Question 3.8.1: Determine all solutions I (t ) for an RLC circuit when L =20...

From (3.8.23) and the given information, we can immediately determine the second-order differential equation that I (t ) satisfies. In particular, since E(t ) = 50sin2t , we have E´(t ) = 100cos2t , and using the values for L, C, and R, I (t ) is a solution to the equation

LI″(t )+RI´(t )+ \frac {1}{C}I (t ) = E´(t )     (3.8.23)

20I″+80I´+100I = 100cos2t          (3.8.25)

Using the substitution x_{1} = I and x_{2}= I´ and multiplying both sides of (3.8.25) by 1/20, the system becomes

\acute{x_{1}}= x_{2}
\acute{x_{2}}=−5x_{1} −4x_{2} +5cos2t

From this, we can write the system in matrix form as

x´=\begin{bmatrix} 0 & 1 \\ -5 & -4 \end{bmatrix} x +\begin{bmatrix} 0 \\5cos2t \end{bmatrix}            (3.8.26)

For the coefficient matrix A in (3.8.26), we compute the eigenvalues and eigenvectors in order to find the complementary solution x_{h} of the system.

Doing so, we find that A has complex eigenvalues and eigenvectors; one eigenvalue-eigenvector pair is

λ=−2+i, v =\begin{bmatrix} -2-i \\5 \end{bmatrix}

Writing

z(t ) = e^{(−2+i)t} \begin{bmatrix} -2-i \\5 \end{bmatrix}

we know from theorem 3.5.2 that the real and imaginary parts of the vector function z(t ) will form two real linearly independent solutions to the homogeneous system x´= Ax. Rewriting z using Euler’s formula,

z(t ) = e^{−2t} (cos t +i sin t ) (\begin{bmatrix} -2\\5 \end{bmatrix}+i\begin{bmatrix} -1\\0\end{bmatrix})

= e^{−2t} \begin{bmatrix} −2cos t +sin t\\5cos t \end{bmatrix}+ ie^{−2t} \begin{bmatrix} −cos t -2sin t\\5sin t \end{bmatrix}

The real and imaginary parts of z are real linearly independent solutions to x´=Ax, so we have determined that the complementary solution to the original system is

x_{h} = c_{1}e^{−2t}\begin{bmatrix} −2cos t +sin t\\5cos t \end{bmatrix} +c_{2}e^{−2t} \begin{bmatrix} −cos t -2sin t\\5sin t \end{bmatrix}

In theory, we are now ready to apply variation of parameters to find a particular solution x_{p}. While we could do so here, the computations get remarkably cumbersome. In the next chapter on higher order differential equations, we will learn that for certain higher order equations, making a good guess at the form of a particular solution provides the simplest approach. In fact, we will even see that keeping certain second-order equations in that form, rather than converting them to systems of first-order equations, often is the best way to proceed.

For now, we will guess a form for x_{p}. Since

b(t ) = \begin{bmatrix} 0\\5cos t \end{bmatrix}

we assume that a particular solution x_{p} has form

x_{p}= \begin{bmatrix} a cos2t +b sin2t\\c cos2t +d sin2t\end{bmatrix}

From this, it follows

\acute{x_{p}}= \begin{bmatrix} -2a cos2t +2b sin2t\\-2c cos2t +2d sin2t\end{bmatrix}

Substituting x_{p} and \acute{x_{p}} for x and x´ in (3.8.26), we have

\begin{bmatrix} -2a cos2t +2b sin2t\\-2c cos2t +2d sin2t\end{bmatrix}=\begin{bmatrix} c cos2t +d sin2t\\-2c cos2t +2d sin2t\end{bmatrix}+\begin{bmatrix} 0\\5cos 2t \end{bmatrix}

Equating the coefficients of sin2t and cos2t in the entries of the vectors in this most recent vector equation, the following system of four linear equations in a,b, c, and d arises:

−2a = d
2b = c
−2c =−5b −4d
2d =−5a −4c +5

Rearranging this system to write it in matrix form and row-reducing, we observe

\begin{bmatrix} -2 & 0&0&-1&0 \\ 0&2&-1&0&0 \\ 0&5&-2&4&0 \\ -5&0&4&2&5 \end{bmatrix}→\begin{bmatrix} 1 & 0&0&0&1/13 \\ 0&1&0&0&8/13 \\ 0&0&1&0&16/13 \\ 0&0&0&1&-2/13 \end{bmatrix}

Thus we conclude that a particular solution is

x_{p} =\begin{bmatrix} 1/13cos2t +8/13sin2t\\16/13cos2t −2/13sin2t \end{bmatrix}

In conjunction with our earlier work to find xh, we have determined that the general solution to the system of first-order differential equations given by (3.8.25) is

x =c_{1} e^{-2 t}\left[\begin{array}{c}-2 \cos t+\sin t \\5 \cos t\end{array}\right]+c_{2} e^{-2 t}\left[\begin{array}{r}-\cos t-2 \sin t \\5 \sin t\end{array}\right]

+\left[\begin{array}{r}1 / 13 \cos 2 t+8 / 13 \sin 2 t \\16 / 13 \cos 2 t-2 / 13 \sin 2 t\end{array}\right]

Recalling that x_{1} = I is the current in the given RLC circuit, we have shown that