About simple differential equations

[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.

 

[HallsofIvy]

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.

 

[jwqwerty]

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.

 

[HallsofIvy]

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.

 

[blue_raver22]

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.