Partial Solutions to 18.03 Problem Set 6.5 Part II

27. If ẋ = y, then by the chain rule \frac{dy}{dt} = ẋ\frac{dy}{dx} = yy′. (We’ve seen this substitution before: see EP p. 167.) Hence ẍ + e^{x} = 0 can be rewritten as ẏ = −e^{x}, ẋ = y. Substituting ẏ = yy′, we get yy′ = −e^{x}. Separating variables, we get ydy = −e^{x}dx, so \frac{y^{2}}{2} = −e^{x} + C_{1}. Hence ẋ = y = \sqrt{2(C_{1} − e^{x})}. (Solving for x in terms of t is trickier. Separating variables:

\frac{dx}{\sqrt{2(C_{1} − e^{x})}} = dt.

This doesn’t have an especially nice solution.)

28. (a) If (a + bi)(x + yi) = (p + qi), then p = ax − by, q = bx + ay. We check that

A(a + bi) [ \begin{matrix}x \\ y \end{matrix} ] = [ \begin{matrix}a & −b \\ b & a \end{matrix} ] [ \begin{matrix}x \\ y \end{matrix} ] = [ \begin{matrix}ax − by \\ bx + ay \end{matrix} ] = [ \begin{matrix}p \\ q \end{matrix} ]

as desired. Also, if z = a + bi and w = c + di, so zw = (ac − bd) + (ad + bc)i, then

A(z)A(w) = [ \begin{matrix}a & −b \\ b & a \end{matrix} ] [ \begin{matrix}c & −d \\ d & c \end{matrix} ] = [ \begin{matrix}ac − bd & −ad − bc \\ ad + bc & ac − bd \end{matrix} ] = A(zw),

and

A(z) + A(w) = [ \begin{matrix}a & −b \\ b & a \end{matrix} ] + [ \begin{matrix}c & −d \\ d & c \end{matrix} ] = [ \begin{matrix}a + c & −(b + d) \\ b + d & a + c \end{matrix} ] = A(z + w).

Convince yourself that A(cos θ + i sin θ) = [ \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ] has the effect of rotating

counterclockwise by an angle θ: for example, A(cos θ + i sin θ) [ \begin{matrix}1 \\ 0 \end{matrix} ] = [ \begin{matrix}cos θ \\ sin θ \end{matrix} ] and

A(cos θ + i sin θ) [ \begin{matrix}0 \\ 1 \end{matrix} ] = [ \begin{matrix}− sin θ \\ cos θ \end{matrix} ] as one would expect. If z = a + bi, then the eigenvalues of A(z) are z = a + bi and \overline{z} = a − bi, and the corresponding eigenvectors are [ \begin{matrix}i \\ 1 \end{matrix} ] and [ \begin{matrix}−i \\ 1 \end{matrix} ] respectively.

Let P = [ \begin{matrix}i & −i \\ 1 & 1 \end{matrix} ] be the eigenvalue matrix, so

P^{−1} = \frac{1}{2i} [ \begin{matrix}1 & i \\ −1 & i \end{matrix} ] = \frac{1}{2} [ \begin{matrix}−i & 1 \\ i & 1 \end{matrix} ] .

Let D = [ \begin{matrix}a + bi & 0 \\ 0 & a − bi \end{matrix} ] be the eigenvector matrix. Then A(z) = P DP^{−1} (see EP p. 454), and

e^{A(z)} = P e^{D}P^{−1}

= \frac{1}{2} [ \begin{matrix}i & −i \\ 1 & 1 \end{matrix} ] [ \begin{matrix}e^{a+bi} & 0 \\ 0 & e^{a−bi} \end{matrix} ] [ \begin{matrix}−i & 1 \\ i & 1 \end{matrix} ]
= [ \frac{e^{a+bi}+e^{a−bi}}{2} −\frac{e^{a+bi}−e^{a−bi}}{2i}_{\frac{e^{a+bi}−e^{a−bi}}{2i}  \frac{e^{a+bi}+e^{a−bi}}{2}} ]
= [ \frac{e^{z}+\overline{e^{z}}}{2} −\frac{e^{z}−\overline{e^{z}}}{2i}_{\frac{e^{z}−\overline{e^{z}}}{2i}  \frac{e^{z}+\overline{e^{z}}}{2}} ]
= A(e^{z}).

(b) We find the general solution to the system u′ = u + v, v′ = v. v = C_{1}e^{t}, so u′ = u + C_{1}e^{t}, so u = C_{1}te^{t} + C_{2}e^{t}. The solution with initial condition x(0) = [ \begin{matrix}1 \\ 0 \end{matrix} ] has C_{1} = 0, C_{2} = 1, so x(t) = [ \begin{matrix}e^{t} \\ 0 \end{matrix} ]. The solution with initial condition x(0) = [ \begin{matrix}0 \\ 1 \end{matrix} ] has

C_{1} = 1, C_{2} = 0, so x(t) = [ \begin{matrix}te^{t} \\ e^{t} \end{matrix} ]. Thus

e^{At} = [ \begin{matrix}e^{t} & te^{t} \\ 0 & e^{t} \end{matrix} ] = e^{t} [ \begin{matrix}1 & t \\ 0 & 1 \end{matrix} ] .

29.