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Someone set us up the bomb (again)
Find solution of initial value problems y ″ + y = u 3 π ( t ) {\displaystyle y''+y=u_{3\pi }(t)} , where y ( 0 ) = 0 {\displaystyle y(0)=0} and y ′ ( 0 ) = 1 {\displaystyle y'(0)=1}
L { y ″ } + L { y } = L { 1 − u 3 π ( t ) } s 2 L { y } − y ′ ( 0 ) − s y ( 0 ) + L { y } = 1 − e − 3 π s s ( s 2 + 1 ) L { y } − 1 = 1 − e − 3 π s s L { y } = 1 s ( s 2 + 1 ) − e − 3 π s s ( s 2 + 1 ) = L { 1 − cos t + sin t } − e − 3 π s L { h ( t ) } = L { 1 − cos t + sin t } − L { u 3 π ( t ) ( 1 − cos ( t − 3 π ) ) } y = 1 − cos t + sin t − u 3 π ( t ) ( 1 + cos t ) = { 1 − cos t + sin t t < 3 π sin t − 2 cos t t ≥ 3 π {\displaystyle {\begin{aligned}{\mathcal {L}}\left\{y''\right\}+{\mathcal {L}}\left\{y\right\}&={\mathcal {L}}\left\{1-u_{3\pi }(t)\right\}\\s^{2}{\mathcal {L}}\left\{y\right\}-y'(0)-s\,y(0)+{\mathcal {L}}\left\{y\right\}&={\frac {1-\mathrm {e} ^{-3\pi \,s}}{s}}\\(s^{2}+1){\mathcal {L}}\left\{y\right\}-1&={\frac {1-\mathrm {e} ^{-3\pi \,s}}{s}}\\{\mathcal {L}}\left\{y\right\}&={\frac {1}{s(s^{2}+1)}}-{\frac {\mathrm {e} ^{-3\pi \,s}}{s(s^{2}+1)}}\\&={\mathcal {L}}\left\{1-\cos {t}+\sin {t}\right\}-\mathrm {e} ^{-3\pi \,s}\,{\mathcal {L}}\left\{h(t)\right\}\\&={\mathcal {L}}\left\{1-\cos {t}+\sin {t}\right\}-{\mathcal {L}}\left\{u_{3\pi }(t)\,\left(1-\cos {\left(t-3\pi \right)}\right)\right\}\\y&=1-\cos {t}+\sin {t}-u_{3\pi }(t)\,(1+\cos {t})\\&={\begin{cases}1-\cos {t}+\sin {t}&t<3\pi \\\sin {t}-2\cos {t}&t\geq 3\pi \end{cases}}\end{aligned}}}
y ″ + 3 y ′ + 2 y = u 2 ( t ) y ( 0 ) = 0 y ′ ( 0 ) = 1 {\displaystyle {\begin{aligned}y''+3y'+2y&=u_{2}(t)&y(0)&=0&y'(0)&=1\end{aligned}}}
L { y ″ } + 3 L { y ′ } + 2 L { y } = L { 1 ⋅ u 2 ( t ) } s 2 L { y } − 1 + 3 s L { y } − 0 + 2 L { y } = e − 2 s s ( s 2 + 3 s + 2 ) L { y } = 1 + e − 2 s s L { y } = 1 ( s + 2 ) ( s + 1 ) + e − 2 s s ( s + 2 ) ( s + 1 ) = L { e − t − e − 2 t } + e − 2 s ( 1 2 L { 1 } + 1 2 L { e − 2 t } − L { e − t } ) y = e − t − e − 2 t + u 2 ( t ) ( 1 2 + e − 2 t 2 − e − t + 2 ) {\displaystyle {\begin{aligned}{\mathcal {L}}\left\{y''\right\}+3{\mathcal {L}}\left\{y'\right\}+2{\mathcal {L}}\left\{y\right\}&={\mathcal {L}}\left\{1\cdot u_{2}(t)\right\}\\s^{2}\,{\mathcal {L}}\left\{y\right\}-1+3s\,{\mathcal {L}}\left\{y\right\}-0+2{\mathcal {L}}\left\{y\right\}&={\frac {\mathrm {e} ^{-2s}}{s}}\\\left(s^{2}+3s+2\right)\,{\mathcal {L}}\left\{y\right\}&=1+{\frac {\mathrm {e} ^{-2s}}{s}}\\{\mathcal {L}}\left\{y\right\}&={\frac {1}{(s+2)(s+1)}}+{\frac {\mathrm {e} ^{-2s}}{s(s+2)(s+1)}}\\&={\mathcal {L}}\left\{\mathrm {e} ^{-t}-\mathrm {e} ^{-2t}\right\}+\mathrm {e} ^{-2s}\,\left({\frac {1}{2}}\,{\mathcal {L}}\left\{1\right\}+{\frac {1}{2}}\,{\mathcal {L}}\left\{\mathrm {e} ^{-2t}\right\}-{\mathcal {L}}\left\{\mathrm {e} ^{-t}\right\}\right)\\y&=\mathrm {e} ^{-t}-\mathrm {e} ^{-2t}+u_{2}(t)\,\left({\frac {1}{2}}+{\frac {\mathrm {e} ^{-2t}}{2}}-\mathrm {e} ^{-t+2}\right)\end{aligned}}}
h ( t ) = y ″ + 2 y ′ + 2 y y ( 0 ) = 0 y ′ ( 0 ) = 1 h ( t ) = { 0 0 < t < π 1 π ≤ t < 2 π 0 2 π < t {\displaystyle {\begin{aligned}h(t)&=y''+2y'+2y&y(0)&=0&y'(0)&=1\\h(t)&={\begin{cases}0&0<t<\pi \\1&\pi \leq t<2\pi \\0&2\pi <t\end{cases}}\end{aligned}}}
Rewrite h ( t ) = 0 + u π ( t ) ( 1 − 0 ) + u 2 π ( t ) ( 0 − 1 ) = u π ( t ) − u 2 π ( t ) {\displaystyle h(t)=0+u_{\pi }(t)\,\left(1-0\right)+u_{2\pi }(t)\,\left(0-1\right)=u_{\pi }(t)-u_{2\pi }(t)}
L { y ″ } + 2 L { y ′ } + 2 L { y } = L { u π ( t ) − u 2 p i ( t ) } s 2 L { y } − 1 + 2 s L { y } − 0 + 2 L { y } = e − π s s − e − 2 π s s ( s 2 + 2 s + 2 ) L { y } = 1 + e − π s s − e − 2 π s s L { y } = 1 ( s 2 + 1 ) 2 + 1 + ( e − π s − e − 2 π s ) ( 1 2 s − 1 2 ( s + 2 ( s + 1 ) 2 + 1 ) ) = L { e − t sin t } + ( e − π s − e − 2 π s ) ( L { 1 2 − 1 2 e − t cos t + e − t sin t } ) = L { e − t sin t } + L { u π ( t ) ( 1 2 − 1 2 ( e − ( t − π ) cos ( t − π ) + e − ( t − π ) sin ( t − π ) ) ) } − L { u 2 π ( t ) ( 1 2 − 1 2 ( e − ( t − 2 π ) cos ( t − 2 π ) + e − ( t − 2 π ) sin ( t − 2 π ) ) ) } y = e − t sin t + u π ( t ) ( 1 2 − 1 2 ( e − ( t − π ) cos ( t − π ) + e − ( t − π ) sin ( t − π ) ) ) − u 2 π ( t ) ( 1 2 − 1 2 ( e − ( t − 2 π ) cos ( t − 2 π ) + e − ( t − 2 π ) sin ( t − 2 π ) ) ) {\displaystyle {\begin{aligned}{\mathcal {L}}\left\{y''\right\}+2{\mathcal {L}}\left\{y'\right\}+2{\mathcal {L}}\left\{y\right\}&={\mathcal {L}}\left\{u_{\pi }(t)-u_{2pi}(t)\right\}\\s^{2}\,{\mathcal {L}}\left\{y\right\}-1+2s\,{\mathcal {L}}\left\{y\right\}-0+2{\mathcal {L}}\left\{y\right\}&={\frac {\mathrm {e} ^{-\pi \,s}}{s}}-{\frac {\mathrm {e} ^{-2\pi \,s}}{s}}\\\left(s^{2}+2s+2\right)\,{\mathcal {L}}\left\{y\right\}&=1+{\frac {\mathrm {e} ^{-\pi \,s}}{s}}-{\frac {\mathrm {e} ^{-2\pi \,s}}{s}}\\{\mathcal {L}}\left\{y\right\}&={\frac {1}{(s^{2}+1)^{2}+1}}+\left(\mathrm {e} ^{-\pi \,s}-\mathrm {e} ^{-2\pi \,s}\right)\,\left({\frac {1}{2s}}-{\frac {1}{2}}\,\left({\frac {s+2}{(s+1)^{2}+1}}\right)\right)\\&={\mathcal {L}}\left\{\mathrm {e} ^{-t}\,\sin {t}\right\}+\left(\mathrm {e} ^{-\pi \,s}-\mathrm {e} ^{-2\pi \,s}\right)\,\left({\mathcal {L}}\left\{{\frac {1}{2}}-{\frac {1}{2}}\,\mathrm {e} ^{-t}\,\cos {t}+\mathrm {e} ^{-t}\,\sin {t}\right\}\right)\\&={\mathcal {L}}\left\{\mathrm {e} ^{-t}\,\sin {t}\right\}+{\mathcal {L}}\left\{u_{\pi }(t)\,\left({\frac {1}{2}}-{\frac {1}{2}}\,\left(\mathrm {e} ^{-(t-\pi )}\,\cos {(t-\pi )}+\mathrm {e} ^{-(t-\pi )}\,\sin {(t-\pi )}\right)\right)\right\}-{\mathcal {L}}\left\{u_{2\pi }(t)\,\left({\frac {1}{2}}-{\frac {1}{2}}\,\left(\mathrm {e} ^{-(t-2\pi )}\,\cos {(t-2\pi )}+\mathrm {e} ^{-(t-2\pi )}\,\sin {(t-2\pi )}\right)\right)\right\}\\y&=\mathrm {e} ^{-t}\,\sin {t}+u_{\pi }(t)\,\left({\frac {1}{2}}-{\frac {1}{2}}\,\left(\mathrm {e} ^{-(t-\pi )}\,\cos {(t-\pi )}+\mathrm {e} ^{-(t-\pi )}\,\sin {(t-\pi )}\right)\right)-u_{2\pi }(t)\,\left({\frac {1}{2}}-{\frac {1}{2}}\,\left(\mathrm {e} ^{-(t-2\pi )}\,\cos {(t-2\pi )}+\mathrm {e} ^{-(t-2\pi )}\,\sin {(t-2\pi )}\right)\right)\end{aligned}}}
, where 
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