- #1
karush
Gold Member
MHB
- 3,269
- 5
$\begin{align*}\displaystyle
y'&=y-5\quad y(0)=y_0\tag{given}\\
y'-y&=-5\\
u(x)&=\exp\int-1\, dx = e^{-t}\\
(e^{-t}y)&=-5e^{-t}\\
e^{-t}y&=-5\int e^{-t} dt = -5e^{-t}+c\\
&y=-5\frac{e^{-t}}{e^{-t}}+\frac{c}{e^{-t}}\\
y&=\color{red}{5+(y_0-5)e^t}
\end{align*}$
this is similar to one I posted before but can't seem to get the book answer (red)
the $y_0$ is ?
y'&=y-5\quad y(0)=y_0\tag{given}\\
y'-y&=-5\\
u(x)&=\exp\int-1\, dx = e^{-t}\\
(e^{-t}y)&=-5e^{-t}\\
e^{-t}y&=-5\int e^{-t} dt = -5e^{-t}+c\\
&y=-5\frac{e^{-t}}{e^{-t}}+\frac{c}{e^{-t}}\\
y&=\color{red}{5+(y_0-5)e^t}
\end{align*}$
this is similar to one I posted before but can't seem to get the book answer (red)
the $y_0$ is ?