Solving $\dfrac{dy}{dt}=y-5$, $y(0)=y_0$

[karush]

$\tiny{b.1.2.2a}$
$\dfrac{dy}{dt}=y-5, \quad y(0)=y_0$$\begin{array}{rll}\displaystyle
y'&=y-5\\
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\dfrac{e^{-t}}{e^{-t}}+\dfrac{c}{e^{-t}}&=-5e+ce^t\\
y(0)&=5+C=y_0\\
\implies C&=y_0-5\\
y&=5+(y_0-5)e^t
\end{array}$

ok I think this is correct... wasn't sure about the 5 scalar
typo's maybe :unsure:

 

[HOI]

I would have done this slightly differently:
From $\frac{dy}{dt}= y- 5$, $\frac{dy}{y- 5}= dt$.

To integrate on the left, let u= y- 5 so du= dy.
$\int\frac{dy}{y- 5}= \int \frac{du}{u}= ln(u)+C= ln(y- 5)+ C$.

$ln(y- 5)+ C= t+ D$ so $ln(y- 5)= t+E$ where E= D- C.

Taking the exponential of both sides, $y- 5= e^{t+ E}= Fe^t$ where $F=e^E$.

$y(x)= Fe^t+ 5$. Since $y(0)= F+ 5= y_0$, $F= y_0- 5$ so $y(x)= (y_0- 5)e^t+ 5$,

That is exactly what you have. Well done!

 

[karush]

I would have done this slightly differently:
From $\frac{dy}{dt}= y- 5$, $\frac{dy}{y- 5}= dt$.

To integrate on the left, let u= y- 5 so du= dy.
$\int\frac{dy}{y- 5}= \int \frac{du}{u}= ln(u)+C= ln(y- 5)+ C$.

$ln(y- 5)+ C= t+ D$ so $ln(y- 5)= t+E$ where E= D- C.

Taking the exponential of both sides, $y- 5= e^{t+ E}= Fe^t$ where $F=e^E$.

$y(x)= Fe^t+ 5$. Since $y(0)= F+ 5= y_0$, $F= y_0- 5$ so $y(x)= (y_0- 5)e^t+ 5$,

That is exactly what you have. Well done!

Well yeah that makes more sense... Haven't see the
$Fe^t$ where $F=e^E$. before

I started to make a PDF of the MHB DE replies
thot I would try to get some input ...
my goal is get a 100 problems listed
also I have a counter that adds up the views
I have acknowledged those who reply
Mahalo

https://dl.orangedox.com/wr9JnddSXWGHrASWF2

 

[HOI]

E is just an undetermined constant so $e^E$ is just another undetermined constant.

 

[topsquark]

E is just an undetermined constant so $e^E$ is just another undetermined constant.

However, notice that we may have \(\displaystyle E \leq 0\) but \(\displaystyle e^E > 0\) for any E. So they aren't quite equivalent.

-Dan

 

[karush]

well that's what I like about MHB
it continues where the textbook left off

there is still a lot of these IVP i need to do..

 

[HOI]

Get busy!