MATH 308 Lecture 17

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Lecture Notes

An Initial Value Problem

$y^{″}-2y^{\prime }=4xy(0)=0y^{\prime }(0)=0$

Solution to homogeneous problem $y^{″}-2y^{\prime }=0$ is $r\in \{0,2\}\to y_h=c_1\cancel{{\mathrm{e}}^{0x}}+c_2{\mathrm{e}}^{2x}$

Try method of undetermined coefficients (could also use method of variation of parameters)

$y_p=Ax+B$, but $c_1\propto B$ is already a solution,so $y_p=x(Ax+B)$

$\begin{array}{rl}{y}_p& =A{x}^2+Bx\\ {y_p}^{\prime }& =2Ax+B\\ {y_p}^{″}& =2A\\ 4x& =y^{″}-2y^{\prime }\\ & =2A-2(2Ax+B)\\ & =(-4A)x+(2A-2B)\\ (A,B)& =(-1,-1)\end{array}$

So $y_p=-x^2-x$ is a particular solution, and our general solution is $y=y_p+y_h=-x^2-x+c_1+c_2{\mathrm{e}}^{2x}$

Plug in initial values and solve for $c_1$ and $c_2$:

$y=-x^2-x-\frac{1}{2}+\frac{1}{2}{\mathrm{e}}^{2x}$

Laplace Transforms

Piecewise functions that need not be continuous but must be piecewise continuous.

Let $f$ be a function on $[0,\mathrm{\infty })$. The Laplace transform of $f$ is the function ${ℒ}\left\{f\right\}$ defined by the integral

${ℒ}\left\{f\right\}(s)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-st}f(t)\mathrm{d}t=\underset{N\to \mathrm{\infty }}{\lim }\underset{0}{\overset{N}{\int }}{\mathrm{e}}^{-st}f(t)\mathrm{d}t$

The domain of ${ℒ}\left\{f\right\}$ is all the values of $s$ for which the integral exists.

The limit exists if $t>M$: $\begin{array}{rlr}|f(t)|& \le k{\mathrm{e}}^{at}|f(t){\mathrm{e}}^{-st}|& \le k{\mathrm{e}}^{(a-s)t}\end{array}$

So the integral $\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{(a-s)t}\mathrm{d}t$ is convergent for $s>a$

Our goal is to be able to take the laplace transform of each side of $y^{″}+y=g(t)$:

${ℒ}\left\{y^{″}\right\}(s)+{ℒ}\left\{y\right\}(s)={ℒ}\{g(t)\}(s)$

Piecewise Continuity

A function is piecewise continuous on an interval $a\le t\le b$ if the interval $[a,b]$ can be partitioned by a finite number of points $a=t_0<t_1<\dots <t_n=b$ so that

1. $f$ is continuous on each open interval $(t_i,t_{i+1})$.
2. $f$ approached a finite limit as the endpoints of each interval are approached from within the subinterval.

Exercises

Find the Laplace transform of $f(t)=1$:

${ℒ}\left\{1\right\}(s)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-st}\mathrm{d}t$

• Does not exist for $s\le 0$

$\begin{array}{rl}{ℒ}\left\{1\right\}(s)& =\underset{N\to \mathrm{\infty }}{\lim }\underset{0}{\overset{N}{\int }}{\mathrm{e}}^{-st}\mathrm{d}t\\ & =\underset{N\to \mathrm{\infty }}{\lim }\frac{{\mathrm{e}}^{-sN}}{-s}-\frac{1}{-s}\\ & =\underset{N\to \mathrm{\infty }}{\lim }-\frac{{\mathrm{e}}^{-sN}}{s}+\frac{1}{s}\\ & =\frac{1}{s}s>0\end{array}$

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Find the Laplace transform of $f(t)=t$:

${ℒ}\left\{t\right\}(s)=\underset{0}{\overset{\mathrm{\infty }}{\int }}t{\mathrm{e}}^{-st}\mathrm{d}t$

• Does not exist for $s\le 0$

$\begin{array}{rl}{ℒ}\left\{1\right\}(s)& =\underset{N\to \mathrm{\infty }}{\lim }\underset{0}{\overset{N}{\int }}t{\mathrm{e}}^{-st}\mathrm{d}t\\ & =\underset{N\to \mathrm{\infty }}{\lim }{\frac{-t{\mathrm{e}}^{-st}}{s}|}_0^N+\underset{0}{\overset{N}{\int }}\frac{{\mathrm{e}}^{-st}}{s}\mathrm{d}t\\ & =\underset{N\to \mathrm{\infty }}{\lim }\frac{-N{\mathrm{e}}^{-sN}}{s}-\frac{{\mathrm{e}}^{-sN}}{s^2}+\frac{1}{s^2}\\ & =\frac{1}{s^2}s>0\end{array}$

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Find the Laplace transform of $f(t)={\mathrm{e}}^{3t}$

${ℒ}\left\{{\mathrm{e}}^{3t}\right\}(s)=\underset{N\to \mathrm{\infty }}{\lim }\underset{0}{\overset{N}{\int }}{\mathrm{e}}^{(3-s)t}\mathrm{d}t$

• $3-s>0$, so integral exists for $s>3$

$\begin{array}{rl}\underset{N\to \mathrm{\infty }}{\lim }\underset{0}{\overset{N}{\int }}{\mathrm{e}}^{(3-s)t}\mathrm{d}t& ={\frac{{\mathrm{e}}^{(3-s)t}}{3-s}|}_0^N\\ & =\frac{1}{s-3}s>3\end{array}$

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Find the Laplace transform of $f(t)={\mathrm{e}}^{5t}\cos t$

${ℒ}\{{\mathrm{e}}^{5t}\cos t\}=\underset{N\to \mathrm{\infty }}{\lim }\underset{0}{\overset{N}{\int }}{\mathrm{e}}^{(5-s)t}\mathrm{d}t$

• Exists for $s>5$

Integrate by parts:

$\begin{array}{rl}\underset{N\to \mathrm{\infty }}{\lim }\underset{0}{\overset{N}{\int }}{\mathrm{e}}^{(5-s)t}\mathrm{d}t& =\underset{N\to \mathrm{\infty }}{\lim }{{\mathrm{e}}^{(5-s)t}\sin t|}_0^N-\underset{0}{\overset{N}{\int }}(5-s){\mathrm{e}}^{(5-s)t}\mathrm{d}t\\ & =\underset{n\to \mathrm{\infty }}{\lim }0-{{\mathrm{e}}^{(5-s)t}\cos t|}_0^N-\underset{0}{\overset{N}{\int }}(5-s{)}^2{\mathrm{e}}^{(5-s)t}\cos t\mathrm{d}t\\ & =-(5-s)-(5-s{)}^2{ℒ}\{{\mathrm{e}}^{5t}\cos t\}\\ & =\frac{s-5}{1+(s-5{)}^2}\end{array}$