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

[bmace]

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.

 

[dextercioby]

Who's f ? Is it a constant, just like a, b, V_0 ? Do you know any method to integrate a first order ODE ?

 

[bmace]

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

 

[dextercioby]

Hmm, do you know the method of the integrating factor ? If so, then first, do some relabeling of functions and constants.

$$V_0 + (a-b) t =: f(t)$$

$$a(2f-1) =: C$$

Now your ODE looks like (assuming $f(t)\neq 0$)

$$\frac{dx(t)}{dt} + \frac{b}{f(t)} x(t) = C$$

Can you find the integrating factor ?

 

[bmace]

the integrating factor should be e^$$\int b/f(t)$$ = f(t)^b/f'(t)

 

[dextercioby]

Yes, but, please, pay attention to the notation used (missing paranthesis).

$$IF = f(t)^{\frac{b}{f'(t)}} = f(t)^{\frac{b}{a-b}}$$

 

[bmace]

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

 

[dextercioby]

Yes, you've got it.

 

[bmace]

thanks really appreciate the help