Undergrad Quick question for Finding EOM with diff eq

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The discussion revolves around finding the equation of motion given the acceleration a = -bv, where b is a constant. The user successfully derived the velocity as v = v0e-bt but encountered issues while integrating to find the position x. They initially arrived at x = (v0/b)e-bt + x0, which differs from the book's answer of x = x0 + (v0/b)(1-e-bt). The discrepancy is attributed to the need for careful handling of integration limits and signs during the integration process. The conversation emphasizes the importance of correctly applying integration techniques in differential equations.
JaDi13
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I have been going through my old books again, and found myself a little stuck. I am not entirely sure if this would be better in this one or diffy eq.
The problem starts with having you find equation of motions when a= -bv, where b is constant and v = v(t)

Using method of separable equations, I was able to integrate the acceleration to get the velocity, v = v0e-bt

The problem is when I try to find the equation for x.
When I integrate v, using u substitution, I get x = (v0/b)e-bt+x0
The books says the answer x = x0 + (v0/b)(1-e-bt)
Also when I take the derivative of the book's answer, I get what I found for v

Am I way off base? Did I miss a step?
 
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JaDi13 said:
I get ##x = (v_{0}/b)e^{-bt}+x_{0}##

because this is wrong ... , the integral ##x=\int_{0}^{t}v_{0}e^{-bt}dt+x_{0}\not=\frac{v_{0}}{b}e^{-bt}+x_{0}##
control the signs! (and also the definite integral ...)
Ssnow
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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