MATH 308 Lecture 19

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Lecture Notes

Section 6.1–6.2

$y''-4y=1\quad y(0)=0\quad y'(0)=1$

Partial Fraction Decomposition from last time.

${\begin{aligned}{\mathcal {L}}\left\{y\right\}&={\frac {s+1}{s(s-2)(s+2)}}\\&={\frac {a}{s}}+{\frac {b}{s-2}}+{\frac {c}{s+2}}\end{aligned}}$

Multiply by each term $q$ in the denominator and take $q=0$

${\begin{aligned}{\frac {s+1}{(s-2)(s+2}}&=a+{\frac {bs}{s-2}}+{\frac {cs}{s+2}}\\-{\frac {1}{4}}&=a&{\text{when}}\ s&=0\\{\frac {s+1}{s(s+2)}}&={\frac {a(s-2)}{s}}+b+{\frac {c(s-2)}{s+2}}\\{\frac {3}{8}}&=b&{\text{when}}\ s&=2\\{\frac {s+1}{s(s-2)}}&={\frac {a(s+2)}{s}}+{\frac {b(s+2)}{s-2}}+c\\-{\frac {1}{8}}&=c&{\text{when}}\ s&=-2\end{aligned}}$

Therefore,

${\begin{aligned}{\mathcal {L}}\left\{y\right\}&=-{\frac {1}{4}}\,{\frac {1}{s}}+{\frac {3}{8}}\,{\frac {1}{s-2}}-{\frac {1}{8}}\,{\frac {1}{s+2}}\\&=-{\frac {1}{4}}\,{\mathcal {L}}\left\{1\right\}+{\frac {3}{8}}\,{\mathcal {L}}\left\{\mathrm {e} ^{2t}\right\}-{\frac {1}{8}}\,{\mathcal {L}}\left\{\mathrm {e} ^{-2t}\right\}\\&={\mathcal {L}}\left\{-{\frac {1}{4}}+{\frac {3}{8}}\,\mathrm {e} ^{2t}-{\frac {1}{8}}\,\mathrm {e} ^{-2t}\right\}\\y&=-{\frac {1}{4}}+{\frac {3}{8}}\,\mathrm {e} ^{2t}-{\frac {1}{8}}\,\mathrm {e} ^{-2t}\end{aligned}}$

Inverse Laplace Transform (Exercise 7)

${\begin{aligned}{\mathcal {L}}\left\{y\right\}&={\frac {3s^{2}+s}{(s+1)(s-1)^{2}}}\\&={\frac {a}{s+1}}+{\frac {b}{(s-1)^{1}}}+{\frac {c}{(s-1)^{2}}}\end{aligned}}$

1. Multiply by $s+1$ and take $s=-1$
2. Multiply by $(s-1)^{2}$ and take $s=1$

${\begin{aligned}{\frac {3s^{2}+s}{(s-1)^{2}}}&=a+{\frac {b(s+1)}{(s-1)^{1}}}+{\frac {c(s+1)}{(s-1)^{2}}}\\{\frac {1}{2}}&=a&{\text{when}}\ s&=-1\\{\frac {3s^{2}+s}{s+1}}&={\frac {a(s-1)^{2}}{s+1}}+b(s-1)+c\\2&=c&{\text{when}}\ s&=1\end{aligned}}$

Taking similar steps won't work for $b$ since taking $s=1$ will cause a division by zero on the "$c$ term".

Multiply by $s$, take $\lim _{s\to \infty }$, and take $s=0$

${\begin{aligned}\lim _{s\to \infty }{\frac {3s^{3}}{s^{3}}}&=3=a+b\\3&={\frac {1}{2}}+b\\b&={\frac {5}{2}}\end{aligned}}$

(I'm not sure what's going on here...)

${\begin{aligned}0&={\frac {1}{2}}-b+2\\b&={\frac {1}{2}}+2\end{aligned}}$

Therefore

${\begin{aligned}{\mathcal {L}}\left\{y\right\}&={\frac {1}{2}}\,{\frac {1}{s+1}}+{\frac {5}{2}}\,{\frac {1}{s-1}}+{\frac {2}{(s-1)^{2}}}\\&={\frac {1}{2}}\,{\mathcal {L}}\left\{\mathrm {e} ^{-t}\right\}+{\frac {5}{2}}\,{\mathcal {L}}\left\{\mathrm {e} ^{t}\right\}+2{\mathcal {L}}\left\{t\,\mathrm {e} ^{t}\right\}\\y&={\frac {\mathrm {e} ^{-t}}{2}}+{\frac {5\mathrm {e} ^{t}}{2}}+2t\,\mathrm {e} ^{t}\end{aligned}}$

Exercise 8

${\begin{aligned}12\mathrm {e} ^{t}&=y''+6y'+5y\quad \quad y(0)=-1\quad y'(0)=7\\12{\mathcal {L}}\left\{\mathrm {e} ^{t}\right\}&={\mathcal {L}}\left\{y''\right\}+6{\mathcal {L}}\left\{y'\right\}+5{\mathcal {L}}\left\{y\right\}\\{\frac {12}{s-1}}&=s^{2}\,{\mathcal {L}}\left\{y\right\}-y'(0)-5y(0)+6s\,{\mathcal {L}}\left\{y\right\}-6y(0)+5{\mathcal {L}}\left\{y\right\}\\{\frac {11-s^{2}+2s}{s-1}}&=\left(s^{2}+6s+5\right)\,{\mathcal {L}}\left\{y\right\}\\{\mathcal {L}}\left\{y\right\}&={\frac {11-s^{2}+2s}{(s-1)(s+5)(s+1)}}={\frac {a}{s-1}}+{\frac {b}{s+5}}+{\frac {c}{s+1}}\\&={\frac {1}{s-1}}-{\frac {1}{s+5}}-{\frac {1}{s+1}}\\&={\mathcal {L}}\left\{\mathrm {e} ^{t}\right\}-{\mathcal {L}}\left\{\mathrm {e} ^{-5t}\right\}-{\mathcal {L}}\left\{\mathrm {e} ^{-t}\right\}\\&={\mathcal {L}}\left\{\mathrm {e} ^{t}-\mathrm {e} ^{-5t}-\mathrm {e} ^{-t}\right\}\\y&=\mathrm {e} ^{t}-\mathrm {e} ^{-5t}-\mathrm {e} ^{-t}\end{aligned}}$

Exercise 9

$w''+w=t^{2}+2\quad w(0)=1\quad w'(0)=-1$

${\begin{aligned}{\mathcal {L}}\left\{w''\right\}+{\mathcal {L}}\left\{w\right\}&={\mathcal {L}}\left\{t^{2}+2\right\}\\s^{2}{\mathcal {L}}\left\{w\right\}-w'(0)-s\,w(0)+{\mathcal {L}}\left\{w\right\}&={\frac {2}{s^{3}}}+{\frac {2}{5}}\\{\mathcal {L}}&={\frac {2+2s^{2}-s^{3}+s^{4}}{s^{3}(s^{2}+1)}}={\frac {a}{s}}+{\frac {b}{s^{2}}}+{\frac {c}{s^{3}}}+{\frac {ds+e}{s^{2}+1}}\end{aligned}}$

We find $c=2$ easily, but lower degrees are harder: multiply by $s$ and take limit to infinity.

$1=a+d$

(out of time; WolframAlpha says the decomposition is ${\frac {2}{s^{3}}}+{\frac {s-1}{s^{2}+1}}$)