MATH 308 Lecture 18

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Lecture Notes

Laplace Transform of Derivatives

${ℒ}\left\{f^{\prime }\right\}(s)=s{ℒ}\left\{f\right\}(s)-f(0)$

Proof.

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

Therefore

${ℒ}\left\{y^{\prime }\right\}=s{ℒ}\left\{y\right\}(s)-y(0)$

Q.E.D.

In General

${ℒ}\left\{f^{(n)}\right\}(s)=s^n{ℒ}\left\{f\right\}(s)-s^{n-1}f(0)-s^{n-2}{f}^{\prime }(0)-\dots -f^{(n-1)}(0)$

Exercise 2

Find the Laplace Transform of ${\mathrm{e}}^{4x}$

$\begin{array}{rl}f(t)& ={\mathrm{e}}^{4t}\\ f^{\prime }(t)& =4{\mathrm{e}}^{4t}=4f(t)\\ {ℒ}\{f^{\prime }(t)\}& =4{ℒ}\{f(t)\}\\ s{ℒ}\{f(t)\}-4{\mathrm{e}}^{4\cdot 0}& =4{ℒ}\{f(t)\}\\ (s-4){ℒ}\{f(t)\}& =1\\ {ℒ}\{f(t)\}& =\frac{1}{s-4}\end{array}$

Exercise 5

$y^{″}-4y=1y(0)=0y^{\prime }(0)=1$

$\begin{array}{rl}{ℒ}\left\{y^{″}\right\}-4{ℒ}\left\{y\right\}& ={ℒ}\left\{1\right\}=\frac{1}{s}\\ s^2{ℒ}\left\{y\right\}-y^{\prime }(0)-sy(0)-4{ℒ}\left\{y\right\}& =\frac{1}{s}\\ (s^2-4){ℒ}\left\{y\right\}-1-0s& =\frac{1}{s}\\ {ℒ}\left\{y\right\}& =\frac{\frac{1}{s}+1}{s^2-4}=\frac{s+1}{s(s-2)(s+2)}\end{array}$

The Laplace transform was good, but the solution is better:

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

We'll learn how to find coefficients $a$, $b$, and $c$ later.

Quiz Time!