Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
It looks like the system is
[tex]x' = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} x + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}[/tex]
Compute the eigenvalues of the coefficient matrix.
[tex]\begin{vmatrix} -1 - \lambda & 5 \\ -1 & 1 - \lambda \end{vmatrix} = \lambda^2 + 4 = 0 \implies \lambda = \pm2i[/tex]
For [tex]\lambda = 2i[/tex], the corresponding eigenvector is [tex]\eta=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^\top[/tex] such that
[tex]\begin{bmatrix} -1 - 2i & 5 \\ -1 & 1 - 2i \end{bmatrix} \begin{bmatrix} \eta_1 \\ \eta_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}[/tex]
Notice that the first row is 1 + 2i times the second row, so
[tex](1+2i) \eta_1 - 5\eta_2 = 0[/tex]
Let [tex]\eta_1 = 1-2i[/tex]; then [tex]\eta_2=1[/tex], so that
[tex]\begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} = 2i \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix}[/tex]
The eigenvector corresponding to [tex]\lambda=-2i[/tex] is the complex conjugate of [tex]\eta[/tex].
So, the characteristic solution to the homogeneous system is
[tex]x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix}[/tex]
The characteristic solution contains [tex]\cos(2t)[/tex] and [tex]\sin(2t)[/tex], both of which are linearly independent to [tex]\cos(t)[/tex] and [tex]\sin(t)[/tex]. So for the nonhomogeneous part, we consider the ansatz particular solution
[tex]x = \cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}[/tex]
Differentiating this and substituting into the ODE system gives
[tex]-\sin(t) \begin{bmatrix} a \\ b \end{bmatrix} + \cos(t) \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \left(\cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}\right) + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}[/tex]
[tex]\implies \begin{cases}a - 5c + d = 1 \\ b - c + d = 0 \\ 5a - b + c = 0 \\ a - b + d = -2 \end{cases} \implies a=\dfrac{11}{41}, b=\dfrac{38}{41}, c=-\dfrac{17}{41}, d=-\dfrac{55}{41}[/tex]
Then the general solution to the system is
[tex]x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix} + \dfrac1{41} \cos(t) \begin{bmatrix} 11 \\ 38 \end{bmatrix} - \dfrac1{41} \sin(t) \begin{bmatrix} 17 \\ 55 \end{bmatrix}[/tex]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
