Question

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Answer 1

\[\boxed{
\newcommand \intl[4] {\int\limits_{#1}^{#2}{#3}{\dd #4}}% integral_a^b{f(x)}dx
\newcommand \Lap [1] {\operatorname{\mathcal L}\left\{\rule{0pt}{2.2ex}#1\right\}} % Laplace transform (p)
\newcommand \Lin [1] {\operatorname{\mathcal L}^{-1}\left\{#1\right\}} % inverse Laplace transform
\newcommand \Ev [3] {\left.{#1}\right|_{#2}^{#3}} % evaluation limits
\newcommand \ps [1] {\left(#1\right)} % dynamic parentheses
}\]

\[
\begin{align}
\intl0x{y(u)}u-y'(x)
&= x\\
\Lap{1\star y}-\Lap{y'}
&= X\\
\frac1p\cdot Y - \ps{pY-y_0}
&=\\
\ps{\frac1p-p}Y
&= X-y_0\\
Y &= \ps{X-y_0}{\frac1{\frac1p-p}}\\
&= \ps{y_0-X}{\frac p{p^2-1}}\\
y(x)&= \ps{y_0\delta(x)-x}\star\cosh x\\
&= \intl0x{(y_0\delta(x-u)-(x-u))\cosh u}u\\
&= y_0\intl0x{\delta(x-u)\cosh u}u-x\intl0x{\cosh u}u
+ \intl0x{u\cosh u}u\\
&= y_0\Ev{\cosh u}{}x-x\Ev{\sinh u}0x
+ \Ev{u\sinh u}0x-\intl0x{\sinh u}u\\
&= y_0\cosh x-x\sinh x
+ x\sinh x-\Ev{\cosh u}0x\\
&= y_0\cosh x - \cosh x + 1\\
&= \ps{y_0-1}\cosh x+1
\end{align}
\]

Answer 2

\[\boxed{
\newcommand \intl[4] {\int\limits_{#1}^{#2}{#3}{\dd #4}}% integral_a^b{f(x)}dx
\newcommand \Lap [1] {\operatorname{\mathcal L}\left\{\rule{0pt}{2.2ex}#1\right\}} % Laplace transform (p)
\newcommand \Lin [1] {\operatorname{\mathcal L}^{-1}\left\{#1\right\}} % inverse Laplace transform
\newcommand \Ev [3] {\left.{#1}\right|_{#2}^{#3}} % evaluation limits
\newcommand \ps [1] {\left(#1\right)} % dynamic parentheses
}
\begin{align}
\intl0x{y(u)}u-y'(x)
= x\\
\Lap{1\star y}-\Lap{y'}
= X\\
\frac1p\cdot Y - \ps{pY-y_0}
=\\
\ps{\frac1p-p}Y
= X-y_0\\
Y = \ps{X-y_0}{\frac1{\frac1p-p}}\\
= \ps{y_0-X}{\frac p{p^2-1}}\\
y(x)= \ps{y_0\delta(x)-x}\star\cosh x\\
= \intl0x{(y_0\delta(x-u)-(x-u))\cosh u}u\\
= y_0\intl0x{\delta(x-u)\cosh u}u-x\intl0x{\cosh u}u
+ \intl0x{u\cosh u}u\\
= y_0\Ev{\cosh u}{}x-x\Ev{\sinh u}0x
+ \Ev{u\sinh u}0x-\intl0x{\sinh u}u\\
= y_0\cosh x-x\sinh x
+ x\sinh x-\Ev{\cosh u}0x\\
= y_0\cosh x - \cosh x + 1\\
= \ps{y_0-1}\cosh x+1
\end{align}\]

Answer 3

\[\boxed{
\newcommand \dd [1] {\,\mathrm d#1} % infinitesimal
\newcommand \intl[4] {\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral_a^b{f(x)}dx
\newcommand \Lap [1] {\operatorname{\mathcal L}\left\{#1\right\}} % Laplace transform (p)
\newcommand \Lin [1] {\operatorname{\mathcal L}^{-1}\left\{#1\right\}} % inverse Laplace transform
\newcommand \Ev [3] {\left.{#1}\right|_{#2}^{#3}} % evaluation limits
\newcommand \ps [1] {\left(#1\right)} % dynamic parentheses
}\]

\[
\begin{align}
\intl0x{y(u)}u-y'(x)
&= x\\
\Lap{1\star y}-\Lap{y'}
&= X\\
\frac1p\cdot Y - \ps{pY-y_0}
&=\\
\ps{\frac1p-p}Y
&= X-y_0\\
Y &= \ps{X-y_0}{\frac1{\frac1p-p}}\\
&= \ps{y_0-X}{\frac p{p^2-1}}\\
y(x)&= \ps{y_0\delta(x)-x}\star\cosh x\\
&= \intl0x{(y_0\delta(x-u)-(x-u))\cosh u}u\\
&= y_0\intl0x{\delta(x-u)\cosh u}u-x\intl0x{\cosh u}u
+ \intl0x{u\cosh u}u\\
&= y_0\Ev{\cosh u}{}x-x\Ev{\sinh u}0x
+ \Ev{u\sinh u}0x-\intl0x{\sinh u}u\\
&= y_0\cosh x-x\sinh x
+ x\sinh x-\Ev{\cosh u}0x\\
&= y_0\cosh x - \cosh x + 1\\
&= \ps{y_0-1}\cosh x+1
\end{align}
\]