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Question Number 2602 by Yozzis last updated on 23/Nov/15
$${Solve}\:{the}\:{following}\:{d}.{e}\:{for}\:{v}\:{in}\:{terms}\:{of}\:{s} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}−{kv}={v}\frac{{dv}}{{ds}}. \\ $$$$ \\ $$
Answered by prakash jain last updated on 24/Nov/15
![((vdv)/(c−kv))=ds −(1/k)[∫ ((c−kv)/(c−kv))dv−∫(c/(c−kv))dv]=ds −(1/k)(v+(c/k)ln (c−kv))+C=s −(v/k)−(c/k^2 )ln (c−kv)+C=s kv+cln (c−kv)−Ck^2 =−sk^2 ln (c−kv)=(1/c)(−kv+Ck^2 −sk^2 ) This is an implicit relation between v and s.](Q2623.png)
$$\frac{{vdv}}{{c}−{kv}}={ds} \\ $$$$−\frac{\mathrm{1}}{{k}}\left[\int\:\frac{{c}−{kv}}{{c}−{kv}}{dv}−\int\frac{{c}}{{c}−{kv}}{dv}\right]={ds} \\ $$$$−\frac{\mathrm{1}}{{k}}\left({v}+\frac{{c}}{{k}}\mathrm{ln}\:\left({c}−{kv}\right)\right)+{C}={s} \\ $$$$−\frac{{v}}{{k}}−\frac{{c}}{{k}^{\mathrm{2}} }\mathrm{ln}\:\left({c}−{kv}\right)+{C}={s} \\ $$$${kv}+{c}\mathrm{ln}\:\left({c}−{kv}\right)−{Ck}^{\mathrm{2}} =−{sk}^{\mathrm{2}} \\ $$$$\mathrm{ln}\:\left({c}−{kv}\right)=\frac{\mathrm{1}}{{c}}\left(−{kv}+{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} \right) \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{an}\:\mathrm{implicit}\:\mathrm{relation}\:\mathrm{between}\:{v}\:\mathrm{and}\:{s}. \\ $$
Commented by Yozzi last updated on 23/Nov/15
$${I}\:{got}\:{up}\:{to}\:{your}\:{result}\:{but}\:{I}'{m}\: \\ $$$${wondering}\:{how}\:{you}'{ll}\:{obtain}\:{explicitly} \\ $$$${v}\:{in}\:{terms}\:{of}\:{s}.\:{Wolfram}\:{Alpha} \\ $$$${gave}\:{an}\:{answer}\:{in}\:{terms}\:{of}\:{something} \\ $$$${called}\:{Lambert}'{s}\:{W}\:{function}.\:{I} \\ $$$${don}'{t}\:{know}\:{that}\:{function}\:{yet}.\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by prakash jain last updated on 23/Nov/15
$$\mathrm{ln}\:\left({c}−{ky}\right)=\frac{\mathrm{1}}{{c}}\left(−{kv}+{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} \right) \\ $$$${c}−{ky}={e}^{\frac{\mathrm{1}}{{c}}\left(−{kv}+{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} \right)} \\ $$$${ky}={c}−\frac{\mathrm{1}}{{c}}{e}^{\frac{\mathrm{1}}{{c}}\left(−{kv}+{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} \right)} \\ $$
Commented by Filup last updated on 24/Nov/15
$${i}\:{want}\:{to}\:{learn}\:{that}\:{function} \\ $$
Commented by 123456 last updated on 24/Nov/15
$$\mathrm{W}\:\mathrm{lambert}\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}\:\mathrm{such}\:\mathrm{that} \\ $$$${ye}^{{y}} ={x}\:\left(\mathrm{or}\:\mathrm{something}\:\mathrm{similiar},\:\mathrm{i}\:\mathrm{dont}\:\mathrm{remember}\:\mathrm{it}\:\mathrm{exact}\right) \\ $$
Commented by prakash jain last updated on 24/Nov/15
![y=W(ye^y ) ye^y =W(ye^y )e^(W(ye^y )) ln (c−kv)=−((kv)/c)+((Ck^2 )/c)−((sk^2 )/c) c−kv=e^(−((kv)/c)) e^((Ck^2 −sk^2 )/c) c−kv=e^1 e^(−1) e^(−((kv)/c)) e^((Ck^2 −sk^2 )/c) (c−kv)=e^(1−((kv)/c)) . e^(((Ck^2 −sk^2 )/c) −1) (c−kv)=e^((c−kv)/c) . e^(((Ck^2 −sk^2 )/c) −1) (c−kv)∙e^(− ((c−kv)/c)) = e^(((Ck^2 −sk^2 )/c) −1) c[−((c−kv)/c)]∙e^(−((c−kv)/c)) =−e^(((Ck^2 −sk^2 )/c) −1) [−((c−kv)/c)]∙e^(−((c−kv)/c)) =((−e^(((Ck^2 −sk^2 )/c) −1) )/c) taking W function on both sides since W(xe^x )=x −((c−kv)/c)=W(((−e^(((Ck^2 −sk^2 )/c) −1) )/c)) −c+kv=cW(−(e^(((Ck^2 −sk^2 )/c) −1) /c)) v=(c/k)W(−(e^(((Ck^2 −sk^2 )/c) −1) /c))+(c/k)](Q2640.png)
$${y}={W}\left({ye}^{{y}} \right) \\ $$$${ye}^{{y}} ={W}\left({ye}^{{y}} \right){e}^{{W}\left({ye}^{{y}} \right)} \\ $$$$\mathrm{ln}\:\left({c}−{kv}\right)=−\frac{{kv}}{{c}}+\frac{{Ck}^{\mathrm{2}} }{{c}}−\frac{{sk}^{\mathrm{2}} }{{c}} \\ $$$${c}−{kv}={e}^{−\frac{{kv}}{{c}}} {e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}} \\ $$$${c}−{kv}={e}^{\mathrm{1}} {e}^{−\mathrm{1}} {e}^{−\frac{{kv}}{{c}}} \:{e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}} \\ $$$$\left({c}−{kv}\right)={e}^{\mathrm{1}−\frac{{kv}}{{c}}} \:.\:{e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}\:−\mathrm{1}} \\ $$$$\left({c}−{kv}\right)={e}^{\frac{{c}−{kv}}{{c}}} \:.\:{e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}\:−\mathrm{1}} \\ $$$$\left({c}−{kv}\right)\centerdot{e}^{−\:\frac{{c}−{kv}}{{c}}} \:=\:{e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}\:−\mathrm{1}} \\ $$$${c}\left[−\frac{{c}−{kv}}{{c}}\right]\centerdot{e}^{−\frac{{c}−{kv}}{{c}}} =−{e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}\:−\mathrm{1}} \\ $$$$\left[−\frac{{c}−{kv}}{{c}}\right]\centerdot{e}^{−\frac{{c}−{kv}}{{c}}} =\frac{−{e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}\:−\mathrm{1}} }{{c}} \\ $$$${taking}\:{W}\:{function}\:{on}\:{both}\:{sides} \\ $$$${since}\:{W}\left({xe}^{{x}} \right)={x} \\ $$$$−\frac{{c}−{kv}}{{c}}={W}\left(\frac{−{e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}\:−\mathrm{1}} }{{c}}\right) \\ $$$$−{c}+{kv}={cW}\left(−\frac{{e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}\:−\mathrm{1}} }{{c}}\right) \\ $$$${v}=\frac{{c}}{{k}}{W}\left(−\frac{{e}^{\frac{{Ck}^{\mathrm{2}} −{sk}^{\mathrm{2}} }{{c}}\:−\mathrm{1}} }{{c}}\right)+\frac{{c}}{{k}} \\ $$$$ \\ $$
Commented by prakash jain last updated on 24/Nov/15
![If f=xe^x W(x)=f^(−1) (x) [f^(−1) is f−inverse]](Q2641.png)
$$\mathrm{If}\:{f}={xe}^{{x}} \\ $$$${W}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\:\:\left[{f}^{−\mathrm{1}} \mathrm{is}\:{f}−\mathrm{inverse}\right] \\ $$
Commented by Yozzi last updated on 24/Nov/15
$${I}\:{see}.\:{Interesting}! \\ $$