Chapter 8: Q35E (page 361)

The system of differential equations for the currents $i_{1} (t)$ and $i_{2} (t)$ in the electrical network shown in Figure 8.3.3 is
$\frac{d}{dt} (\begin{matrix} i_{1} \\ i_{2} \end{matrix}) = (\begin{matrix} - (R_{1} + R_{2}) / L_{2} & R_{2} / L_{2} \\ R_{2} / L_{1} & - R_{2} / L_{1} \end{matrix}) (\begin{matrix} i_{1} \\ i_{2} \end{matrix}) + (\begin{matrix} E / L_{2} \\ 0 \end{matrix})$
Use variation of parameters to solve the system if
localid="1668106421663" $R_{1} = 8 Ω R_{2} = 3 Ω, L_{1} = 1 h, L_{2} = 1 h, E (t) = 100 sint V, i_{1} (0) = 0,$ and $i_{2} (0) = 0$ .

Short Answer

Expert verified

The general solution of the given matrix is

$X = c_{1} (\begin{matrix} 1 \\ 3 \end{matrix}) e^{- 2 t} + c_{2} (\begin{matrix} 3 \\ - 1 \end{matrix}) e^{- 12 t} + (\begin{matrix} \frac{332}{29} \sin t & - \frac{76}{29} \cos t \\ \frac{276}{29} \sin t & - \frac{168}{29} \cos t \end{matrix})$

Step by step solution

Definition of Eigenvalue.

Eigenvalues, also known as characteristic roots, characteristic values, appropriate values, or latent roots, are a specific set of scalars associated with a linear system of equations (i.e., a matrix equation).
Each eigenvalue has an eigenvector that corresponds to it (or, in general, a corresponding right eigenvector and a corresponding left eigenvector; there is no analogous distinction between left and right for eigenvalues).

Substitute the known values

Given that,

$\frac{d}{dt} (\begin{matrix} i_{1} \\ i_{2} \end{matrix}) = (\begin{matrix} - \frac{(R_{1} + R_{2})}{L_{2}} & \frac{R_{2}}{L_{2}} \\ \frac{R_{2}}{L_{1}} & - \frac{R_{2}}{L_{1}} \end{matrix}) (\begin{matrix} i_{1} \\ i_{2} \end{matrix}) + (\begin{matrix} \frac{E}{L_{2}} \\ 0 \end{matrix})$

Also given role="math" localid="1668098019014" $R_{1} = 8 Ω R_{2} = 3 Ω, L_{1} = 1 h, L_{2} = 1 h, E (t) = 100 sint v, i_{1} (0) = 0$ . and $i_{2} (0) = 0$

Substituting these values into our given system the system becomes

$\frac{d}{dt} (\begin{matrix} i_{1} \\ i_{2} \end{matrix}) = (\begin{matrix} - 11 & 3 \\ 3 & - 3 \end{matrix}) (\begin{matrix} i_{1} \\ i_{2} \end{matrix}) + (\begin{matrix} 100 \sin t \\ 0 \end{matrix})$

let $X = (\begin{matrix} i_{1} \\ i_{2} \end{matrix})$

$X' = (\begin{matrix} - 11 & 3 \\ 3 & - 3 \end{matrix}) X + (\begin{matrix} 100 \sin t \\ 0 \end{matrix})$

where $A = (\begin{matrix} - 11 & 3 \\ 3 & - 3 \end{matrix})$ , and $F (t) = (\begin{matrix} 100 \sin t \\ 0 \end{matrix})$ .

Determine the eigenvalues of the coefficient matrix

Obtaining the eigenvalues of the coefficient matrix,

$\det (A - λI) = 0 \to |\begin{matrix} - 11 - λ & 3 \\ 3 & - 3 - λ \end{matrix}| = 0$

$(- 11 - λ) (- 3 - λ) - 9 = 0$

$33 + 11 λ + 3 λ + λ^{2} - 9 = 0$

$λ^{2} + 14 λ + 24 = 0$

$(λ + 2) (λ + 12) = 0$

So our eigenvalues are $λ_{1} = - 2$ and $λ_{2} = - 12$ .

For $λ_{1} = - 2$ :

$(A + 2 I | 0) = (\begin{matrix} - 11 - (- 2) & 3 & 0 \\ 3 & - 3 - (- 2) & 0 \end{matrix})$

$= (\begin{matrix} - 9 & 3 & 0 \\ 3 & - 1 & 0 \end{matrix})$

Apply row operation $3 R_{2} + R_{1} \to R_{2}$ :

$= (\begin{matrix} - 9 & 3 & 0 \\ 0 & 0 & 0 \end{matrix})$

$- 9 k_{1} + 3 k_{2} = 0 \to k_{1} = \frac{1}{3} k_{2}$

Choosing $k_{2} = 3$ yields $k_{1} = 1$ . This gives an eigenvector and a corresponding solution

vector

$K_{1} = (\begin{matrix} 1 \\ 3 \end{matrix}), X_{1} = (\begin{matrix} 1 \\ 3 \end{matrix}) e^{- 2 t}$

For $λ_{2} = - 12$ :

$(A + 12 I | 0) = (\begin{matrix} - 11 - (- 12) & 3 & 0 \\ 3 & - 3 - (- 12) & 0 \end{matrix})$

$= (\begin{matrix} 1 & 3 & 0 \\ 3 & 9 & 0 \end{matrix})$

Apply row operation $R_{2} - 3 R_{1} \to R_{2}$ :

$= (\begin{matrix} 1 & 3 & 0 \\ 0 & 0 & 0 \end{matrix})$

so here we have

$k_{1} + 3 k_{2} = 0 \to k_{1} = 3 k_{2}$

choosing $k_{2} = - 1$ yields $k_{1} = 3$ . This gives an eigenvector and a corresponding solution

vector

$K_{2} = (\begin{matrix} 3 \\ - 1 \end{matrix}), X_{2} = (\begin{matrix} 3 \\ - 1 \end{matrix}) e^{- 12 t}$

hence, the complementary solution is

$X_{c} = c_{1} X_{1} + c_{2} X_{2}$

$= c_{1} (\begin{matrix} 1 \\ 3 \end{matrix}) e^{- 2 t} + c_{2} (\begin{matrix} 3 \\ - 1 \end{matrix}) e^{- 12 t}$

Step 4:Determine the particular solution.

The entries in $X_{1}$ form the first column of $ϕ (t)$ , and the entries in $X_{2}$ form the second

column of $ϕ (t)$ . Therefore,

$ϕ (t) = (\begin{matrix} e^{- 2 t} & 3 e^{- 12 t} \\ 3 e^{- 2 t} & - e^{- 12 t} \end{matrix})$

To make sure that $ϕ (t)$ is an invertible matrix by checking the determinant, where

$|\begin{matrix} ϕ (t) \end{matrix}| = (\begin{matrix} e^{- 2 t} & 3 e^{- 12 t} \\ 3 e^{- 2 t} & - e^{- 12 t} \end{matrix})$

$= - e^{- 2 t} e^{- 12 t} - 9 e^{- 12 t} e^{- 2 t}$

$= - e^{- 14 t} - 9 e^{- 14 t}$

$= - 10 e^{- 14 t} \neq 0$

Since the determinant does not equal zero, the matrix is in fact, an invertible matrix.

So now,

$(t) = - \frac{1}{10 e^{- 14 t}} (\begin{matrix} - e^{- 12 t} & - 3 e^{- 12 t} \\ - 3 e^{- 2 t} & e^{- 2 t} \end{matrix}) = (\begin{matrix} \frac{1}{10} e^{2 t} & \frac{3}{10} e^{2 t} \\ \frac{3}{10} e^{12 t} & \frac{1}{10} e^{12 t} \end{matrix})$

Obtaining the particular solution,

$X_{p} = ϕ (t) \int ϕ^{- 1} (t) F (t) dt$

role="math" localid="1668101302545" $= (\begin{matrix} e^{- 2 t} & 3 e^{- 12 t} \\ 3 e^{- 2 t} & - e^{- 12 t} \end{matrix}) \int (\begin{matrix} \frac{1}{10} e^{2 t} & \frac{3}{10} e^{2 t} \\ \frac{3}{10} e^{12 t} & - \frac{1}{10} e^{12 t} \end{matrix}) (\begin{matrix} 100 sint \\ 0 \end{matrix}) dt$

role="math" localid="1668101386187" $= (\begin{matrix} e^{- 2 t} & 3 e^{- 12 t} \\ 3 e^{- 2 t} & - e^{- 12 t} \end{matrix}) \int (\begin{matrix} 10 e^{2 t} sint \\ 30 e^{12 t} sint \end{matrix}) dt$

Evaluate the integral

$\int e^{2 t} \sin tdt$

By using integration by parts twice in a row.

First time:

$u = \sin t dv = e^{2 t}$

$du = \cos t v = \frac{e^{2 t}}{2}$

where we will get

$\int e^{2 t} \sin tdt = uv - \int vdu$

$= \frac{1}{2} e^{2 t} \sin t - \frac{1}{2} \int e^{2 t} \cos tdt$

Second time:

$u = \cos t dv = e^{2 t}$

$du = - \sin t v = \frac{e^{2 t}}{2}$

where we will get

$\int e^{2 t} \sin tdt = \frac{1}{2} e^{2 t} \sin t - \frac{1}{2} [\frac{1}{2} e^{2 t} \cos t + \frac{1}{2} \int e^{2 t} \sin tdt]$

$= \frac{1}{2} e^{2 t} \sin t - \frac{1}{4} e^{2 t} \cos t - \frac{1}{4} \int e^{2 t} \sin tdt$

Notice that the integral $\int e^{2 t}$ appears again on the right side of the equation, we

can solve for it,

$\int e^{2 t} \sin tdt + \frac{1}{4} \int e^{2 t} \sin tdt$

$= \frac{1}{2} e^{2 t} \sin t - \frac{1}{4} e^{2 t} \cos t \frac{5}{4} \int e^{2 t} \sin tdt$

$= \frac{1}{2} e^{2 t} \sin t - \frac{1}{4} e^{2 t} \cos t \int e^{2 t} \sin tdt$

role="math" localid="1668103363776" $= \frac{2}{5} e^{2 t} \sin t - \frac{1}{5} e^{2 t} \cos t$

Evaluate the integral

To evaluate the integral

$\int e^{12 t} \sin tdt$

we will use integration by parts twice in a row.

First time:

$u = \sin t dv = e^{12 t}$

$du = \cos t v = \frac{e^{12 t}}{12}$

where we will get

$\int e^{12 t} \sin tdt = uv - \int vdu$

$= \frac{1}{2} e^{12 t} \sin t - \frac{1}{12} \int e^{12 t} \cos tdt$

Second time:

$u = \cos t dv = e^{12 t}$

$du = - \sin t v = \frac{e^{12 t}}{12}$

where we will get

$\int e^{12 t} \sin tdt = \frac{1}{12} e^{12 t} \sin t - \frac{1}{12} [\frac{1}{12} e^{12 t} \cos t + \frac{1}{12} \int e^{12 t} \sin tdt]$

$= \frac{1}{12} e^{12 t} \sin t - \frac{1}{144} e^{2 t} \cos t - \frac{1}{144} \int e^{12 t} \sin tdt$

Notice that the integral $\int e^{12 t}$ appears again on the right side of the equation, we

can solve for it,

$\int e^{12 t} \sin tdt + \frac{1}{144} \int e^{12 t} \sin tdt = \frac{1}{12} e^{12 t} sint - \frac{1}{144} e^{2 t} \cos t \frac{145}{144} \int e^{12 t} \sin tdt$

role="math" localid="1668103670737" $= \frac{1}{12} e^{12 t} \sin t - \frac{1}{144} e^{12 t} \cos t \int e^{12} \sin tdt$

$\frac{12}{145} e^{12 t} \sin t - \frac{1}{145} e^{12 t} \cos t$

Determine the general solution

To obtaining the particular solution,

$X_{P} = (\begin{matrix} e^{- 2 t} & 3 e^{- 12 t} \\ 3 e^{- 2 t} & - e^{- 12 t} \end{matrix}) \int (\begin{matrix} 10 e^{2 t} \sin t \\ 30 e^{12 t} \sin t \end{matrix}) dt$

$= (\begin{matrix} e^{- 2 t} & 3 e^{- 12 t} \\ 3 e^{- 2 t} & - e^{- 12 t} \end{matrix}) (\begin{matrix} 10 [\begin{matrix} \frac{2}{5} e^{2 t} \end{matrix} \sin t - \frac{1}{5} e^{2 t} \cos t] \\ 30 [\begin{matrix} \frac{12}{145} e^{2 t} \end{matrix} \sin t - \frac{1}{145} e^{12 t} \cos t] \end{matrix})$

$= (\begin{matrix} e^{- 2 t} & 3 e^{- 12 t} \\ 3 e^{- 2 t} & - e^{- 12 t} \end{matrix}) (\begin{matrix} 2 e^{2 t} [\begin{matrix} 2 \end{matrix} \sin t - \cos t] \\ \frac{6}{29} e^{12 t} [\begin{matrix} 12 \end{matrix} \sin t - \cos t] \end{matrix})$

role="math" localid="1668104715389" $= (\begin{matrix} 4 \sin t - 2 \cos t + \frac{216}{29} \sin t - \frac{18}{29} \cos t \\ 12 \sin t - 6 \cos t - \frac{72}{29} \sin t + \frac{6}{29} \cos t \end{matrix})$

$= (\begin{matrix} \frac{332}{29} \sin t - \frac{76}{29} \cos t \\ \frac{276}{29} \sin t - \frac{168}{29} \cos t \end{matrix})$

Finally, the general solution of the system is

$X = X_{c} + X_{p}$

$= c_{1} (\begin{matrix} 1 \\ 3 \end{matrix}) e^{- 2 t} + c_{2} (\begin{matrix} 3 \\ - 1 \end{matrix}) e^{- 12 t} + (\begin{matrix} \frac{332}{29} \sin t - \frac{76}{29} \cos t \\ \frac{276}{29} \sin t - \frac{168}{29} \cos t \end{matrix})$

Therefore, the general solution is $X = c_{1} (\begin{matrix} 1 \\ 3 \end{matrix}) e^{- 2 t} + c_{2} (\begin{matrix} 3 \\ - 1 \end{matrix}) e^{- 12 t} + (\begin{matrix} \frac{332}{29} \sin t - \frac{76}{29} \cos t \\ \frac{276}{29} \sin t - \frac{168}{29} \cos t \end{matrix})$

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Short Answer

Step by step solution

Definition of Eigenvalue.

Substitute the known values

Determine the eigenvalues of the coefficient matrix

Step 4:Determine the particular solution.

Evaluate the integral

Evaluate the integral

Determine the general solution

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