Solving this Differential

BCHL85

dv/(g - kv^2/m) = dt
Then write g - kv^2/m = (A - v).(B + v) ...
I guess you could do it from now

∫dv/(g-kv²/m) = ∫1.dt
t = ∫1/(g+v√(k/m))(g-v√(k/m)) dv

Using partial fractions.
1/(√g+v√(k/m))(√g-v√(k/m)) = A/(√g+v√(k/m)) + B/(√g-v√(k/m))
1 = A(√g-v√(k/m)) + B(√g+v√(k/m))
Let v√(k/m)) = √g
1 = 0 + B(2√g)
B = 1/2√g

Let v√(k/m)) = -√g
1 = A(√g--√g)
A = 1/2√g

t = ∫1/(√g+v√(k/m))(√g-v√(k/m)) dv
=> t = ∫1/2√g[1/(√g+v√(k/m)) + 1/(√g-v√(k/m)]dv
=> t = 1/2√g∫[1/(√g+v√(k/m)) + 1/(√g-v√(k/m)]dv
=> t = 1/2√g[√(m/k)ln(√g+v√(k/m)-√(m/k)ln(√g-v√(k/m)]

Reply 7

OP

Its FP 1 yea. Right, thanks Widowmaker. I got something similar. How does one make v the subject now?

Reply 8

Widowmaker

∫dv/(g-kv²/m) = ∫1.dt
t = ∫1/(g+v√(k/m))(g-v√(k/m)) dv

Using partial fractions.
1/(√g+v√(k/m))(√g-v√(k/m)) = A/(√g+v√(k/m)) + B/(√g-v√(k/m))
1 = A(√g-v√(k/m)) + B(√g+v√(k/m))
Let v√(k/m)) = √g
1 = 0 + B(2√g)
B = 1/2√g

Let v√(k/m)) = -√g
1 = A(√g--√g)
A = 1/2√g

t = ∫1/(√g+v√(k/m))(√g-v√(k/m)) dv
=> t = ∫1/2√g[1/(√g+v√(k/m)) + 1/(√g-v√(k/m)]dv
=> t = 1/2√g∫[1/(√g+v√(k/m)) + 1/(√g-v√(k/m)]dv
=> t = 1/2√g[√(m/k)ln(√g+v√(k/m)-√(m/k)ln(√g-v√(k/m)]

Making v the subject
2t√g = [√(m/k)ln(√g+v√(k/m)-√(m/k)ln(√g-v√(k/m)]
2t√g = √(m/k)ln(√g+v√(k/m)/(√g-v√(k/m))
2t√g.√(k/m) = ln(√g+v√(k/m)/(√g-v√(k/m))
2t√(kg/m) = ln(√g+v√(k/m)/(√g-v√(k/m))
e^{2t√(kg/m)} = (√g+v√(k/m)/(√g-v√(k/m))

Let u = √(k/m) to make simplifying and typing easier.
e^{2t√(gu)} = (√g+vu)/(√g-vu)
e^{2t√(gu)} = g - v²u²
v²u² = g-e^{2t√(gu)}
v² = (g-e^{2t√(gu)})/u²
v = √(g-e^{2t√(gu)})/u

No putting u = √(k/m) back into equation.
v = √(g-e^{2t√(gk/m)})/√(k/m)
v = √(g-e^{2t√(gk/m)}) x √(m/k)
v = √(m/k(g-e^{2t√(gk/m)})

Reply 9

samd

10

Vazzyb

Its FP 1 yea. Right, thanks Widowmaker. I got something similar. How does one make v the subject now?

It isn't in Edexcel FP1. I am fairly sure.

Reply 10

OP

Yea, I think that this maybe a bit beyond that level. Thanks very much widowmaker. Btw, have you got your offer from Cambridge? ( not clear in your sig )

Reply 11

Vazzyb

Yea, I think that this maybe a bit beyond that level. Thanks very much widowmaker. Btw, have you got your offer from Cambridge? ( not clear in your sig )

Nah, they make offers after interviews which are held in december.

Reply 12

OP

Yea I thought so lol. ok cheers.

Reply 13

OP

Widowmaker

=> t = 1/2√g∫[1/(√g+v√(k/m)) + 1/(√g-v√(k/m)]dv
=> t = 1/2√g[√(m/k)ln(√g+v√(k/m)-√(m/k)ln(√g-v√(k/m)]

Sorry, but could you please explain this step to me. Thanks.

Reply 14

samd

10

He has integrated.

y = 1/(ax + b)
∫ydx = (1/a)ln(ax + b)

In this case: a = √(m/k)

Reply 15

=> t = 1/2√g∫[1/(√g+v√(k/m)) + 1/(√g-v√(k/m)]dv
=> t = 1/2√g[√(m/k)ln(√g+v√(k/m)-√(m/k)ln(√g-v√(k/m)]

Reply 16

OP

ahh yes, how stupid of me. Thanks very much, again :smile:

.

Reply 17

OP