Science:Math Exam Resources/Courses/MATH215/December 2011/Question 05 (b)

MATH215 December 2011

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Question 05 (b)
Given the function $f(t)={\begin{cases}0,&t<1\\2\pi ,&1\leq t<2\\\pi ,&t\geq 2\end{cases}}$ (b) Solve the initial value problem $y''-5y'+6y=f(t),\quad y(0)=1,\quad y'(0)=0.$

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
What technique did we use in part (a)? How can we exploit what we did there here?

Hint 2
When computing inverse Laplace transforms, try to put things into a form similar to what appears in the chart at the end of the exam.

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. Given that in part (a) of this problem we were asked to compute the Laplace transform of f(t), it stands to reason that using Laplace transforms will be a good way to solve this problem. We begin by taking the Laplace transform of the entire differential equation. We can do this because the Laplace Transform is a linear operator, i.e. ${\mathcal {L}}(af(t)+bg(t))=a{\mathcal {L}}(f(t))+b{\mathcal {L}}(g(t)).$ Taking the Laplace transform, we get ${\mathcal {L}}(y'')-5{\mathcal {L}}(y')+6{\mathcal {L}}(y)={\mathcal {L}}(f(t)).$ Define the following ${\begin{aligned}F(s)&={\mathcal {L}}(f(t))\\Y(s)&={\mathcal {L}}(y(t)).\end{aligned}}$ From part (a), we have that $F(s)={\frac {\pi }{s}}\left(2\exp(-s)-\exp(-2s)\right).$ If we look at the table at the back of the exam we can retrieve the Laplace transforms for derivatives (of course we could also compute these if we needed to). We see that the general rule for the Laplace transform of the nth derivative of a function, g(t) is ${\mathcal {L}}(g^{(n)}(t))=s^{n}G(s)-s^{n-1}g(0)-\dots -g^{(n-1)}(0)$ and therefore we can write ${\begin{aligned}{\mathcal {L}}(y')&=sY(s)-y(0)=sY-1\\{\mathcal {L}}(y'')&=s^{2}Y(s)-sy(0)-y'(0)=s^{2}Y-s\end{aligned}}$ where we have used that $\displaystyle {}y(0)=1$ and $\displaystyle {}y'(0)=0$ as given in the question. Substituting these expressions into the original differential equation we get, $\displaystyle {}s^{2}Y(s)-s-5(sY(s)-1)+6Y=F(s).$ Simplifying this expression we get $\displaystyle {}(s^{2}-5s+6)Y(s)=F(s)+s-5$ which we can solve for Y(s)to get $Y(s)={\frac {F(s)-4s}{(s-3)(s-2)}}={\frac {F(s)}{(s-3)(s-2)}}+{\frac {s-5}{(s-3)(s-2)}}$ where we have factored the quadratic. While it may look like we have solved the problem, we need to present the solution in terms of y(t), the original function. To do this we need to compute the inverse Laplace transform of Y(s). While we could attempt to do this manually using the formula for inverse Laplace transforms, we should attempt to put it in a form where we can use the chart at the end of the exam. Once again, the inverse Laplace transform is a linear operator so we have $y(t)={\mathcal {L}}^{-1}(Y(s))={\mathcal {L}}^{-1}\left({\frac {F(s)}{(s-3)(s-2)}}\right)+{\mathcal {L}}^{-1}\left({\frac {s-5}{(s-3)(s-2)}}\right)$ and we can compute these two quantities separately. We will start with the second inverse Laplace transform. Looking at our chart at the end of the exam we see an inverse Laplace transform for functions of the form ${\mathcal {L}}^{-1}\left({\frac {1}{s-a}}\right)=\exp(at).$ However, our problem has a product of linear terms in the denominator and this suggests that we split our problem up using partial fraction decomposition. Doing this we get ${\frac {s-5}{(s-3)(s-2)}}={\frac {3}{(s-2)}}-{\frac {2}{(s-3)}}.$ We now have this in a form where we can apply the inverse Laplace transform but we will do that at the end. We now focus on the first term, ${\frac {F(s)}{(s-3)(s-2)}}={\frac {{\frac {\pi }{s}}\left(2\exp(-s)-\exp(-2s)\right)}{(s-3)(s-2)}}.$ Now there are several ways we can proceed. One way is to consider $G(s)={\frac {1}{(s-3)(s-2)}}$ and then our term looks like F(s)G(s). We could then invert this using the convolution identity ${\mathcal {L}}^{-1}(F(s)G(s))=\int _{0}^{t}f(t-\tau )g(t)dt.$ While this is valid, it could lead to confusion since f(t) is piecewise defined and so we would have to be careful about the different cases that come out of the integration. Another way we can do the inversion is to include the 1/s terms in F(s) along with the other linear terms in the denominator of G(s). In this way we write ${\frac {F(s)}{(s-3)(s-2)}}={\frac {\pi \left(2\exp(-s)-\exp(-2s)\right)}{s(s-3)(s-2)}}=2\pi \exp(-s)G(s)-\pi \exp(-2s)G(s)$ where we now define $G(s)={\frac {1}{s(s-3)(s-2)}}.$ The advantage of this form is that the table at the end of the exam lists ${\mathcal {L}}^{-1}\left(\exp(-cs)F(s)\right)=u_{c}(t)f(t-c)$ where u_c is the Heaviside function defined as $u_{c}={\begin{cases}0,&t<c\\1,&t\geq {}c\end{cases}}.$ We can use this identity if we once again write G(s) using partial fraction decomposition. Doing this we get $G(s)={\frac {1}{s(s-3)(s-2)}}={\frac {1}{6s}}+{\frac {1}{3(s-3)}}-{\frac {1}{2(s-2)}}.$ We can now return to the full problem and actually compute the various inverse Laplace transforms using these identities. ${\begin{aligned}y(t)={\mathcal {L}}^{-1}(Y(s))&={\mathcal {L}}^{-1}\left({\frac {F(s)}{(s-3)(s-2)}}\right)+{\mathcal {L}}^{-1}\left({\frac {s-5}{(s-3)(s-2)}}\right)\\&=2\pi {\mathcal {L}}^{-1}(\exp(-s)G(s))-\pi {\mathcal {L}}^{-1}(\exp(-2s)G(s))+3{\mathcal {L}}^{-1}\left({\frac {1}{s-2}}\right)-2{\mathcal {L}}^{-1}\left({\frac {1}{s-3}}\right)\\&=2\pi {}u_{1}g(t-1)-\pi {}u_{2}g(t-2)+3\exp(2t)-2\exp(3t)\end{aligned}}$ where $g(t)={\mathcal {L}}^{-1}G(s)={\mathcal {L}}^{-1}\left({\frac {1}{6s}}+{\frac {1}{3(s-3)}}-{\frac {1}{2(s-2)}}\right)={\frac {1}{6}}+{\frac {1}{3}}\exp(3t)-{\frac {1}{2}}\exp(2t).$ Notes: Had we decided to use the convolution identity we would get the same results, we would just have to either explicitly write down the different cases or we would need to independently identify the different Heaviside functions that are appearing. There are two handy things we can confirm to at least give us some confidence in our solution. Firstly, if we substitute in t=0 then we can easily see that both of the initial conditions are satisfied. Secondly, if t<1 then the source term f(t) contributes nothing and so we have $\displaystyle {}y(t)=3\exp(2t)-2\exp(3t)$ which we can verify, by direct substitution or otherwise, solves the homogeneous problem. Finally, if reading the Heaviside notation is awkward, we present the final solution in terms of cases: $y(t)={\begin{cases}3\exp(2t)-2\exp(3t),&t<1\\{\frac {\pi }{3}}+{\frac {2\pi }{3}}\exp(3(t-1))-\pi \exp(2(t-1))+3\exp(2t)-2\exp(3t),&1\leq {}t<2\\{\frac {\pi }{6}}+{\frac {2\pi }{3}}\exp(3(t-1))-\pi \exp(2(t-1))-{\frac {\pi }{3}}\exp(3(t-2))+{\frac {\pi }{2}}\exp(2(t-2))+3\exp(2t)-2\exp(3t),&t\geq {}2\end{cases}}.$

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