Science:Math Exam Resources/Courses/MATH215/December 2013/Question 01 (f)

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 01 (f)
Find the Laplace transform $X(s)$ of the solution to the initial value problem $x'''+x=0,\quad x(0)=1,x'(0)=2,x''(0)=3$ (but do not compute $x(t)$).

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Hint
How does the n-th derivative, x(n)(t) transform for an arbitrary function x(t)?

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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. The Laplace transform of the 3rd derivative $\displaystyle x'''(t)$ is given by ${\mathcal {L}}(x'''(t))=s^{3}X(s)-s^{2}x(0)-sx'(0)-x''(0)$ Plugging this into the equation, and using that the Laplace transform is linear, we obtain $0={\mathcal {L}}(x'''(t)+x(t))=s^{3}X(s)-s^{2}x(0)-sx'(0)-x''(0)+X(s)$ Upon using $\displaystyle x(0)=1,x'(0)=2,x''(0)=3$, we have $\displaystyle (s^{3}+1)X(s)=s^{2}+2s+3$ so $X(s)={\frac {s^{2}+2s+3}{s^{3}+1}}$