Question 7.3.3: Find a general solution of the system dx1/dt = 4x1 - 3x2, dx...

Find a general solution of the system

\frac{dx_{1}}{dt} = 4x_{1}  – 3x_{2},

\frac{dx_{2}}{dt} = 3x_{1} + 4x_{2}.                      (21)

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The coefficient matrix

A = \begin{bmatrix} 4 & -3 \\ 3 & 4 \end{bmatrix}

has the characteristic equation

|A – λI| = \begin{vmatrix} 4-\lambda & -3 \\ 3 & 4-\lambda \end{vmatrix} = (4 – λ)² + 9 =0

and hence has the complex conjugate eigenvalues λ = 4 – 3i and \bar{\lambda } = 4 + 3i . Substituting λ = 4 – 3i in the eigenvector equation (A – λI)v = 0, we get the equation

\left [ \begin{matrix} \textbf{A} – (4 – 3i) · \textbf{I} \end{matrix} \right ] \textbf{v} = \begin{bmatrix} 3i & -3 \\ 3 & 3i \end{bmatrix} \left [ \begin{matrix} a \\ b \end{matrix} \right ] = \left [ \begin{matrix} 0 \\ 0 \end{matrix} \right ]

for an associated eigenvalue v = [a   b ]^{T} . Division of each row by 3 yields the two scalar equations

ia – b = 0,
a + ib = 0,

each of which is satisfied by a = 1 and b = i . Thus, v = [1    i]^{T} is a complex eigenvector associated with the complex eigenvalue λ = 4 – 3i . The corresponding complex-valued solution x(t) = ve^{λ_{t}}  of  x^{′ } = Ax is then

x(t) = \left [ \begin{matrix} 1 \\ i \end{matrix} \right ] e^{(4-3i)t} =\left [ \begin{matrix} 1 \\ i \end{matrix} \right ] e^{4t} (\cos 3t  –  i  \sin 3t) = e^{4t} \left [ \begin{matrix} \cos 3t  –  i \sin 3t \\ i  \cos 3t  +  \sin 3t \end{matrix} \right ] .

The real and imaginary parts of x(t) are the real-valued solutions

x_{1}(t) = e^{4t} \left [ \begin{matrix}\cos 3t \\ \sin 3t \end{matrix} \right ]  and x_{2}(t) = e^{4t} \left [ \begin{matrix} -\sin 3t \\ \cos 3t \end{matrix} \right ]

A real-valued general solution of x^{′} = Ax is then given by

\textbf{x}(t) = c_{1}  \textbf{x}_{1}(t) + c_{2}  \textbf{x}_{2}(t) = e^{4t} \left [ \begin{matrix} c_{1} \cos 3t  –  c_{2}  \sin 3t \\ c_{1} \sin 3t  + c_{2}  \cos 3t \end{matrix} \right ] .

Finally, a general solution of the system in (21) in scalar form is

x_{1}(t) = e^{4t} (c_{1}  \cos 3t  –  c_{2}  \sin 3t ),

x_{2}(t) = e^{4t} (c_{1}  \sin 3t  +  c_{2}  \cos 3t ).

Figure 7.3.4 shows some typical solution curves of the system in (21). Each appears to spiral counterclockwise as it emanates from the origin in the x_{1}x_{2}-plane. Actually, because of the factor e^{4t} in the general solution, we see that

  • Along each solution curve, the point (x_{1}(t), x_{2}(t)) approaches the origin as t → – \infty,whereas
  • The absolute values of x_{1}(t)  and  x_{2}(t) both increase without bound as t → + \infty.
3.4

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