- #1
Prove It
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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.