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[SOLVED] 1.2.2a initial value problem

karush

Well-known member
Jan 31, 2012
2,886
$\begin{align*}\displaystyle
y'&=y-5\quad y(0)=y_0\tag{given}\\
y'-y&=-5\\
u(x)&=\exp\int-1\, dx = e^{-t}\\
(e^{-t}y)&=-5e^{-t}\\
e^{-t}y&=-5\int e^{-t} dt = -5e^{-t}+c\\
&y=-5\frac{e^{-t}}{e^{-t}}+\frac{c}{e^{-t}}\\
y&=\color{red}{5+(y_0-5)e^t}
\end{align*}$
this is similiar to one I posted before but can't seem to get the book answer (red)
the $y_0$ is ???
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\begin{align*}\displaystyle
y'&=y-5\quad y(0)=y_0\tag{given}\\
y'-y&=-5\\
u(x)&=\exp\int-1\, dx = e^{-t}\\
(e^{-t}y)&=-5e^{-t}\\
e^{-t}y&=-5\int e^{-t} dt = -5e^{-t}+c\\
&y=-5\frac{e^{-t}}{e^{-t}}+\frac{c}{e^{-t}}\\
y&=\color{red}{5+(y_0-5)e^t}
\end{align*}$
this is similiar to one I posted before but can't seem to get the book answer (red)
the $y_0$ is ???
Note that:

\(\displaystyle -5\int e^{-u}\,du=5e^{-u}+C\)

And so after integrating, you should have:

\(\displaystyle e^{-t}y=5e^{-t}+C\)

Hence:

\(\displaystyle y(t)=5+Ce^{t}\)

Now, we are given:

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

And so:

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

Does that make sense?
 

karush

Well-known member
Jan 31, 2012
2,886
Ok

I had the C thing not happening


I noticed we get a lot of views on these!