Science:Math Exam Resources/Courses/MATH215/December 2013/Question 04 (c)

MATH215 December 2013

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Question 04 (c)

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): $g(t)={\begin{cases}0,\quad t<1\\e^{t-1},\quad t\geq 1\end{cases}}$ (c) Calculate the inverse Laplace transform of Y(s) obtained in (b) to find the solution, y(t), to the IVP in (b).

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 can start by trying to find the inverse Laplace transform of $1/(s-1)^{2}$ and $1/(s-1)^{3}$. These inverse transforms can be derived by looking at ${\mathcal {L}}(e^{at})$ and ${\mathcal {L}}(t^{n}e^{at})$.

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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 have $Y(s)={\frac {e^{-s}}{(s-1)^{3}}}+{\frac {1}{(s-1)^{2}}}$ and we need to find $y(t)={\mathcal {L}}^{-1}(Y(s))$. The most important thing here is finding a function $f(t)$ such that ${\mathcal {L}}(f(t))=1/(s-a)^{n}$. The key identity to realize is that ${\mathcal {L}}(t^{n}f(t))=(-1)^{n}{\frac {d^{n}}{ds^{n}}}F(s)$. With this knowledge, since ${\mathcal {L}}(e^{t})=1/(s-1)$, we have ${\mathcal {L}}(te^{t})=(-1)(-1/(s-1)^{2})=1/(s-1)^{2}$ and ${\mathcal {L}}(t^{2}e^{t})=2/(s-1)^{3}$ (note the factor of 2!). Then $Y(s)=e^{-s}{\mathcal {L}}({\frac {1}{2}}t^{2}e^{t})+{\mathcal {L}}(te^{t})$. We use one more identity that ${\mathcal {L}}(u(t-a)f(t-a))=e^{-as}F(s)$ which allows us to invert the first term (using a=1) on the right-hand side. We find $y(t)=u(t-1){\frac {1}{2}}(t-1)^{2}e^{t-1}+te^{t}$.

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