Science:Math Exam Resources/Courses/MATH215/December 2011/Question 06 (c)

MATH215 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q7 (d)  •  

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  •  April 2013  •  December 2011  •  December 2013  •  

Question 06 (c)
Let $A=\left({\begin{array}{cc}3&1\\-4&-1\end{array}}\right)$ (c) Find the solution to the initial value problem $\mathbf {x} '=\left({\begin{array}{cc}3&1\\-4&-1\end{array}}\right)\mathbf {x} +\left({\begin{array}{c}-2e^{2t}\\1\end{array}}\right)$ and $\mathbf {x} (0)={\begin{pmatrix}1\\-5\end{pmatrix}}$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
From part (b), the homogeneous solution was found. Can you find a particular solution?

Hint 2
How would you find the particular solution if this were a scalar first order differential equation? How can you modify that method for a vector solution?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. As there are constants and exponentials in this first-order linear ODE, we will use undetermined coefficients, postulating that the particular solution is $x_{p}(t)=\left({\begin{array}{c}B_{1}\\B_{2}\end{array}}\right)e^{2t}+\left({\begin{array}{c}C_{1}\\C_{2}\end{array}}\right).$ so that $x_{p}'(t)=\left({\begin{array}{c}2B_{1}\\2B_{2}\end{array}}\right)e^{2t}.$ Substituting our solution into the system we get, ${\begin{aligned}&Ax+{\begin{pmatrix}-2e^{2t}\\1\end{pmatrix}}\\&={\begin{pmatrix}3&1\\-4&-1\end{pmatrix}}{\begin{pmatrix}B_{1}e^{2t}+C_{1}\\B_{2}e^{2t}+C_{2}\end{pmatrix}}+{\begin{pmatrix}-2e^{2t}\\1\end{pmatrix}}\\&={\begin{pmatrix}3B_{1}+B_{2}-2\\-4B_{1}-B_{2}\end{pmatrix}}e^{2t}+{\begin{pmatrix}3C_{1}+C_{2}\\-4C_{1}-C_{2}+1\end{pmatrix}}.\end{aligned}}$ This must equal $\displaystyle {}x_{p}'$ and so we have ${\begin{aligned}{\begin{pmatrix}3B_{1}+B_{2}-2\\-4B_{1}-B_{2}\end{pmatrix}}e^{2t}+{\begin{pmatrix}3C_{1}+C_{2}\\-4C_{1}-C_{2}+1\end{pmatrix}}&={\begin{pmatrix}2B_{1}\\2B_{2}\end{pmatrix}}e^{2t}\\{\begin{pmatrix}B_{1}+B_{2}-2\\-4B_{1}-3B_{2}\end{pmatrix}}e^{2t}+{\begin{pmatrix}3C_{1}+C_{2}\\-4C_{1}-C_{2}+1\end{pmatrix}}&=\mathbf {0} \end{aligned}}$ This gives us the system of equations ${\begin{aligned}B_{1}+B_{2}-2&=0\\-4B_{1}-3B_{2}&=0\end{aligned}}$ (based on the $e^{2t}$ terms) and ${\begin{aligned}3C_{1}+C_{2}&=0\\-4C_{1}-C_{2}&=-1\end{aligned}}$ (based on comparing the constant vectors). Looking at the first equation of the first system of equations, we have $B_{1}=2-B_{2}$ so that in the second equation, ${\begin{aligned}-4(2-B_{2})-3B_{2}&=0\\B_{2}&=8.\end{aligned}}$ Using the first equation again, we get $\displaystyle {}B_{1}=-6.$ Looking at the first equation of the second system of equations, we have $C_{2}=-3C_{1}$. From the second equation we get ${\begin{aligned}-4C_{1}-(-3C_{1})&=-1\\C_{1}&=1.\end{aligned}}$ and hence $\displaystyle {}C_{2}=-3$ after substituting back into the first equation. The general solution to the ODE system will be $u_{p}+u_{h}$ where $u_{h}$ is our homogeneous solution from part (b). Thus, $x(t)={\begin{pmatrix}1\\-3\end{pmatrix}}+{\begin{pmatrix}-6\\8\end{pmatrix}}e^{2t}+{\begin{pmatrix}cte^{t}+be^{t}\\-2cte^{t}+(c-2b)e^{t}\end{pmatrix}}.$ If $\displaystyle {}x(0)=(1,-5)^{T}$ then ${\begin{pmatrix}1\\-5\end{pmatrix}}={\begin{pmatrix}1-6+b\\-3+8+(c-2b))\end{pmatrix}}.$ The first component tells us that $b=6$ while using this value of b in the second component gives $c=2$. Therefore, the solution is $x(t)=\left({\begin{array}{c}1\\-3\end{array}}\right)+\left({\begin{array}{c}-6\\8\end{array}}\right)e^{2t}+\left({\begin{array}{c}2te^{t}+6e^{t}\\-4te^{t}-10e^{t}\end{array}}\right).$