Science:Math Exam Resources/Courses/MATH215/December 2013/Question 05 (b)

MATH215 December 2013

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

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  •  April 2013  •  December 2011  •  December 2013  •  

Question 05 (b)

Now find the general solution of the nonhomogenous system (with matrix $A$ as above): ${\vec {x}}'=A{\vec {x}}+\left[{\begin{array}{c}25e^{3t}\\0\end{array}}\right]$ Hint: to find a particular solution, you may use either the method of variation of parameters, or the method of undetermined coefficients; for the latter, try ${\vec {x}}_{p}(t)=\left[{\begin{array}{c}(at+b)e^{3t}\\(ct+d)e^{3t}\end{array}}\right]$

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
If you use undetermined coefficients like in the hint, substitute ${\vec {x_{p}}}$ into the equation, being careful with all your terms. Be sure to include the nonhomogeneous term in your resulting equation. You need to solve for a, b, c, and d.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We need a particular solution. To obtain it, we will use the method of undetermined coefficients. If ${\vec {x}}_{p}={\begin{bmatrix}ate^{3t}+be^{3t}\\cte^{3t}+de^{3t}\end{bmatrix}}$ then ${\vec {x_{p}}}'={\begin{bmatrix}(a+3b)e^{3t}+3ate^{3t}\\(c+3d)e^{3t}+3cte^{3t}\end{bmatrix}}$ and $Ax_{p}+{\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}={\begin{bmatrix}3cte^{3t}+(3d+25)e^{3t}\\(2a+c)te^{3t}+(2b+d)e^{3t}\end{bmatrix}}$ Therefore if ${\vec {x_{p}}}'=A{\vec {x_{p}}}+{\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}$ we get ${\begin{bmatrix}(a+3b)e^{3t}+3ate^{3t}\\(c+3d)e^{3t}+3cte^{3t}\end{bmatrix}}={\begin{bmatrix}3cte^{3t}+(3d+25)e^{3t}\\(2a+c)te^{3t}+(2b+d)e^{3t}\end{bmatrix}}$. In comparing the coefficients of $e^{3t}$ from row 1, the coefficients of $te^{3t}$ from row 1, the coefficients of $e^{3t}$ from row 2, and the coefficients of $te^{3t}$ from row 2, the following equations must hold: ${\begin{aligned}a+3b&=3d+25\\3a&=3c\\c+3d&=2b+d\\3c&=2a+c\end{aligned}}$ We can turn this into a matrix equation and row-reduce: ${\begin{bmatrix}1&3&0&-3\\1&0&-1&0\\0&2&-1&-2\\1&0&-1&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}25\\0\\0\\0\end{bmatrix}}$ Subtracting row 2 from row 4 and row 1 from row 2 yields: ${\begin{bmatrix}1&1&1&-1\\0&-3&-1&3\\0&2&-1&-2\\0&0&0&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}25\\-25\\0\\0\end{bmatrix}}$ We next multiply the third row by 3 and then add the second row twice: ${\begin{bmatrix}1&1&1&-1\\0&-3&-1&3\\0&0&-5&0\\0&0&0&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}25\\-25\\-50\\0\end{bmatrix}}$ Next, divide the third row by -5 and then add the result to the second row, and subtract it from the first row: ${\begin{bmatrix}1&1&0&-1\\0&-3&0&3\\0&0&1&0\\0&0&0&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}15\\-15\\10\\0\end{bmatrix}}$ In our last step we divide the second row by -3 and then subtract the result from the first row: ${\begin{bmatrix}1&0&0&0\\0&1&0&-1\\0&0&1&0\\0&0&0&0\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}={\begin{bmatrix}10\\5\\10\\0\end{bmatrix}}$ We see that d is a free parameter, and since we only need one particular we choose d=0 for simplicity. Then we read off a = 10, b - d = 5, c = 10. Putting this together we find a particular solution ${\vec {x_{p}}}={\begin{bmatrix}10te^{3t}+5e^{3t}\\10te^{3t}\end{bmatrix}}$. The general solution is the homogeneous solution (which we found in part (a)) plus a particular solution: $x(t)=C_{1}{\begin{bmatrix}1\\1\end{bmatrix}}e^{3t}+C_{2}{\begin{bmatrix}3\\-2\end{bmatrix}}e^{-2t}+{\begin{bmatrix}10te^{3t}+5e^{3t}\\10te^{3t}\end{bmatrix}}$.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This problem can also be done using variation of parameters. Given the homogeneous solution $C_{1}{\begin{bmatrix}1\\1\end{bmatrix}}e^{3t}+C_{2}{\begin{bmatrix}3\\-2\end{bmatrix}}e^{-2t}$, we form a fundamental matrix ${\mathcal {X}}={\begin{bmatrix}e^{3t}&3e^{-2t}\\e^{3t}&-2e^{-2t}\end{bmatrix}}$ using independent solutions of the homogeneous problem as columns. Observe that ${\mathcal {X}}'(t)=A{\mathcal {X}}(t)$. For the nonhomogeneous equation, we will try to find a solution of the form ${\vec {x}}(t)={\mathcal {X}}(t){\vec {g}}(t)$ where ${\vec {x}}'(t)=A{\vec {x}}+{\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}.$ With this solution form, the equation reads: ${\vec {x}}'(t)={\mathcal {X}}'(t){\vec {g}}(t)+{\mathcal {X}}(t){\vec {g}}'(t)=A{\vec {x}}(t)+{\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}$ ${\vec {x}}'(t)=A\underbrace {{\mathcal {X}}{\vec {g}}(t)} _{={\vec {x}}(t)}+{\mathcal {X}}(t){\vec {g}}'(t)=A{\vec {x}}(t)+{\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}$ so that ${\mathcal {X}}{\vec {g}}'={\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}$. We can now work on solving for ${\vec {g}}$ (note the primed in ${\vec {g}}'$). ${\vec {g}}'={\mathcal {X}}^{-1}{\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}={\frac {1}{e^{3t}(-2e^{-2t})-e^{3t}(3e^{-2t})}}{\begin{bmatrix}2e^{-2t}&-3e^{-2t}\\-e^{3t}&e^{3t}\end{bmatrix}}={\frac {e^{-t}}{5}}{\begin{bmatrix}2e^{-2t}&3e^{-2t}\\e^{3t}&-e^{3t}\end{bmatrix}}{\begin{bmatrix}25e^{3t}\\0\end{bmatrix}}={\begin{bmatrix}10\\5e^{5t}\end{bmatrix}}$. Integrating gives ${\vec {g}}={\begin{bmatrix}10t+K_{1}\\e^{5t}+K_{2}\end{bmatrix}}$ for arbitrary $K_{1},K_{2}$. We finally have a general solution ${\vec {x}}(t)={\mathcal {X}}(t){\vec {g}}(t)={\begin{bmatrix}10te^{3t}+3e^{3t}+K_{1}e^{3t}+3K_{2}e^{-2t}\\10te^{3t}-2e^{3t}+K_{1}e^{3t}-2K_{2}e^{-2t}\end{bmatrix}}={\begin{bmatrix}10te^{3t}+3e^{3t}\\10te^{3t}-2e^{3t}\end{bmatrix}}+K_{1}{\begin{bmatrix}1\\1\end{bmatrix}}e^{3t}+K_{2}{\begin{bmatrix}3\\-2\end{bmatrix}}e^{-2t}$. This is in agreement with the first solution. Observe that based on the numbers in that solution, a=c=10, and b-d=5.