MATH 308 Lecture 17

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Lecture Notes

An Initial Value Problem

$y''-2y'=4x\quad y(0)=0\quad y'(0)=0$

Solution to homogeneous problem $y''-2y'=0$ is $r\in \{0,2\}\rightarrow y_{h}=c_{1}\,{\cancel {\mathrm {e} ^{0x}}}+c_{2}\,\mathrm {e} ^{2x}$

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

$y_{p}=A\,x+B$, but $c_{1}\propto B$ is already a solution,so $y_{p}=x\,(A\,x+B)$

${\begin{aligned}y_{p}&=A\,x^{2}+B\,x\\y_{p}'&=2A\,x+B\\y_{p}''&=2A\\4x&=y''-2y'\\&=2A-2(2A\,x+B)\\&=(-4A)\,x+(2A-2B)\\(A,B)&=(-1,-1)\end{aligned}}$

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

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

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

The limit exists if $t>M$: ${\begin{aligned}\left|f(t)\right|&\leq k\mathrm {e} ^{a\,t}\left|f(t)\,\mathrm {e} ^{-s\,t}\right|&\leq k\,\mathrm {e} ^{(a-s)t}\end{aligned}}$

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

Our goal is to be able to take the laplace transform of each side of $y''+y=g(t)$:

${\mathcal {L}}\left\{y''\right\}(s)+{\mathcal {L}}\left\{y\right\}(s)={\mathcal {L}}\left\{g(t)\right\}(s)$

Piecewise Continuity

A function is piecewise continuous on an interval $a\leq t\leq b$ if the interval $\left[a,b\right]$ 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 $\left(t_{i},t_{i+1}\right)$.
2. $f$ approached a finite limit as the endpoints of each interval are approached from within the subinterval.

Exercises

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

${\mathcal {L}}\left\{1\right\}(s)=\int _{0}^{\infty }\mathrm {e} ^{-s\,t}\,\mathrm {d} t$

• Does not exist for $s\leq 0$

${\begin{aligned}{\mathcal {L}}\left\{1\right\}(s)&=\lim _{N\to \infty }\int _{0}^{N}\mathrm {e} ^{-s\,t}\,\mathrm {d} t\\&=\lim _{N\to \infty }{\frac {\mathrm {e} ^{-s\,N}}{-s}}-{\frac {1}{-s}}\\&=\lim _{N\to \infty }-{\frac {\mathrm {e} ^{-s\,N}}{s}}+{\frac {1}{s}}\\&={\frac {1}{s}}\quad s>0\end{aligned}}$

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

${\mathcal {L}}\left\{t\right\}(s)=\int _{0}^{\infty }t\,\mathrm {e} ^{-s\,t}\,\mathrm {d} t$

• Does not exist for $s\leq 0$

${\begin{aligned}{\mathcal {L}}\left\{1\right\}(s)&=\lim _{N\to \infty }\int _{0}^{N}t\,\mathrm {e} ^{-s\,t}\,\mathrm {d} t\\&=\lim _{N\to \infty }\left.{\frac {-t\,\mathrm {e} ^{-s\,t}}{s}}\right|_{0}^{N}+\int _{0}^{N}{\frac {\mathrm {e} ^{-s\,t}}{s}}\,\mathrm {d} t\\&=\lim _{N\to \infty }{\frac {-N\,\mathrm {e} ^{-s\,N}}{s}}-{\frac {\mathrm {e} ^{-s\,N}}{s^{2}}}+{\frac {1}{s^{2}}}\\&={\frac {1}{s^{2}}}\quad s>0\end{aligned}}$

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

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

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

${\begin{aligned}\lim _{N\to \infty }\int _{0}^{N}\mathrm {e} ^{(3-s)t}\,\mathrm {d} t&=\left.{\frac {\mathrm {e} ^{(3-s)t}}{3-s}}\right|_{0}^{N}\\&={\frac {1}{s-3}}\quad s>3\end{aligned}}$

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

${\mathcal {L}}\left\{\mathrm {e} ^{5t}\,\cos {t}\right\}=\lim _{N\to \infty }\int _{0}^{N}\mathrm {e} ^{(5-s)t}\,\mathrm {d} t$

• Exists for $s>5$

Integrate by parts:

${\begin{aligned}\lim _{N\to \infty }\int _{0}^{N}\mathrm {e} ^{(5-s)t}\,\mathrm {d} t&=\lim _{N\to \infty }\left.\mathrm {e} ^{(5-s)t}\,\sin {t}\right|_{0}^{N}-\int _{0}^{N}(5-s)\,\mathrm {e} ^{(5-s)t}\,\mathrm {d} t\\&=\lim _{n\to \infty }0-\left.\mathrm {e} ^{(5-s)t}\,\cos {t}\right|_{0}^{N}-\int _{0}^{N}(5-s)^{2}\,\mathrm {e} ^{(5-s)t}\,\cos {t}\,\mathrm {d} t\\&=-(5-s)-(5-s)^{2}\,{\mathcal {L}}\left\{\mathrm {e} ^{5t}\,\cos {t}\right\}\\&={\frac {s-5}{1+(s-5)^{2}}}\end{aligned}}$