Question

laplace transform

y'+2y=e^-t, y(0)=1

Answer 1

Take the Laplace transform of both sides:
\[\mathscr{L}\left\{y'+2y\right\}=\mathscr{L}\left\{e^{-t}\right\}\]
Recall \(\mathscr{L}\left\{y'\right\}=s\mathscr{L}\left\{y\right\}-y(0)\):
\[s\mathscr{L}\left\{y\right\}-y(0)+2\mathscr{L}\left\{y\right\}=\mathscr{L}\left\{e^{-t}\right\}\\
(s+2)\mathscr{L}\left\{y\right\}-1=\frac{1}{s+1}\\
\mathscr{L}\left\{y\right\}=\frac{\frac{1}{s+1}+1}{s+2}\\
\mathscr{L}\left\{y\right\}=\frac{\frac{s+2}{s+1}}{s+2}\\
\mathscr{L}\left\{y\right\}=\frac{1}{s+1}\\
y=\mathscr{L}^{-1}\left\{\frac{1}{s+1}\right\}\]

Answer 2

thank u