Solving for x(t): Find x Given x(0)=-V0

  • Thread starter bmace
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In summary, the conversation is about finding the function x(t) given a specific initial condition and an ODE involving constants a, b, V0, and f. The conversation goes on to discuss methods for solving the ODE, including the use of an integrating factor. The final solution involves finding the integrating factor and using it to rearrange the ODE into a form that can be easily integrated.
  • #1
bmace
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Homework Statement


Find x=x(t) given x(0) = -V0

(V0 + (a - b)t)(dx/dt) = (V0 + (a - b)t)a(2f-1) -bx

a and b are lambda in and lambda out


The Attempt at a Solution



Honestly don't know where to start that's why I came and asked it here .
The only thing I can think of is to make some kind of substitution somewhere.
 
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  • #2
Who's f ? Is it a constant, just like a, b, V_0 ? Do you know any method to integrate a first order ODE ?
 
  • #3
yeah f is another constant, I've taken diff eq before, this is for a mathematical physics class, I just can't seem to get it down to a recognizable form that I know how to differentiate
 
  • #4
Hmm, do you know the method of the integrating factor ? If so, then first, do some relabeling of functions and constants.

[tex] V_0 + (a-b) t =: f(t) [/tex]

[tex] a(2f-1) =: C [/tex]

Now your ODE looks like (assuming [itex] f(t)\neq 0 [/itex])

[tex] \frac{dx(t)}{dt} + \frac{b}{f(t)} x(t) = C [/tex]

Can you find the integrating factor ?
 
  • #5
the integrating factor should be e^[tex]\int b/f(t)[/tex] = f(t)^b/f'(t)
 
  • #6
Yes, but, please, pay attention to the notation used (missing paranthesis).

[tex] IF = f(t)^{\frac{b}{f'(t)}} = f(t)^{\frac{b}{a-b}} [/tex]
 
  • #7
alright think I've got it Cf(t)^1/f'(t) + Kf(t)^(-b/f'(t))
where K is the constant of integration and from there just got to plug in the intial conditions and solve for K
 
  • #9
thanks really appreciate the help
 

1. How do you solve for x(t) given x(0)=-V0?

To solve for x(t), we use the equation x(t) = x(0) + V0*t, where x(0) is the initial position and V0 is the initial velocity. We plug in the values for x(0) and V0 and solve for x(t).

2. What is the significance of x(0)=-V0 in this equation?

The initial position (x(0)) being equal to negative initial velocity (-V0) means that the object is starting at a position that is opposite to the direction of its initial velocity. This can be represented as a negative displacement in the equation.

3. Can this equation be used for any type of motion?

Yes, this equation can be used for any type of motion as long as the acceleration is constant. It is commonly used in physics to calculate the position of an object at a given time.

4. How does the value of V0 affect the value of x(t)?

The value of V0 affects the slope of the position-time graph. A higher initial velocity will result in a steeper slope, indicating a faster rate of change in position over time. A lower initial velocity will result in a less steep slope, indicating a slower rate of change in position over time.

5. Can this equation be used to solve for x(t) at any given time?

Yes, this equation can be used to solve for x(t) at any given time as long as the values for x(0) and V0 are known. Additionally, this equation can be rearranged to solve for other variables such as time or velocity.

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