Solve second order differential equation with Heaviside function using Laplace transform

Question

The equation is:

$$y'' + 3y = u_4(t)\cos(5(t-4)), \quad y(0) = 0, \quad y'(0) = -2$$ Here $u_4$ is the Heaviside function with activation switch at $t=4$.

I can get all the way to the partial fraction decomposition but then we get strange coefficients like $45/999$ for $S$ and so on!

Can someone help me? That surely can't be right!

First I get $L(y)$ which ends up being $$L(y) = e^{-4s}\frac{s}{(s^2+25)(s^2+3)}-\frac{2}{s^2+3}$$

Then I try doing partial fraction decomposition on the first bit by making $(s/((s^2+25)(s^2+3)) $equal to$ (As+B)/(s^2+25) + (Cs+D)/(s^2+3)$ but then I get the awkward coefficients..

Answer 1

We are given:

$$y'' + 3y = u_4(t) \cos(5(t-4)) = u(t-4) \cos(5(t-4)), y(0) = 0, y'(0) = -2$$

We have:

• $\mathscr{L} (y''(t)) = s^2 y(s) - s y(0) - y'(0) = s^2 y(s) +2$
• $\mathscr{L} (3y(t)) = 3 y(s)$
• $\mathscr{L} (u_4(t)\cos(5(t-4))) = \dfrac{e^{-4 s} s}{s^2+25}$

From this we write:

$$(s^2 + 3)y(s) = \frac{s~e^{-4 s}}{s^2+25} - 2$$

Solving for $y(s)$ yields:

$y(s) = \dfrac{1}{s^2+3}\left( \dfrac{e^{-4 s} s}{s^2+25}-2\right) = -\dfrac{2}{s^2+3} +\dfrac{e^{-4 s} s}{\left(s^2+3\right) \left(s^2+25\right)}$, so

$$y(s) = -\dfrac{2}{s^2+3}+ e^{-4 s}\left( \dfrac{s}{22 \left(s^2+3\right)}-\dfrac{s}{22 \left(s^2+25\right)}\right)$$

The inverse Laplace transform, using a Table of Laplace Transforms is given by:

$$\mathscr{L}^{-1}(y(s)) = y(t) = -\frac{2 \sin \left(\sqrt{3} t\right)}{\sqrt{3}} + \frac{1}{22} u(t-4) \left(\cos \left(\sqrt{3} (t-4)\right)-\cos (5 (t-4))\right)$$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm y}''\pars{t} + 3{\rm y}\pars{t} = \Theta\pars{t - 4}\cos\pars{5\bracks{t - 4}}\,,\qquad{\rm y}\pars{0} = 0\,, \quad {\rm y}'\pars{0} = -2}$

\begin{align} {\rm y}\pars{t}&= -\,{2\root{3} \over 3}\,\sin\pars{\root{3}t} +\int_{-\infty}^{\infty}{\rm G}\pars{t,t'} \Theta\pars{t' - 4}\cos\pars{5\bracks{t' - 4}}\,\dd t' \\[3mm]&=-\,{2\root{3} \over 3}\,\sin\pars{\root{3}t} +\int_{4}^{\infty}{\rm G}\pars{t,t'} \cos\pars{5\bracks{t' - 4}}\,\dd t' \end{align} where $\ds{{\rm G}\pars{t,t'}}$ satisfies $$ \pars{\partiald[2]{}{t} + 3}{\rm G}\pars{t,t'} = 0\quad\mbox{if}\quad t \not= t' \quad\mbox{and}\quad\left\vert% \begin{array}{rcl} {\rm G}\pars{0,t'} & = & 0\,,\quad \forall\ t' \\[3mm] \left.\partiald{{\rm G}\pars{t,t'}}{t}\right\vert_{t\ =\ 0} & = & 0\,, \quad \forall\ t' \\[3mm] \left.\partiald{{\rm G}\pars{t,t'}}{t}\right\vert_{t\ =\ t'^{-}}^{t\ =\ t'^{+}} & = & 1 \end{array}\right. $$

Then $$ {\rm G}\pars{t,t'} =\left\lbrace% \begin{array}{lcl} 0 & \mbox{if} & t < t' \\[3mm] {\root{3} \over 3}\,\sin\pars{\root{3}\bracks{t - t'}} & \mbox{if} & t > t' \end{array}\right. $$

\begin{align} {\rm y}\pars{t}&=-\,{2\root{3} \over 3}\,\sin\pars{\root{3}t} \\[3mm]&\phantom{=}\mbox{}+\int_{4}^{\infty}\Theta\pars{t - t'}{\root{3} \over 3}\, \sin\pars{\root{3}\bracks{t - t'}} \cos\pars{5\bracks{t' - 4}}\,\dd t' \\[3mm]&=-\,{2\root{3} \over 3}\,\sin\pars{\root{3}t} \\[3mm]&\phantom{=}\mbox{}+\Theta\pars{t - 4}\,{\root{3} \over 3}\ \underbrace{\int_{4}^{t}\sin\pars{\root{3}\bracks{t - t'}} \cos\pars{5\bracks{t' - 4}}\,\dd t'} _{\ds{{\root{3} \over 22}\braces{\cos\pars{\root{3}\bracks{t - 4}} - \cos\pars{5\bracks{t - 4}}}}} \end{align}

$$\color{#00f}{% {\rm y}\pars{t}=-\,{2\root{3} \over 3}\,\sin\pars{\root{3}t} +\Theta\pars{t - 4}\,{\cos\pars{\root{3}\bracks{t - 4}} - \cos\pars{5\bracks{t - 4}} \over 22}} $$