Thanks to everyone who will bare with me and read this attempt and further validate it correct and improve upon it or otherwise correct its wrong.
The question: $$ y''(t)+2y'(t)+y(t)=u\left( t-\frac { \pi }{ 3 } \right) sin(t) $$
While y(0)=0 and y'(0)=1
Some notation:
$$ u\left( t-\frac { \pi }{ 3 } \right) \left\{ 0\quad ,\quad t\quad <\frac {\pi }{3}; 1\quad,\quad t\quad>\frac { \pi }{ 3 } \right\} $$
Attempt:
Using trigonometry:
sin(t-pi/3)=sin(t)cos(pi/3)-cos(t)sin(pi/3)
therefore:
2sin(t-pi/3)=sqrt(3)sin(t)-cos(t) eqn. 1
also since:
cos(t-pi/3)=cos(t)cos(pi/3)-sin(t)sin(pi/3)
then:
2cos(t-pi/3)=sqrt(3)cos(t)-sin(t) eqn. 2
now we calculate eqn. 2 +sqrt(3)*eqn. 1
2sqrt(3)sin(t-pi/3)+2cos(t-pi/3)=3sin(t)+sin(t)
Finally we arrive at the conclusion that:
sin(t)=(sqrt(3)/2)sin(t-pi/3)+0.5cos(t-pi/3)..... p=sqrt(3)/2
so
$$ y''(t)+2y'(t)+y(t)=u\left( t-\frac { \pi }{ 3 } \right) \left( \frac { \sqrt { 3 } }{ 2 } \sin { \left( t-\frac { \pi }{ 3 } \right) } +\frac { 1 }{ 2 } \cos { \left( t-\frac { \pi }{ 3 } \right) } \right) $$ $$ L \left( y''(t)+2y'(t)+y(t)=u\left( t-\frac { \pi }{ 3 } \right) \left( \frac { \sqrt { 3 } }{ 2 } \sin { \left( t-\frac { \pi }{ 3 } \right) } +\frac { 1 }{ 2 } \cos { \left( t-\frac { \pi }{ 3 } \right) } \right) \right) $$
$$ s^{ 2 }Y-sy(0)-y'(0)+2sY-2y(0)+Y={ e }^{ -\frac { \pi }{ 3 } s }\left( \frac { \sqrt { 3 } }{ 2 } \left( \frac { 1 }{ { s }^{ 2 }+1 } \right) +\frac { 1 }{ 2 } \left( \frac { s }{ { s }^{ 2 }+1 } \right) \right) $$
We can simplify: $$ Y={ e }^{ -\frac { \pi }{ 3 } s }\left( \frac { \sqrt { 3 } }{ 2 } \left( \frac { 1 }{ \left( { s }^{ 2 }+1 \right) { \left( s+1 \right) }^{ 2 } } \right) +\frac { 1 }{ 2 } \left( \frac { s }{ \left( { s }^{ 2 }+1 \right) { \left( s+1 \right) }^{ 2 } } \right) \right) +\frac { 1 }{ { \left( s+1 \right) }^{ 2 } } $$
Using partial fractions methods we can derive that: $$ Y={ e }^{ -\frac { \pi }{ 3 } s }\left( \frac { \sqrt { 3 } }{ 2 } \left( \frac { 1 }{ 2 } \left( \frac { 1 }{ { \left( s+1 \right) }^{ 2 } } \right) +\frac { 1 }{ 2 } \left( \frac { 1 }{ s+1 } \right) -\frac { 1 }{ 2 } \left( \frac { s }{ { s }^{ 2 }+1 } \right) \right) +\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } \left( \frac { 1 }{ { s }^{ 2 }+1 } \right) -\frac { 1 }{ 2 } \left( \frac { 1 }{ { \left( s+1 \right) }^{ 2 } } \right) \right) \right) +\frac { 1 }{ { \left( s+1 \right) }^{ 2 } } $$
Finally (unnnnnnnhhhhh) we can inverse Laplace to get:
$$ y(t)={ e }^{ -t }t+u\left( t-\frac { \pi }{ 3 } \right) \left( \frac { \sqrt { 3 } }{ 4 } \left( { e }^{ -\left( t-\frac { \pi }{ 3 } \right) }+{ e }^{ -\left( t-\frac { \pi }{ 3 } \right) }\left( t-\frac { \pi }{ 3 } \right) +\cos { \left( t-\frac { \pi }{ 3 } \right) } \right) +\frac { 1 }{ 4 } \left( \sin { \left( t-\frac { \pi }{ 3 } \right) -{ e }^{ -\left( t-\frac { \pi }{ 3 } \right) }\left( t-\frac { \pi }{ 3 } \right) } \right) \right) $$
Alright, I couldn't find anything to check my work so could someone help me please? thanks in advance again your help is appreciated.
P.S.:(1)This material is new to me and I am no expert, hence why I am asking. (2!)I apologise in advance for my lack of skill in using this site's math-typing function and formats. (3)I apologise for all the grammer and spelling mistakes above.
First, \begin{align} \mathscr{L}\{y'\} & = \int_{0}^{\infty}e^{-st}y'(t)dt \\ & = e^{-st}y(t)|_{t=0}^{\infty}+s\int_{0}^{\infty}e^{-st}y(t)dt \\ & = -y(0)+s\mathscr{L}\{y\} = s\mathscr{L}\{y\} \end{align} Then, using the above, the transform of the second derivative is \begin{align} \mathscr{L}\{y''\} & =\int_{0}^{\infty}e^{-st}y''(t)dt \\ & = e^{-st}y'(t)|_{t=0}^{\infty}+s\int_{0}^{\infty}e^{-st}y'(t)dt \\ & = -y'(0)+s\mathscr{L}\{y'\} \\ & = -1 + s^{2}\mathscr{L}\{y\} \end{align} Therefore, \begin{align} \mathscr{L}\{y''+2y'+y\} & =s^{2}\mathscr{L}\{y\}+2s\mathscr{L}\{y\})+\mathscr{L}\{y\}-1 \\ & = (s+1)^{2}\mathscr{L}\{y\}-1. \end{align} And this must equal \begin{align} \mathscr{L}\left\{u(t-\frac{\pi}{3})\sin(t)\right\} & = \int_{\pi/3}^{\infty}e^{-st}\sin(t)dt \\ & = \int_{\pi/3}^{\infty}\frac{e^{-st+it}-e^{-st-it}}{2i}dt \\ & = \frac{1}{2i}\left[\frac{1}{i-s}e^{-st+it}-\frac{1}{i+s}e^{-st-it}\right]_{t=\pi/3}^{\infty} \\ & = \frac{e^{-s\pi/3}}{2i}\left[\frac{e^{i\pi/3}}{s-i}+\frac{e^{-i\pi/3}}{s+i}\right] \\ & = \frac{e^{-s\pi/3}}{2i}\left[\frac{\frac{1}{2}+i\frac{\sqrt{3}}{2}}{s-i}+\frac{\frac{1}{2}-i\frac{\sqrt{3}}{2}}{s+i}\right] \\ & = \frac{e^{-s\pi/3}}{2i}\left[\frac{s-\sqrt{3}}{(s-i)(s+i)}\right] \end{align} Therefore, $$ \mathscr{L}\{y\}=\frac{1}{(s+1)^{2}}+\frac{e^{-s\pi/3}}{2i(s+1)^{2}}\left[\frac{s-\sqrt{3}}{(s-i)(s+i)}\right]. $$ We have slightly different Laplace transforms.