/sci/ - Science & Math

C1 is a constant, what's so hard about that?

>>2527797

Nothing, I do get why it's there and what it does. I just don't get where the hell does it come from. I'm used to adding a constant to the end when I use indefinite integrals. If this one is multiplying by the constant I assume there's an actual logical process to applying these things, and one just doesn't go about trying in random places to see where a constant would have the appropriate role

>2527773
du/u = 2dx | int
ln(u) = 2x + c
u = e^(2x + c)=+/-e^c · e^(2x)
C1=:+/-e^c

oops:

du/u = 2dx | int
ln(u) = 2x + c
u = +/-e^(2x + c)=+/-e^c · e^(2x)
C1=:+/-e^c

>>2527827

God damn it, e^C = C1.

Thanks brovolone, nobody ever told me that

>>2527844

a constant is a constant is a constant

Notice that C =: +/- e^c =/= 0.
This is because you divided by u in the first step, but u=0 is also a solution to u'=2u.

Also notice that u'=Au is the only homogenous linear first order ordinary differential equation, that is, A can also be a matrix. You should just "see" the solution since alot of differential equation systems can be brought to that form (e.g. the harmonic oscilator)
also, the answer was probably there as well:
http://en.wikipedia.org/wiki/Linear_differential_equation

>>2527872

But C=/=0 since I divided by u in the first step, that I get. But how come u = 0 is also a solution to u'=2u?

And where does the +/-e^c comes from? None of my books show it, wolframalpha also doesn't say anything

>>2527897
+/- e^(C) comes from distributing the exponential expression to each e. Namely, e^(n+C) = e^(n) times e^(C). Now the +/- arises from removing the absolute value of u when you applied the natural logarithm to it. Now that we have u = +/-e^(C)*e^(n), we can substitute C1 = +/-e^(C). QED

>>2527897
When you divide by u you assume that u=/=0. Thus you have to check the case u=0 seperately.
Inserting u(x)=0 (everywhere) you see that is is obviously a solution to u'=Au, regardless of what A is.
Wait a sec, I'll plot the solutions for ya.

Here the constant c=+/-e^c0 goes takes values between -3 and 10.

Here I plot only the solutions which araise from e^c, you see that they are stricly positive of course.
Only including +/- and the zero solution gives Ce^2x