Solve-the-following-d-e-for-v-in-terms-of-s-c-kv-v-dv-ds-

Question Number 2602 by Yozzis last updated on 23/Nov/15

S o l v e t h e f o l l o w i n g d . e f o r v i n t e r m s o f s

c - k v = v \frac{d v}{d s} .

Answered by prakash jain last updated on 24/Nov/15

\frac{v d v}{c - k v} = d s

- \frac{1}{k} [\int \frac{c - k v}{c - k v} d v - \int \frac{c}{c - k v} d v] = d s

- \frac{1}{k} (v + \frac{c}{k} \ln (c - k v)) + C = s

- \frac{v}{k} - \frac{c}{k^{2}} \ln (c - k v) + C = s

k v + c \ln (c - k v) - {C k}^{2} = - {s k}^{2}

\ln (c - k v) = \frac{1}{c} (- k v + {C k}^{2} - {s k}^{2})

This is an implicit relation between v and s .

Commented by Yozzi last updated on 23/Nov/15

I g o t u p t o y o u r r e s u l t b u t I^{'} m

w o n d e r i n g h o w {y o u}^{'} l l o b t a i n e x p l i c i t l y

v i n t e r m s o f s . W o l f r a m A l p h a

g a v e a n a n s w e r i n t e r m s o f s o m e t h i n g

c a l l e d {L a m b e r t}^{'} s W f u n c t i o n . I

{d o n}^{'} t k n o w t h a t f u n c t i o n y e t .

Commented by prakash jain last updated on 23/Nov/15

\ln (c - k y) = \frac{1}{c} (- k v + {C k}^{2} - {s k}^{2})

c - k y = e^{\frac{1}{c} (- k v + {C k}^{2} - {s k}^{2})}

k y = c - \frac{1}{c} e^{\frac{1}{c} (- k v + {C k}^{2} - {s k}^{2})}

Commented by Filup last updated on 24/Nov/15

i w a n t t o l e a r n t h a t f u n c t i o n

Commented by 123456 last updated on 24/Nov/15

W lambert is a function such that

{y e}^{y} = x (or something similiar, i dont remember it exact)

Commented by prakash jain last updated on 24/Nov/15

y = W ({y e}^{y})

{y e}^{y} = W ({y e}^{y}) e^{W ({y e}^{y})}

\ln (c - k v) = - \frac{k v}{c} + \frac{{C k}^{2}}{c} - \frac{{s k}^{2}}{c}

c - k v = e^{- \frac{k v}{c}} e^{\frac{{C k}^{2} - {s k}^{2}}{c}}

c - k v = e^{1} e^{- 1} e^{- \frac{k v}{c}} e^{\frac{{C k}^{2} - {s k}^{2}}{c}}

(c - k v) = e^{1 - \frac{k v}{c}} . e^{\frac{{C k}^{2} - {s k}^{2}}{c} - 1}

(c - k v) = e^{\frac{c - k v}{c}} . e^{\frac{{C k}^{2} - {s k}^{2}}{c} - 1}

(c - k v) \cdot e^{- \frac{c - k v}{c}} = e^{\frac{{C k}^{2} - {s k}^{2}}{c} - 1}

c [- \frac{c - k v}{c}] \cdot e^{- \frac{c - k v}{c}} = - e^{\frac{{C k}^{2} - {s k}^{2}}{c} - 1}

[- \frac{c - k v}{c}] \cdot e^{- \frac{c - k v}{c}} = \frac{- e^{\frac{{C k}^{2} - {s k}^{2}}{c} - 1}}{c}

t a k i n g W f u n c t i o n o n b o t h s i d e s

s i n c e W ({x e}^{x}) = x

- \frac{c - k v}{c} = W (\frac{- e^{\frac{{C k}^{2} - {s k}^{2}}{c} - 1}}{c})

- c + k v = c W (- \frac{e^{\frac{{C k}^{2} - {s k}^{2}}{c} - 1}}{c})

v = \frac{c}{k} W (- \frac{e^{\frac{{C k}^{2} - {s k}^{2}}{c} - 1}}{c}) + \frac{c}{k}

Commented by prakash jain last updated on 24/Nov/15

If f = {x e}^{x}

W (x) = f^{- 1} (x) [f^{- 1} is f - inverse]

Commented by Yozzi last updated on 24/Nov/15

I s e e . I n t e r e s t i n g!

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