Solve a differential equation by a Laplace transform

Question

I have this differential equation

$\frac{d^4 y}{dx^4} - y = f(x)$

Where $y $ and the first three derivatives of $y$ disappear at $x = 0$.

I have shown that

$$\bar{y}(s)= \int_{0}^{\infty} f(\xi) \frac{e^{-s \xi}}{s^4 -1} d\xi $$

which is the first part of the question.

I need to show that

$$ y(x) = \frac{1}{2} \int_{0}^{x} f(\xi) [\sinh(x - \xi) - \sin(x - \xi)] d\xi$$

By breaking $\frac{1}{s^4 -1}$ into partial fractions we have

$$\frac{1}{s^4 -1} = \frac{1}{2} \int_{0}^{\infty} (\sinh x - \sin x ) e^{-sx} dx $$

And we have a double integral. How do I get to the solution from here?

Answer 1

First, we note that the partial fraction expansion of $\frac{1}{s^4-1}$ is given by

$$\frac{1}{s^4-1}=\frac14\left(\frac{1}{s-1}-\frac{1}{s+1}+\frac{i}{s-i}-\frac{i}{s+i}\right) \tag 1$$

Second, using the convolution theorem, it is easy to see that the inverse Laplace Transform of $G(s)=\frac{F(s)}{s-s_0}$ is

$$\mathscr{L}^{-1}\left(G\right)(t)=\int_0^t f(t')e^{-s_0(t'-t)}\,dt' \tag 2$$

Using the results in $(1)$ and $(2)$, we find that the inverse Laplace Transform of $Y(s)=\frac{F(s)}{s^4-1}$

$$\begin{align} y(t)&=\frac14\int_0^t f(t')\left(e^{t-t'}-e^{-(t-t')}+ie^{i(t-t')}-ie^{-i(t-t')}\right)\,dt'\\\\ &=\frac12\int_0^t f(t')\left(\sinh(t-t')-\sin(t-t') \right)\,dt' \end{align}$$

as was to be shown!

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Note we could have proceeded directly by using the residue theorem to evaluate the inversion integral

$$y(t)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{F(s)}{s^4-1}e^{st}\,ds$$

where $c>1$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{y}\pars{x} & = \int_{1^{+} - \infty\ic}^{1^{+} + \infty\ic} {\hat{\mrm{f}}\pars{s} \over s^{4} - 1}\expo{xs}\,{\dd s \over 2\pi\ic} = \int_{1^{+} - \infty\ic}^{1^{+} + \infty\ic} \bracks{\int_{0}^{\infty}\mrm{f}\pars{\xi}\expo{-s\xi}\,\dd\xi} {\expo{xs} \over s^{4} - 1}\,{\dd s \over 2\pi\ic} \\[5mm] & = \int_{0}^{\infty}\mrm{f}\pars{\xi}\bracks{% \int_{1^{+} - \infty\ic}^{1^{+} + \infty\ic} {\expo{\pars{x - \xi}s} \over s^{4} - 1}\,{\dd s \over 2\pi\ic}}\dd\xi \\[5mm] & = \int_{0}^{\infty}\mrm{f}\pars{\xi}\braces{\bracks{x - \xi > 0} \bracks{{\expo{\pars{x - \xi}\pars{1}} \over 4 \times 1^{3}} + {\expo{\pars{x - \xi}\ic} \over 4 \times \ic^{3}} + {\expo{\pars{x - \xi}\pars{-1}} \over 4 \times \pars{-1}^{3}} + {\expo{\pars{x - \xi}\pars{-\ic}} \over 4 \times \pars{-\ic}^{3}}}}\dd\xi \\[5mm] & = \int_{0}^{x}\mrm{f}\pars{\xi} \bracks{{\expo{x - \xi} - \expo{-\pars{x - \xi}}\over 4} - {\expo{\pars{x - \xi}\ic} - \expo{-\pars{x - \xi}\ic} \over 4\ic}}\dd\xi \\[5mm] & = \bbx{{1 \over 2}\int_{0}^{x}\mrm{f}\pars{\xi} \bracks{\vphantom{\Large A}\sinh\pars{x - \xi} - \sin\pars{x - \xi}}\dd\xi} \end{align}