Respuesta :
In matrix form, the ODE is given by
[tex]\underbrace{\begin{bmatrix}x'\\y'\end{bmatrix}}_{\vec x'}=\underbrace{\begin{bmatrix}5&4\\1&8\end{bmatrix}}_A\underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{\vec x}+t\underbrace{\begin{bmatrix}1\\-1\end{bmatrix}}_{\vec f}[/tex]
a. Move [tex]A\vec x[/tex] to the left side and multiply both sides by the integrating factor, the matrix exponential of [tex]-A[/tex], [tex]e^{-At}[/tex]:
[tex]e^{-At}\vec x'-Ae^{-At}\vec x=te^{-At}\vec f[/tex]
Condense the left side as the derivative of a product:
[tex]\left(e^{-At}\vec x\right)=te^{-At}\vec f[/tex]
Integrate both sides and multipy by [tex]e^{At}[/tex] to solve for [tex]\vec x[/tex]:
[tex]e^{-At}\vec x=\displaystyle\left(\int te^{-At}\,\mathrm dt\right)\vec f\implies\vec x=\displaystyle e^{At}\left(\int te^{-At}\,\mathrm dt\right)\vec f[/tex]
Finding [tex]e^{\pm At}[/tex] requires that we diagonalize [tex]A[/tex].
[tex]A[/tex] has eigenvalues 4 and 9, with corresponding eigenvectors [tex]\begin{bmatrix}-4&1\end{bmatrix}^\top[/tex] and [tex]\begin{bmatrix}1&1\end{bmatrix}^\top[/tex] (explanation for this in part (b)), so we have
[tex]A=\begin{bmatrix}-4&1\\1&1\end{bmatrix}\begin{bmatrix}4&0\\0&9\end{bmatrix}\begin{bmatrix}-4&1\\1&1\end{bmatrix}^{-1}[/tex]
[tex]\implies A^n=\begin{bmatrix}-4&1\\1&1\end{bmatrix}\begin{bmatrix}4^n&0\\0&9^n\end{bmatrix}\begin{bmatrix}-4&1\\1&1\end{bmatrix}^{-1}[/tex]
[tex]\implies A^n=\dfrac15\begin{bmatrix}4^{n+1}+9^n&4\cdot9^n-4^{n+1}\\9^n-4^n&4^n+4\cdot9^n\end{bmatrix}[/tex]
[tex]\implies e^{\pm At}=\dfrac15\begin{bmatrix}4e^{\pm4t}+e^{\pm9t}&4e^{\pm9t}-4e^{\pm4t}\\e^{\pm9t}-e^{\pm4t}&e^{\pm4t}+4e^{\pm9t}\end{bmatrix}[/tex]
[tex]\implies\vec x=\dfrac15e^{At}\begin{bmatrix}C_1\\C_2\end{bmatrix}-\dfrac1{216}\begin{bmatrix}72t+20\\-36t-7\end{bmatrix}[/tex]
b. Find the eigenvalues of [tex]A[/tex]:
[tex]\det(A-\lambda I_2)=\begin{vmatrix}5-\lambda&4\\1&8-\lambda\end{vmatrix}=\lambda^2-13\lambda+36=0[/tex]
[tex]\implies(\lambda-4)(\lambda-9)=0\implies\lambda_1=4,\lambda_2=9[/tex]
Let [tex]\vec\eta=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^\top[/tex] and [tex]\vec\theta=\begin{bmatrix}\theta_1&\theta_2\end{bmatrix}^\top[/tex] be the corresponding eigenvectors.
For [tex]\lambda_1=4[/tex], we have
[tex]\begin{bmatrix}1&4\\1&4\end{bmatrix}\begin{bmatrix}\eta_1\\\eta_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}[/tex]
which means we can pick [tex]\eta_1=-4[/tex] and [tex]\eta_2=1[/tex].
For [tex]\lambda_2=9[/tex], we have
[tex]\begin{bmatrix}-4&4\\1&-1\end{bmatrix}\begin{bmatrix}\theta_1\\\theta_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}[/tex]
so we pick [tex]\theta_1=\theta_2=1[/tex].
Then the characteristic solution to the system is
[tex]\vec x_c=C_1e^{\lambda_1t}\vec\eta+C_2e^{\lambda_2t}\vec\theta[/tex]
[tex]\vec x_c=C_1e^{4t}\begin{bmatrix}-4\\1\end{bmatrix}+C_2e^{9t}\begin{bmatrix}1\\1\end{bmatrix}[/tex]
c. Now we find the particular solution with undetermined coefficients.
The nonhomogeneous part of the ODE is a linear function, so we can start with assuming a particular solution of the form
[tex]\vec x_p=\vec at+\vec b\implies\vec x_p'=\vec a[/tex]
Substituting these into the system gives
[tex]\begin{bmatrix}a_1\\a_2\end{bmatrix}=\begin{bmatrix}5&4\\1&8\end{bmatrix}\left(\begin{bmatrix}a_1\\a_2\end{bmatrix}t+\begin{bmatrix}b_1\\b_2\end{bmatrix}\right)+\begin{bmatrix}1\\-1\end{bmatrix}t[/tex]
[tex]\begin{bmatrix}a_1\\a_2\end{bmatrix}=\begin{bmatrix}5&4\\1&8\end{bmatrix}\begin{bmatrix}a_1t+b_1\\a_2t+b_2\end{bmatrix}+\begin{bmatrix}t\\-t\end{bmatrix}[/tex]
[tex]\begin{bmatrix}a_1\\a_2\end{bmatrix}=\begin{bmatrix}(5a_1+4a_2+1)t+(5b_1+4b_2)\\(a_1+8a_2-1)t+(b_1+8b_2)\end{bmatrix}[/tex]
[tex]\implies\begin{cases}5a_1+4a_2=-1\\5b_1+4b_2=a_1\\a_1+8a_2=1\\b_1+8b_2=a_2\end{cases}\implies a_1=-\dfrac13,a_2=\dfrac16,b_1=-\dfrac5{54},b_2=\dfrac7{216}[/tex]
Put everything together to get a solution
[tex]\vec x=\vec x_c+\vec x_p[/tex]
that should match the solution in part (a).
