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.