MHB Understanding Initial Value Problems: Solving for y in y' = y-5 with y(0) = y0

[karush]

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

 

[MarkFL]

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

Note that:

$$-5\int e^{-u}\,du=5e^{-u}+C$$

And so after integrating, you should have:

$$e^{-t}y=5e^{-t}+C$$

Hence:

$$y(t)=5+Ce^{t}$$

Now, we are given:

$$y(0)=5+C=y_0\implies C=y_0-5$$

And so:

$$y(t)=5+(y_0-5)e^{t}$$

Does that make sense?

 

[karush]

Ok

I had the C thing not happeningI noticed we get a lot of views on these!