Science:Math Exam Resources/Courses/MATH215/April 2014/Question 01 (e)

MATH215 April 2014

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Question 01 (e)
Find the general solution of the differential equation system: $\displaystyle {\frac {d{\vec {y}}(t)}{dt}}={\begin{pmatrix}7&3\\5&-7\end{pmatrix}}{\vec {y}}(t)$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How do the eigenvalues and corresponding eigenvectors of the matrix ${\begin{pmatrix}7&3\\5&-7\end{pmatrix}}$ help to solve the problem?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To find the general solution for this system, we need to determine the eigenvalues and corresponding eigenvectors for $A={\begin{pmatrix}7&3\\5&-7\end{pmatrix}}.$ We find the eigenvalues by solving the characteristic equation, $\det(A-\lambda I)=0$. This gives the equation $(7-\lambda )(-7-\lambda )-(5)(3)=\lambda ^{2}-64=0$, and so $\lambda _{1}=8$ and $\lambda _{2}=-8$ are the eigenvalues for $A$. To find the eigenvector corresponding to $\lambda _{1}=8$, we solve $(A-8I){\vec {v}}_{1}=0$ for ${\vec {v}}_{1}$, i.e. we solve ${\begin{pmatrix}-1&3\\5&-15\end{pmatrix}}{\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}$ which implies ${\vec {v_{1}}}={\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}={\begin{pmatrix}3\\1\end{pmatrix}}.$ Analogously for the second eigenvector ${\vec {v}}_{2}$ we solve $(A+8I){\vec {v}}_{2}=0$, i.e. we solve ${\begin{pmatrix}15&3\\5&1\end{pmatrix}}{\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}$ which yields ${\vec {v_{2}}}={\begin{pmatrix}1\\-5\end{pmatrix}}$. The solutions ${\vec {y}}_{1}(t)=e^{8t}{\vec {v}}_{1}$ and ${\vec {y}}_{2}(t)=e^{-8t}{\vec {v}}_{2}$ form a fundamental set of solutions for the system, and therefore the general solution is an arbitrary linear combination of ${\vec {y}}_{1}$ and ${\vec {y}}_{2}:$ ${\vec {y}}(t)=c_{1}e^{8t}{\begin{pmatrix}3\\1\end{pmatrix}}+c_{2}e^{-8t}{\begin{pmatrix}1\\-5\end{pmatrix}}$ where $c_{1},c_{2}$ are arbitrary real constants.

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