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Question Number 207317 by mr W last updated on 11/May/24
solve for y (1/(y′))+(1/(y′′))=1
$${solve}\:{for}\:{y} \\ $$$$\frac{\mathrm{1}}{{y}'}+\frac{\mathrm{1}}{{y}''}=\mathrm{1} \\ $$
Answered by Berbere last updated on 11/May/24
y′=z (1/z)+(1/(z′))=1⇒z′=(z/(z−1))⇒(((z−1)/z))dz=dx ⇒z−ln(z)=x+c⇒(e^z /z)=e^(x+c) ⇒ze^(−z) =e^(−x−c) ⇒−ze^(−z) =−e^(−x−c) ⇒z=−W(−e^(−x−c) ) (dy/dx)=−W(−e^(−x−c) );Let t=−e^(−x−c) ⇒dt=e^(−x−c) W,Lambert function ∫−W(−e^(−x−c) )=∫((W(t))/t)dt= W(t)e^(W(t)) =t⇒W′(1+W(t))e^(W(t)) =1 W′(t)=(1/((1+W(t))e^(W(t)) ))=((W(t))/(t(1+W(t)))) ⇒((W(t))/t)=W′(1+W(t))⇒∫((W(t))/t)dt=∫W′(1+W(t))dt =W(t)+(1/2)W^2 (t)+c_2 ; y=W(−e^(−x−c) )+(1/2)(W(−e^(−x−c) ))^2 +c_2 ;c,c_2 ∈R
$${y}'={z} \\ $$$$\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{z}'}=\mathrm{1}\Rightarrow{z}'=\frac{{z}}{{z}−\mathrm{1}}\Rightarrow\left(\frac{{z}−\mathrm{1}}{{z}}\right){dz}={dx} \\ $$$$\Rightarrow{z}−{ln}\left({z}\right)={x}+{c}\Rightarrow\frac{{e}^{{z}} }{{z}}={e}^{{x}+{c}} \Rightarrow{ze}^{−{z}} ={e}^{−{x}−{c}} \\ $$$$\Rightarrow−{ze}^{−{z}} =−{e}^{−{x}−{c}} \Rightarrow{z}=−{W}\left(−{e}^{−{x}−{c}} \right) \\ $$$$\frac{{dy}}{{dx}}=−{W}\left(−{e}^{−{x}−{c}} \right);{Let}\:{t}=−{e}^{−{x}−{c}} \Rightarrow{dt}={e}^{−{x}−{c}} \\ $$$${W},{Lambert}\:{function} \\ $$$$\int−{W}\left(−{e}^{−{x}−{c}} \right)=\int\frac{{W}\left({t}\right)}{{t}}{dt}= \\ $$$${W}\left({t}\right){e}^{{W}\left({t}\right)} ={t}\Rightarrow{W}'\left(\mathrm{1}+\mathrm{W}\left({t}\right)\right){e}^{{W}\left({t}\right)} =\mathrm{1} \\ $$$${W}'\left({t}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{W}\left({t}\right)\right){e}^{{W}\left({t}\right)} }=\frac{{W}\left({t}\right)}{{t}\left(\mathrm{1}+{W}\left({t}\right)\right)} \\ $$$$\Rightarrow\frac{{W}\left({t}\right)}{{t}}={W}'\left(\mathrm{1}+{W}\left({t}\right)\right)\Rightarrow\int\frac{{W}\left({t}\right)}{{t}}{dt}=\int{W}'\left(\mathrm{1}+{W}\left({t}\right)\right){dt} \\ $$$$={W}\left({t}\right)+\frac{\mathrm{1}}{\mathrm{2}}{W}^{\mathrm{2}} \left({t}\right)+{c}_{\mathrm{2}} ; \\ $$$${y}={W}\left(−{e}^{−{x}−{c}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left({W}\left(−{e}^{−{x}−{c}} \right)\right)^{\mathrm{2}} +{c}_{\mathrm{2}} ;{c},{c}_{\mathrm{2}} \in\mathbb{R} \\ $$$$ \\ $$
Answered by mr W last updated on 12/May/24
say p=y′=(dy/dx) y′′=(dp/dx)=p(dp/dy) (1/p)+(dy/(pdp))=1 dy=(p−1)dp ⇒y=(p^2 /2)−p+C p^2 −2p+1=2y+C_1 p=1±(√(2y+C_1 )) (dy/dx)=1±(√(2y+C_1 )) (dy/( 1±(√(2y+C_1 ))))=dx say u^2 =2y+C_1 dy=udu ((udu)/(1±u))=dx case 1: ((udu)/(1+u))=dx (1−(1/(1+u)))du=dx u−ln ∣1+u∣=x+C_2 −1 1+u−ln ∣1+u∣=x+C_2 (1+u)e^((1+u)) =e^((x+C_2 )) 1+u=W(e^((x+C_2 )) ) u=W(e^((x+C_2 )) )−1 2y+C_1 =W(e^((x+C_2 )) )−1 ⇒y=(1/2)W(e^((x+C_2 )) )+C_1 case 2: ((udu)/(1−u))=dx ((1/(1−u))−1)du=dx −ln ∣u−1∣−u=x+C_2 ln ∣u−1∣+u−1=−(x+C_2 ) (u−1)e^(u−1) =e^(−(x+C_2 )) u−1=W(e^(−(x+C_2 )) ) y=(1/2)W(e^(−(x+C_2 )) )+C_1 summary: y=(1/2)W(e^(±(x+C_2 )) )+C_1
$${say}\:{p}={y}'=\frac{{dy}}{{dx}} \\ $$$${y}''=\frac{{dp}}{{dx}}={p}\frac{{dp}}{{dy}} \\ $$$$\frac{\mathrm{1}}{{p}}+\frac{{dy}}{{pdp}}=\mathrm{1} \\ $$$${dy}=\left({p}−\mathrm{1}\right){dp} \\ $$$$\Rightarrow{y}=\frac{{p}^{\mathrm{2}} }{\mathrm{2}}−{p}+{C} \\ $$$${p}^{\mathrm{2}} −\mathrm{2}{p}+\mathrm{1}=\mathrm{2}{y}+{C}_{\mathrm{1}} \\ $$$${p}=\mathrm{1}\pm\sqrt{\mathrm{2}{y}+{C}_{\mathrm{1}} } \\ $$$$\frac{{dy}}{{dx}}=\mathrm{1}\pm\sqrt{\mathrm{2}{y}+{C}_{\mathrm{1}} } \\ $$$$\frac{{dy}}{\:\mathrm{1}\pm\sqrt{\mathrm{2}{y}+{C}_{\mathrm{1}} }}={dx} \\ $$$${say}\:{u}^{\mathrm{2}} =\mathrm{2}{y}+{C}_{\mathrm{1}} \\ $$$${dy}={udu} \\ $$$$\frac{{udu}}{\mathrm{1}\pm{u}}={dx} \\ $$$${case}\:\mathrm{1}: \\ $$$$\frac{{udu}}{\mathrm{1}+{u}}={dx} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du}={dx} \\ $$$${u}−\mathrm{ln}\:\mid\mathrm{1}+{u}\mid={x}+{C}_{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{1}+{u}−\mathrm{ln}\:\mid\mathrm{1}+{u}\mid={x}+{C}_{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{u}\right){e}^{\left(\mathrm{1}+{u}\right)} ={e}^{\left({x}+{C}_{\mathrm{2}} \right)} \\ $$$$\mathrm{1}+{u}={W}\left({e}^{\left({x}+{C}_{\mathrm{2}} \right)} \right) \\ $$$${u}={W}\left({e}^{\left({x}+{C}_{\mathrm{2}} \right)} \right)−\mathrm{1} \\ $$$$\mathrm{2}{y}+{C}_{\mathrm{1}} ={W}\left({e}^{\left({x}+{C}_{\mathrm{2}} \right)} \right)−\mathrm{1} \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}}{W}\left({e}^{\left({x}+{C}_{\mathrm{2}} \right)} \right)+{C}_{\mathrm{1}} \\ $$$${case}\:\mathrm{2}: \\ $$$$\frac{{udu}}{\mathrm{1}−{u}}={dx} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}−\mathrm{1}\right){du}={dx} \\ $$$$−\mathrm{ln}\:\mid{u}−\mathrm{1}\mid−{u}={x}+{C}_{\mathrm{2}} \\ $$$$\mathrm{ln}\:\mid{u}−\mathrm{1}\mid+{u}−\mathrm{1}=−\left({x}+{C}_{\mathrm{2}} \right) \\ $$$$\left({u}−\mathrm{1}\right){e}^{{u}−\mathrm{1}} ={e}^{−\left({x}+{C}_{\mathrm{2}} \right)} \\ $$$${u}−\mathrm{1}={W}\left({e}^{−\left({x}+{C}_{\mathrm{2}} \right)} \right) \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}{W}\left({e}^{−\left({x}+{C}_{\mathrm{2}} \right)} \right)+{C}_{\mathrm{1}} \\ $$$${summary}: \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}{W}\left({e}^{\pm\left({x}+{C}_{\mathrm{2}} \right)} \right)+{C}_{\mathrm{1}} \\ $$

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