and

b_{2} = - 1 .

Finally,

a_{2}

can be arbitrary and still satisfy the equations. We are looking for just a single solution so presumably the simplest one is when

a_{2} = 0 .

Therefore,

\vec{x} = \vec{a} e^{t} + \vec{b} t e^{t} + \vec{c} t + \vec{d} = [\begin{matrix} ^{1} /_{2} \\ 0 \end{matrix}] e^{t} + [\begin{matrix} 0 \\ - 1 \end{matrix}] t e^{t} + [\begin{matrix} 0 \\ - 1 \end{matrix}] t + [\begin{matrix} 0 \\ - 1 \end{matrix}] = [\begin{matrix} \frac{1}{2} e^{t} \\ - t e^{t} - t - 1 \end{matrix}] .

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That is,

x_{1} = \frac{1}{2} e^{t},

x_{2} = - t e^{t} - t - 1 .

We would add this to the complementary solution to get the general solution of the problem. Notice that both

\vec{a} e^{t}

and

\vec{b} t e^{t}

were really needed.

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Exercise 3.9.2.

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Check that

x_{1}

and

x_{2}

solve the problem. Try setting

a_{2} = 1

and check we get a solution as well. What is the difference between the two solutions we obtained (one with

a_{2} = 0

and one with

a_{2} = 1

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As you can see, other than the handling of conflicts, undetermined coefficients works exactly the same as it did for single equations. However, the computations can get out of hand pretty quickly for systems. The equation we considered was pretty simple.

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Subsection 3.9.2 First order variable coefficient

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Subsubsection 3.9.2.1 Variation of parameters

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Just as for a single equation, there is the method of variation of parameters. For constant coefficient systems, it is essentially the same thing as the integrating factor method we discussed earlier. However, this method works for any linear system, even if it is not constant coefficient, provided we somehow solve the associated homogeneous problem.

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Suppose we have the equation

\begin{matrix} (3.10) & {\vec{x}}^{'} = A (t) \vec{x} + \vec{f} (t) . \end{matrix}

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Further, suppose we solved the associated homogeneous equation

{\vec{x}}^{'} = A (t) \vec{x}

and found a fundamental matrix solution

X (t) .

The general solution to the associated homogeneous equation is

X (t) \vec{c}

for a constant vector

\vec{c} .

Just like for variation of parameters for single equation we try the solution to the nonhomogeneous equation of the form

{\vec{x}}_{p} = X (t) \vec{u} (t),

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where

\vec{u} (t)

is a vector-valued function instead of a constant. We substitute

{\vec{x}}_{p}

into (3.10) to obtain

\underset{{\vec{x}}_{p}^{'} (t)}{\underset{⏟}{X^{'} (t) \vec{u} (t) + X (t) {\vec{u}}^{'} (t)}} = \underset{A (t) {\vec{x}}_{p} (t)}{\underset{⏟}{A (t) X (t) \vec{u} (t)}} + \vec{f} (t) .

🔗

But

X (t)

is a fundamental matrix solution to the homogeneous problem. So

X^{'} (t) = A (t) X (t),

and

X^{'} (t) \vec{u} (t) + X (t) {\vec{u}}^{'} (t) = X^{'} (t) \vec{u} (t) + \vec{f} (t) .

🔗

Hence

X (t) {\vec{u}}^{'} (t) = \vec{f} (t) .

If we compute

{[X (t)]}^{- 1},

then

{\vec{u}}^{'} (t) = {[X (t)]}^{- 1} \vec{f} (t) .

We integrate to obtain

\vec{u}

and we have the particular solution

{\vec{x}}_{p} = X (t) \vec{u} (t) .

Let us write this as a formula

{\vec{x}}_{p} = X (t) \int {[X (t)]}^{- 1} \vec{f} (t) d t .

🔗

A

is a constant matrix and

X (t) = e^{t A},

then

{[X (t)]}^{- 1} = e^{- t A} .

We obtain a solution

{\vec{x}}_{p} = e^{t A} \int e^{- t A} \vec{f} (t) d t,

which is precisely what we got using the integrating factor method.

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Example 3.9.4.

🔗

Find a particular solution to

\begin{matrix} (3.11) & {\vec{x}}^{'} = \frac{1}{t^{2} + 1} [\begin{matrix} t & - 1 \\ 1 & t \end{matrix}] \vec{x} + [\begin{matrix} t \\ 1 \end{matrix}] (t^{2} + 1) . \end{matrix}

🔗

Here

A = \frac{1}{t^{2} + 1} [\begin{matrix} t & - 1 \\ 1 & t \end{matrix}]

is most definitely not constant. Perhaps by a lucky guess, we find that

X = [\begin{matrix} 1 & - t \\ t & 1 \end{matrix}]

solves

X^{'} (t) = A (t) X (t) .

Once we know the complementary solution we can easily find a solution to (3.11). First we find

{[X (t)]}^{- 1} = \frac{1}{t^{2} + 1} [\begin{matrix} 1 & t \\ - t & 1 \end{matrix}] .

🔗

Next we know a particular solution to (3.11) is

\begin{aligned} {\vec{x}}_{p} & = X (t) \int {[X (t)]}^{- 1} \vec{f} (t) d t \\ = [\begin{array}{c} 1 & - t \\ t & 1 \end{array}] \int \frac{1}{t^{2} + 1} [\begin{array}{c} 1 & t \\ - t & 1 \end{array}] [\begin{array}{c} t \\ 1 \end{array}] (t^{2} + 1) d t \\ = [\begin{array}{c} 1 & - t \\ t & 1 \end{array}] \int [\begin{array}{c} 2 t \\ - t^{2} + 1 \end{array}] d t \\ = [\begin{array}{c} 1 & - t \\ t & 1 \end{array}] [\begin{array}{c} t^{2} \\ - \frac{1}{3} t^{3} + t \end{array}] \\ = [\begin{array}{c} \frac{1}{3} t^{4} \\ \frac{2}{3} t^{3} + t \end{array}] . \end{aligned}

🔗

Adding the complementary solution we find the general solution to (3.11):

\vec{x} = [\begin{matrix} 1 & - t \\ t & 1 \end{matrix}] [\begin{matrix} c_{1} \\ c_{2} \end{matrix}] + [\begin{matrix} \frac{1}{3} t^{4} \\ \frac{2}{3} t^{3} + t \end{matrix}] = [\begin{matrix} c_{1} - c_{2} t + \frac{1}{3} t^{4} \\ c_{2} + (c_{1} + 1) t + \frac{2}{3} t^{3} \end{matrix}] .

🔗

Exercise 3.9.3.

🔗

Check that

x_{1} = \frac{1}{3} t^{4}

and

x_{2} = \frac{2}{3} t^{3} + t

really solve (3.11).

🔗

In the variation of parameters, just like in the integrating factor method we can obtain the general solution by adding in constants of integration. That is, we will add

X (t) \vec{c}

for a vector of arbitrary constants. But that is precisely the complementary solution.

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Subsection 3.9.3 Second order constant coefficients

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Subsubsection 3.9.3.1 Undetermined coefficients

🔗

We have already seen a simple example of the method of undetermined coefficients for second order systems in Section 3.6. This method is essentially the same as undetermined coefficients for first order systems. There are some simplifications that we can make, as we did in Section 3.6. Let the equation be

{\vec{x}}^{″} = A \vec{x} + \vec{F} (t),

🔗

where

A

is a constant matrix. If

\vec{F} (t)

is of the form

{\vec{F}}_{0} \cos (ω t),

then as two derivatives of cosine is again cosine we can try a solution of the form

{\vec{x}}_{p} = \vec{c} \cos (ω t),

🔗

and we do not need to introduce sines.

🔗

If the

\vec{F}

is a sum of cosines, note that we still have the superposition principle. If

\vec{F} (t) = {\vec{F}}_{0} \cos (ω_{0} t) + {\vec{F}}_{1} \cos (ω_{1} t),

then we would try

\vec{a} \cos (ω_{0} t)

for the problem

{\vec{x}}^{″} = A \vec{x} + {\vec{F}}_{0} \cos (ω_{0} t),

and we would try

\vec{b} \cos (ω_{1} t)

for the problem

{\vec{x}}^{″} = A \vec{x} + {\vec{F}}_{1} \cos (ω_{1} t) .

Then we sum the solutions.

🔗

However, if there is duplication with the complementary solution, or the equation is of the form

{\vec{x}}^{″} = A {\vec{x}}^{'} + B \vec{x} + \vec{F} (t),

then we need to do the same thing as we do for first order systems.

🔗

You will never go wrong with putting in more terms than needed into your guess. You will find that the extra coefficients will turn out to be zero. But it is useful to save some time and effort.

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Subsubsection 3.9.3.2 Eigenvector decomposition

🔗

If we have the system

{\vec{x}}^{″} = A \vec{x} + \vec{f} (t),

🔗

we can do eigenvector decomposition, just like for first order systems.

🔗

Let

λ_{1}, λ_{2}, \dots, λ_{n}

be the eigenvalues and

{\vec{v}}_{1}, {\vec{v}}_{2}, \dots, {\vec{v}}_{n}

be eigenvectors. Again form the matrix

E = [{\vec{v}}_{1} {\vec{v}}_{2} \dots {\vec{v}}_{n}] .

Write

\vec{x} (t) = {\vec{v}}_{1} ξ_{1} (t) + {\vec{v}}_{2} ξ_{2} (t) + \dots + {\vec{v}}_{n} ξ_{n} (t) .

🔗

Decompose

\vec{f}

in terms of the eigenvectors

\vec{f} (t) = {\vec{v}}_{1} g_{1} (t) + {\vec{v}}_{2} g_{2} (t) + \dots + {\vec{v}}_{n} g_{n} (t),

🔗

where, again,

\vec{g} = E^{- 1} \vec{f} .

🔗

We plug in and as before we obtain

\begin{aligned} \overset{{\vec{x}}^{″}}{\overset{⏞}{{\vec{v}}_{1} ξ_{1}^{″} + {\vec{v}}_{2} ξ_{2}^{″} + \dots + {\vec{v}}_{n} ξ_{n}^{″}}} & = \overset{A \vec{x}}{\overset{⏞}{A ({\vec{v}}_{1} ξ_{1} + {\vec{v}}_{2} ξ_{2} + \dots + {\vec{v}}_{n} ξ_{n})}} + \overset{\vec{f}}{\overset{⏞}{{\vec{v}}_{1} g_{1} + {\vec{v}}_{2} g_{2} + \dots + {\vec{v}}_{n} g_{n}}} \\ = A {\vec{v}}_{1} ξ_{1} + A {\vec{v}}_{2} ξ_{2} + \dots + A {\vec{v}}_{n} ξ_{n} + {\vec{v}}_{1} g_{1} + {\vec{v}}_{2} g_{2} + \dots + {\vec{v}}_{n} g_{n} \\ = {\vec{v}}_{1} λ_{1} ξ_{1} + {\vec{v}}_{2} λ_{2} ξ_{2} + \dots + {\vec{v}}_{n} λ_{n} ξ_{n} + {\vec{v}}_{1} g_{1} + {\vec{v}}_{2} g_{2} + \dots + {\vec{v}}_{n} g_{n} \\ = {\vec{v}}_{1} (λ_{1} ξ_{1} + g_{1}) + {\vec{v}}_{2} (λ_{2} ξ_{2} + g_{2}) + \dots + {\vec{v}}_{n} (λ_{n} ξ_{n} + g_{n}) . \end{aligned}

🔗

We identify the coefficients of the eigenvectors to get the equations

\begin{aligned} ξ_{1}^{″} & = λ_{1} ξ_{1} + g_{1}, \\ ξ_{2}^{″} & = λ_{2} ξ_{2} + g_{2}, \\ ⋮ \\ ξ_{n}^{″} & = λ_{n} ξ_{n} + g_{n} . \end{aligned}

🔗

Each one of these equations is independent of the others. We solve each equation using the methods of Chapter 2. We write

\vec{x} (t) = {\vec{v}}_{1} ξ_{1} (t) + {\vec{v}}_{2} ξ_{2} (t) + \dots + {\vec{v}}_{n} ξ_{n} (t),

and we are done; we have a particular solution. We find the general solutions for

ξ_{1}

through

ξ_{n},

and again

\vec{x} (t) = {\vec{v}}_{1} ξ_{1} (t) + {\vec{v}}_{2} ξ_{2} (t) + \dots + {\vec{v}}_{n} ξ_{n} (t)

is the general solution (and not just a particular solution).

🔗

Example 3.9.5.

🔗

Let us do the example from Section 3.6 using this method. The equation is

{\vec{x}}^{″} = [\begin{matrix} - 3 & 1 \\ 2 & - 2 \end{matrix}] \vec{x} + [\begin{matrix} 0 \\ 2 \end{matrix}] \cos (3 t) .

🔗

The eigenvalues are

- 1

and

- 4,

with eigenvectors

[\begin{matrix} 1 \\ 2 \end{matrix}]

and

[\begin{matrix} 1 \\ - 1 \end{matrix}] .

Therefore

E = [\begin{matrix} 1 & 1 \\ 2 & - 1 \end{matrix}]

and

E^{- 1} = \frac{1}{3} [\begin{matrix} 1 & 1 \\ 2 & - 1 \end{matrix}] .

Therefore,

[\begin{matrix} g_{1} \\ g_{2} \end{matrix}] = E^{- 1} \vec{f} (t) = \frac{1}{3} [\begin{matrix} 1 & 1 \\ 2 & - 1 \end{matrix}] [\begin{matrix} 0 \\ 2 \cos (3 t) \end{matrix}] = [\begin{matrix} \frac{2}{3} \cos (3 t) \\ \frac{- 2}{3} \cos (3 t) \end{matrix}] .

🔗

So after the whole song and dance of plugging in, the equations we get are

ξ_{1}^{″} = - ξ_{1} + \frac{2}{3} \cos (3 t), ξ_{2}^{″} = - 4 ξ_{2} - \frac{2}{3} \cos (3 t) .

🔗

For each equation we use the method of undetermined coefficients. We try

C_{1} \cos (3 t)

for the first equation and

C_{2} \cos (3 t)

for the second equation. We plug in to get

\begin{aligned} - 9 C_{1} \cos (3 t) & = - C_{1} \cos (3 t) + \frac{2}{3} \cos (3 t), \\ - 9 C_{2} \cos (3 t) & = - 4 C_{2} \cos (3 t) - \frac{2}{3} \cos (3 t) . \end{aligned}

🔗

We solve each of these equations separately. We get

- 9 C_{1} = - C_{1} +^{2} /_{3}

and

- 9 C_{2} = - 4 C_{2} -^{2} /_{3} .

And hence

C_{1} =^{- 1} /_{12}

and

C_{2} =^{2} /_{15} .

So our particular solution is

\vec{x} = [\begin{matrix} 1 \\ 2 \end{matrix}] (\frac{- 1}{12} \cos (3 t)) + [\begin{matrix} 1 \\ - 1 \end{matrix}] (\frac{2}{15} \cos (3 t)) = [\begin{matrix} ^{1} /_{20} \\ ^{- 3} /_{10} \end{matrix}] \cos (3 t) .

🔗

This solution matches what we got previously in Section 3.6.

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Exercises 3.9.4 Exercises

🔗

3.9.4.

🔗

Find a particular solution to

x^{'} = x + 2 y + 2 t,

y^{'} = 3 x + 2 y - 4,

using integrating factor method,
using eigenvector decomposition,
using undetermined coefficients.

🔗

3.9.5.

🔗

Find the general solution to

x^{'} = 4 x + y - 1,

y^{'} = x + 4 y - e^{t},

using integrating factor method,
using eigenvector decomposition,
using undetermined coefficients.

🔗

3.9.6.

🔗

Find the general solution to

x_{1}^{″} = - 6 x_{1} + 3 x_{2} + \cos (t),

x_{2}^{″} = 2 x_{1} - 7 x_{2} + 3 \cos (t),

using eigenvector decomposition,
using undetermined coefficients.

🔗

3.9.7.

🔗

Find the general solution to

x_{1}^{″} = - 6 x_{1} + 3 x_{2} + \cos (2 t),

x_{2}^{″} = 2 x_{1} - 7 x_{2} + 3 \cos (2 t),

using eigenvector decomposition,
using undetermined coefficients.

🔗

3.9.8.

🔗

Take the equation

{\vec{x}}^{'} = [\begin{matrix} \frac{1}{t} & - 1 \\ 1 & \frac{1}{t} \end{matrix}] \vec{x} + [\begin{matrix} t^{2} \\ - t \end{matrix}] .

Check that ${\vec{x}}_{c} = c_{1} [\begin{matrix} t \sin t \\ - t \cos t \end{matrix}] + c_{2} [\begin{matrix} t \cos t \\ t \sin t \end{matrix}]$ is the complementary solution.
Use variation of parameters to find a particular solution.