MHB Douglas' questions about Laplace Transforms

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Douglas seeks clarification on his Laplace Transform solutions, noting discrepancies in syntax and results compared to provided solutions. In his first question, he miscalculated coefficients in the partial fraction decomposition, leading to an incorrect form of the solution. For the second question, he wonders if his answer is equivalent to the provided solution, emphasizing the importance of the Heaviside function's placement. In the third question, he questions the sign difference in a term and the formatting requirements for submission in Weblearn. Overall, he aims to ensure his answers align with expected formats to avoid issues with electronic marking.
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Hi, I got 2 questions extremely close to the solution provided by weblearn, I was wondering you could explain if my answers were right but in the wrong syntax or where I made a mistake (whenever you have the time).

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Question 1/
d^2y/dx^2 + 7*dy/dx + 12*y = -36H(t-2) , y(0) = 0 y'(0) = 0

Find y(t)?

My answer was y(t) = -1/3 - 9*exp(-4*t) + 12exp(-3*t)

The solution was y(t) = -3H(t-2) [ 1 - 4exp(-3*(t-2)) + 3exp(-4*(t-2))

Partway through my working out I rearranged the equation to get Y(s) = -36/ s*(s^2 + 7s + 12)
Which I solved using partial fractions. A= -1/3 B=-9 C= 12

Where does the shift come in and how can I recognise it? The Heaviside on the right hand side was only to a constant, 36 not a function f(t)?

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Question 2/

Inverse Laplace of 2(s^2 +48s -36) / s( s^2 - 36) e^-2s

The answer I got was 2 + 16sinh(6(t-2)) H(t-2)

The solution on weblearn was 2H(t-2)( 1 + 8sinh( 6(t-2)) )

- In weblearn does the Heaviside component always need to be written at the front of the answer to be registered?
- Am I right by thinking both answers are the same just with 2 taken as a factor? Does weblearn only register a single format for the answer?

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Question 3/

L[ f(t) = -2t + 11 + H(t-4) ( 8t^2 - 32t) ]

My answer with the exact syntax used -
(-2/s^2) + (11/s) + (exp(-4*s) * ((16/s^3) - (32/s^2) ) )

Weblearn Solution:
-2/s^2 + 11/s + [ 16/s^3 + 32/s^2 ] * e^-4s

The only difference I can see is the + sign in front of the last term and that e^-4s is written at the end and not the front.

- Why is it positive and not negative?
- Will weblearn register the answer correctly if the exponent is in front or does it need to be at the end of the brackets?

- Are square brackets registered on weblearn?
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I know most of these questions are about the syntax but it's better to clarify now rather than towards the end of semester. It's mostly because electronic marking can be a bit finicky sometimes.

Thanks in advance

Douglas

1.
$\displaystyle \begin{align*} y''(t) + 7y'(t) + 12y(t) &= -36H(t - 2) \textrm{ with } y(0) = 0, y'(0) = 0 \\ \mathcal{L} \left\{ y''(t) + 7y'(t) + 12y(t) \right\} &= \mathcal{L} \left\{ -36H(t-2) \right\} \\ s^2\,Y(s) + s\,y(0) + y'(0) + 7 \left[ s\,Y(s) - y(0) \right] + 12Y(s) &= -\frac{36}{s}e^{-2s} \\ s^2\,Y(s) + 7s\,Y(s) + 12Y(s) &= -\frac{36}{s}e^{-2s} \\ Y(s) \left( s^2 + 7s + 12 \right) &= -\frac{36}{s}e^{-2s} \\ Y(s) \left( s + 3 \right) \left( s + 4 \right) &= -\frac{36}{s} e^{-2s} \\ Y(s) &= -\frac{36}{s \left( s + 3 \right) \left( s + 4 \right) } e^{-2s} \end{align*}$

So now applying partial fractions

$\displaystyle \begin{align*} \frac{A}{s} + \frac{B}{s + 3} + \frac{C}{s + 4} &\equiv \frac{36}{s \left( s + 3 \right) \left( s + 4 \right) } \\ \frac{A \left( s + 3 \right) \left( s + 4 \right) + B\,s \left( s + 4 \right) + C \, s \left( s + 3 \right) }{s \left( s + 3 \right) \left( s + 4 \right) } &\equiv \frac{36}{s \left( s + 3 \right) \left( s + 4 \right) } \\ A \left( s + 3 \right) \left( s + 4 \right) + B \, s \left( s + 4 \right) + C\,s \left( s + 3 \right) &\equiv 36 \end{align*}$

Let $\displaystyle \begin{align*} s = 0 \end{align*}$ to find $\displaystyle \begin{align*} 12A = 36 \implies A = 3 \end{align*}$. Let $\displaystyle \begin{align*} s = -3 \end{align*}$ to find $\displaystyle \begin{align*} -3B = 36 \implies B = -12 \end{align*}$. Let $\displaystyle \begin{align*} s = -4 \end{align*}$ to find $\displaystyle \begin{align*} 4C = 36 \implies C = 9 \end{align*}$. So that means $\displaystyle \begin{align*} \frac{36}{s \left( s + 3 \right) \left( s + 4 \right) } \equiv \frac{3}{s} - \frac{12}{s + 3} + \frac{9}{s + 4} \end{align*}$. So for starters, a mistake is that you have 1/3 as one of the coefficients instead of 3. Anyway...

$\displaystyle \begin{align*} Y(s) &= -\frac{36}{s \left( s + 3 \right) \left( s + 4 \right) } e^{-2s} \\ Y(s) &= \left( -\frac{3}{s} + \frac{12}{s + 3} - \frac{9}{s + 4} \right) \, e^{-2s} \end{align*}$

and now we apply the rule $\displaystyle \begin{align*} \mathcal{L}^{-1} \left\{ e^{-a\,s} \, F(s) \right\} = f(t-a)\,H(t-a) \end{align*}$. Notice that $\displaystyle \begin{align*} \mathcal{L}^{-1} \left\{ -\frac{3}{s} + \frac{12}{s + 3} - \frac{9}{s + 4} \right\} = -3 + 12e^{-3t} - 9e^{-4t} \end{align*}$, so that means

$\displaystyle \begin{align*} Y(s) &= \left( -\frac{3}{s} + \frac{12}{s + 3} - \frac{9}{s + 4} \right) \, e^{-2s} \\ y(t) &= \mathcal{L}^{-1} \left\{ \left( -\frac{3}{s} + \frac{12}{s + 3} - \frac{9}{s + 4} \right) \, e^{-2s} \right\} \\ y(t) &= \left[ -3 + 12e^{-3 \left( t - 2 \right) } - 9 e^{-4 \left( t - 2 \right) } \right] \, H(t - 2) \end{align*}$2. If you had entered:

( 2 + 16sinh(6(t-2)) ) H(t-2)

you would have gotten the answer correct. Yes, what the solution gives is the same as this with a factor of 2 taken out.3. Starting with $\displaystyle \begin{align*} 8t^2 - 32t &= 8 \left( t^2 - 4t \right) \end{align*}$, let $\displaystyle \begin{align*} u = t - 4 \implies t = u + 4 \end{align*}$, then we have

$\displaystyle \begin{align*} 8 \left( t^2 - 4t \right) &= 8 \left[ \left( u + 4 \right) ^2 - 4 \left( u + 4 \right) \right] \\ &= 8 \left( u^2 + 8u + 16 - 4u - 16 \right) \\ &= 8 \left( u^2 + 4u \right) \\ &= 8 \left[ \left( t - 4 \right) ^2 + 4 \left( t - 4 \right) \right] \end{align*}$

This is why there is a + instead of a - where your answer differ. Continuing we have...

$\displaystyle \begin{align*} \mathcal{L} \left\{ -2t + 11 + H(t - 4) \, \left( 8t^2 - 32t \right) \right\} &= \mathcal{L} \left\{ -2t + 11 + 8H(t-4) \, \left[ \left( t - 4 \right) ^2 + 4 \left( t - 4 \right) \right] \right\} \\ &= -\frac{2}{s^2} + \frac{11}{s} + 8e^{-4s} \, \mathcal{L} \left\{ t^2 + 4t \right\} \\ &= -\frac{2}{s^2} + \frac{11}{s} + 8e^{-4s} \left( \frac{2}{s^3} + \frac{4}{s^2} \right) \end{align*}$

To answer your other question, Weblearn uses the same syntax as Maple, and all brackets (even nested brackets) should be entered as round brackets.
 
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So in your case, your final answer could have been entered into Weblearn as:

(-2/s^2) + (11/s) + (8*exp(-4*s)) * ( (2/s^3) + (4/s^2) )

I hope that helps.

 
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