Science:Math Exam Resources/Courses/MATH215/December 2013/Question 04 (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 04 (b)
Let $Y(s)={\mathcal {L}}[y]$ be the Laplace transform of the solution $y(t)$ to the following initial value problem: ${\begin{cases}y''-2y'+y=g(t),\\y(0)=0,y'(0)=1\end{cases}}$ where $g(t)$ is the function defined in (a). Determine $Y(s)$.

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
You'll need to take the Laplace transform of the entire equation, and then solve for $Y(s)$. Pay attention to how derivatives transform. You'll need to use your result in (a).

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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. We take the Laplace transform of the equation: ${\mathcal {L}}(y''-2y'+y)={\mathcal {L}}(g(t))$ $\underbrace {s^{2}Y(s)-sy(0)-y'(0)} _{{\mathcal {L}}(y'')}-2\underbrace {(sY(s)-y(0))} _{{\mathcal {L}}(y')}+\underbrace {Y(s)} _{{\mathcal {L}}(y)}=\underbrace {\frac {e^{-s}}{s-1}} _{{\mathcal {L}}(g(t))}$ $s^{2}Y(s)-0-1-2(sY(s)-0)+Y(s)=(s^{2}-2s+1)Y(s)-1={\frac {e^{-s}}{s-1}}$ Since $\displaystyle s^{2}-2s+1=(s-1)^{2}$ we arrive at our final answer $Y(s)={\frac {e^{-s}}{(s-1)^{3}}}+{\frac {1}{(s-1)^{2}}}$

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