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

• MHB
• karush
In summary, the author suggests that there is an undetermined constant, $e^E$, which is greater than zero for any E.
karush
Gold Member
MHB
$\tiny{b.1.2.2a}$
$\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:

Last edited:
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!

Country Boy said:
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

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

Country Boy said:
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

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..

Get busy!

## 1. What is the meaning of $\dfrac{dy}{dt}=y-5$ in this context?

This is a differential equation that represents the rate of change of a function $y$ with respect to time $t$. The function $y$ is equal to itself minus 5, indicating that the rate of change is dependent on the current value of $y$.

## 2. How do I solve this differential equation?

This is a first-order linear differential equation, which can be solved using the method of separation of variables. This involves isolating the $y$ and $t$ terms on opposite sides of the equation and then integrating both sides. The initial condition $y(0)=y_0$ can then be used to solve for the constant of integration.

## 3. What is the significance of the initial condition $y(0)=y_0$?

The initial condition represents the starting point of the function $y$ at time $t=0$. It is used to determine the specific solution to the differential equation, as the constant of integration obtained from the integration process will be different for different initial conditions.

## 4. Can this differential equation be solved analytically?

Yes, this differential equation can be solved analytically using the method of separation of variables. However, for more complex differential equations, it may not always be possible to find an analytical solution and numerical methods may be used instead.

## 5. What are some real-life applications of this type of differential equation?

This type of differential equation can be used to model various physical and biological phenomena, such as population growth, chemical reactions, and electrical circuits. It can also be used in economics and finance to model changes in interest rates or stock prices over time.

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