MATH 308 Lecture 19

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Lecture Notes

Section 6.1–6.2

$y^{″}-4y=1y(0)=0y^{\prime }(0)=1$

Partial Fraction Decomposition from last time.

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

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

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

Therefore,

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

Inverse Laplace Transform (Exercise 7)

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

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

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

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 $\underset{s\to \mathrm{\infty }}{\lim }$, and take $s=0$

$\begin{array}{rl}\underset{s\to \mathrm{\infty }}{\lim }\frac{3s^3}{s^3}& =3=a+b\\ 3& =\frac{1}{2}+b\\ b& =\frac{5}{2}\end{array}$

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

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

Therefore

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

Exercise 8

$\begin{array}{rl}12{\mathrm{e}}^t& =y^{″}+6y^{\prime }+5yy(0)=-1y^{\prime }(0)=7\\ 12{ℒ}\left\{{\mathrm{e}}^t\right\}& ={ℒ}\left\{y^{″}\right\}+6{ℒ}\left\{y^{\prime }\right\}+5{ℒ}\left\{y\right\}\\ \frac{12}{s-1}& =s^2{ℒ}\left\{y\right\}-y^{\prime }(0)-5y(0)+6s{ℒ}\left\{y\right\}-6y(0)+5{ℒ}\left\{y\right\}\\ \frac{11-s^2+2s}{s-1}& =(s^2+6s+5){ℒ}\left\{y\right\}\\ {ℒ}\left\{y\right\}& =\frac{11-s^2+2s}{(s-1)(s+5)(s+1)}=\frac{a}{s-1}+\frac{b}{s+5}+\frac{c}{s+1}\\ & =\frac{1}{s-1}-\frac{1}{s+5}-\frac{1}{s+1}\\ & ={ℒ}\left\{{\mathrm{e}}^t\right\}-{ℒ}\left\{{\mathrm{e}}^{-5t}\right\}-{ℒ}\left\{{\mathrm{e}}^{-t}\right\}\\ & ={ℒ}\{{\mathrm{e}}^t-{\mathrm{e}}^{-5t}-{\mathrm{e}}^{-t}\}\\ y& ={\mathrm{e}}^t-{\mathrm{e}}^{-5t}-{\mathrm{e}}^{-t}\end{array}$

Exercise 9

$w^{″}+w=t^2+2w(0)=1w^{\prime }(0)=-1$

$\begin{array}{rl}{ℒ}\left\{w^{″}\right\}+{ℒ}\left\{w\right\}& ={ℒ}\{t^2+2\}\\ s^2{ℒ}\left\{w\right\}-w^{\prime }(0)-sw(0)+{ℒ}\left\{w\right\}& =\frac{2}{s^3}+\frac{2}{5}\\ {ℒ}& =\frac{2+2s^2-s^3+s^4}{s^3(s^2+1)}=\frac{a}{s}+\frac{b}{s^2}+\frac{c}{s^3}+\frac{ds+e}{s^2+1}\end{array}$

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 $\frac{2}{s^3}+\frac{s-1}{s^2+1}$)