transformed of Laplace

Question

$$y'+2y=\left\{\begin{array}{ll}t, &0\leq t< 1\\0,&t\geq1\end{array}\right.;\hspace{5mm}y(0)=0$$ what I do not understand is, at the moment of applying the transformed, $$\mathscr{L}\{t-t\mathscr{U}(t-1)\}$$ I do not understand how to determine Laplace's transformation of the previous result

Answer 1

$$y'+2y=\left\{\begin{array}{ll}t, &0\leq t< 1\\0,&t\geq1\end{array}\right.;\hspace{5mm}y(0)=0$$ For $0 \le t\le 1$ $$y'+2y=t$$ $$(ye^{2t})'=te^{2t}$$ $$y=e^{-2t}\int te^{2t}dt$$ $$y=e^{-2t}(te^{2t}/2-e^{2t}/4 +K)$$ Since $y(0)=0 \to K=\dfrac 14$ $$\boxed{\text{ 1) }y(t)=\frac t2-\frac 14 +\frac {e^{-2t}}4}$$

For $t \ge 1$ : $$y'+2t=0$$ $$(ye^{2t})'=0$$ $$y=K_2e^{-2t}$$ We must have from first equation we deduce the value of y(1): $$\text{ 1) }y(t)=\frac t2-\frac 14 +\frac {e^{-2t}}4$$ $$y(1)=\frac {e^{-2}+1}{4}$$ Then we deduce the value of the constante $K_2$: $$ y=K_2e^{-2t} \to y(1)=K_2e^{-2} \to K_2=\frac {e^{2}+1}{4}$$ Therefore for $t \ge 1$ $$\boxed{y(t)=\left(\frac {e^{2}+1}{4}\right)e^{-2t}}$$

Answer 2

Let $f(t)=\left\{\begin{array}{ll}t, &0\leq t< 1\\0,&t\geq1\end{array}\right.$. Then the DE becomes $$ y'+2y=f(t) $$ and hence $$ \mathscr{L}\{y'\}+2\mathscr{L}\{y\}=\mathscr{L}\{f(t)\}. \tag{1}$$ Note $$ \mathscr{L}\{y'\}=sY(s)-y(0),\mathscr{L}\{y\}=Y(s) $$ and $$ \mathscr{L}\{f(t)\}=\int_0^\infty e^{-st}f(t)dt=\int_0^1e^{-st}tdt=\frac{1-e^{-s}(1+s)}{s^2}. $$ So (1) becomes $$ sY(s)-2Y(s)=\frac{1-e^{-s}(1+s)}{s^2} $$ and hence $$ Y(s)=\frac{e^{-s}(e^s-s-1)}{s^2(s+2)}=\frac{1}{s^2(s+2)}-\frac{e^{-s}(s+1)}{s^2(s+2)}. $$ Note $$ \mathscr{L}^{-1}\left\{\frac{1}{s^2(s+2)}\right\}=\frac14(e^{-2t}+2t-1). $$ Let $$ G(s)=\frac{s+1}{s^2(s+2)} $$ and then $g(t)=\mathscr{L}^{-1}\{G(s)\}=\frac14(-e^{-2t}+2t+1)$ and $$ \mathscr{L}^{-1}\left\{\frac{e^{-s}(s+1)}{s^2(s+2)}\right\}=\mathscr{L}^{-1}\left\{e^{-s}G(s)\right\}=u(t-1)g(t-1)=\frac14u(t-1)(-e^{-2(t-1)}+2t-1) $$ Thus $$ y(t)=\mathscr{L}^{-1}\left\{\frac{1}{s^2(s+2)}\right\}-\mathscr{L}^{-1}\left\{\frac{e^{-s}(s+1)}{s^2(s+2)}\right\}=\frac14(e^{-2t}+2t-1)-\frac14u(t-1)(-e^{-2(t-1)}+2t-1).$$