• jwqwerty
In summary, the conversation discusses how to solve the differential equation dy/dt = y-b by integrating both sides and finding a general solution. However, there is a debate about whether to replace the constant c1 with a new constant c2, as it can take on a negative, positive, or zero value. It is argued that c1 cannot be 0, but this would mean that the solution misses the possibility of y= b for all t. Ultimately, it is determined that the value of c1 (or c2) depends on additional information provided in the problem.
jwqwerty
when i solve dy/dt= y-b

(1/y-b)(dy/dt)=1
d(ln│y-b│)/dt=1

when i integrate both sides respect to t,
ln│y-b│=t+c (c is a constant)
y=±e^(at+c)+b
=±c1*e^at + b (c1 is a constant)

then the book replaces ±c1 with c2 (constant)
but isn't it wrong to do so? Because c2 can't show that it can have two answers.

There aren't "two solutions". There is a single "general solution" which, for some given "initial value", gives a single solution. $e^{at+c}= e^c e^{at}$. Since c is an arbitrary constant, so is $e^c$ except that it must be positive since e to any power is positive- your "$c_1$ must be a positive number. Putting "$\pm$" gets you negative values but is still not quite enough since $\pm ce^{at}$ still cannot be 0. Yet, y= b for all t certainly is a solution so your $y(t)= \pm c_1e^{at}+b$ is not complete- it misses y identically 0. That can be fixed by replacing "$\pm c_1$" with $c_2$ which can be a postive, negative, or 0.

HallsofIvy said:
That can be fixed by replacing "$\pm c_1$" with $c_2$ which can be a postive, negative, or 0.

So do you mean that c1 means a negative number or positive number or 0 and identiying only one of them? By the way i do not get why c1 can be 0. since c1= e^c1 >0, c1 can't be 0.

Last edited:
jwqwerty said:
So do you mean that c1 means a negative number or positive number or 0 and identiying only one of them?
I mean that c1 can be any single real number, negative, positive, or 0. Which number depends on what additional information (such as an initial value) you are given.

By the way i do not get why c1 can be 0. since c1= e^c1 >0, c1 can't be 0.
Yes, that is exactly what is wrong with your solution. You can't have c1 equal to 0 so you cannot get y= b. But y= b for all x clearly is a solution.

It is not necessarily wrong to replace ±c1 with c2 in this case. The constant c2 represents the combination of both c1 and the ± sign, so it can still represent the two possible solutions of the equation. However, it is important to note that c1 and c2 may not necessarily have the same numerical value. In this case, c2 may have a different numerical value for each of the two solutions. Additionally, it is important to specify the ± sign when using c2 to avoid confusion and ensure that both solutions are accounted for.

## What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives. It is used to describe the relationship between a quantity and how it changes over time or space.

## What is a simple differential equation?

A simple differential equation is a differential equation that involves only one independent variable and its derivatives. It is often used to model simple physical systems and can be solved analytically.

## What is the order of a differential equation?

The order of a differential equation is the highest derivative present in the equation. For example, a first-order differential equation contains only first derivatives, while a second-order differential equation contains second derivatives.

## What are initial conditions in a differential equation?

Initial conditions in a differential equation refer to the values of the dependent variable and its derivatives at a specific point in the independent variable. These conditions are used to find a particular solution to the differential equation.

## What are boundary conditions in a differential equation?

Boundary conditions in a differential equation refer to the values of the dependent variable at the boundaries of the domain. These conditions are used to find a general solution to the differential equation, which can then be used to determine a specific solution based on initial conditions.

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