Simple Linear DiffEq, not understanding the book

[Feodalherren]

Homework Statement

Use separation of variables to solve the following problem for t ≥ 0

Relevant Equations

dx/dt + 2x = 20
x(0) = 3

The way I want to solve it is the way that I always want to solve separable linear diffEqs:

after some trivial algebra and an easy integral I end up with

t = (-1/2) ln (20-2x) +C

Easy enough, solve for x(t) yields

x(t) = 10 - (1/2)e^(-2t) + C

Solve for C when x(0) = 3 yields

C = -13/2

But the book does something really weird and integrates the x-side from 3 to x(t) giving them the result

ln(20-2x(t)) - ln(20-2*3) = -2t

finally yielding

x(t) = 10 - 7e^(-2t)

I've never seen it done like this. What am I doing wrong? I feel like my method ought to work and I ought to end up with the same answer.

 

[pasmith]

Your solution for x(t) is wrong. If <br /> t = -\tfrac12 \ln(20 - 2x) + C then <br /> \ln(20 - 2x) = 2(C - t) and exponentiating gives <br /> 20 - 2x = e^{2C}e^{-2t}. Try again from here.

 

[Feodalherren]

Ah, of course! Thank you very much.

I vaguely remembered from my DiffEq class that you can 'play around' with C because it's just some constant. But of course you can't just ignore the algebraic rules for it, duh! Once I saw what you did with e^2C I remembered what 'play around with C' means i.e in this case it's still just some constant so call e^2C = C1 and forget about the exponents and other clutter.
I haven't seen a differential equation in quite some time but I got a new job as a Control Systems Engineer so I'm going through my textbook to try to refresh my memory on PID controllers.

Thank you!