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Question Number 116375 by mnjuly1970 last updated on 03/Oct/20

... nice calculus... ordinary differential equation(o.d.e) y(d^2 y/dx^2 ) −((dy/dx))^2 =y^2 (lny) ... find : general solution ..m.n.1970..

$$\:\:\:\:\:\:\:...\:\:{nice}\:\:{calculus}... \\ $$$$\:{ordinary}\:{differential} \\ $$$${equation}\left({o}.{d}.{e}\right) \\ $$$$\:\:\: \\ $$$$\:\:\:{y}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} ={y}^{\mathrm{2}} \left({lny}\right)\:\:... \\ $$$$\:\:\:{find}\::\:\:{general}\:\:{solution} \\ $$$$\:\:\:\:\:..{m}.{n}.\mathrm{1970}.. \\ $$

Answered by Olaf last updated on 03/Oct/20

((y((d^2 y/dx^2 ))−((dy/dx))((dy/dx)))/y^2 ) = lny (d/dx)(((((dy/dx)))/y)) = lny y = e^u (d/dx)(((((du/dx))e^u )/e^u )) = u (d^2 u/dx^2 ) = u (d^2 u/dx^2 ) − u = 0 r^2 −1 = 0 r = ±1 u = Ae^x +Be^(−x) y = e^u = e^(Ae^x +Be^(−x) )

$$\frac{{y}\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)−\left(\frac{{dy}}{{dx}}\right)\left(\frac{{dy}}{{dx}}\right)}{{y}^{\mathrm{2}} }\:=\:\mathrm{ln}{y} \\ $$$$\frac{{d}}{{dx}}\left(\frac{\left(\frac{{dy}}{{dx}}\right)}{{y}}\right)\:=\:\mathrm{ln}{y} \\ $$$${y}\:=\:{e}^{{u}} \\ $$$$\frac{{d}}{{dx}}\left(\frac{\left(\frac{{du}}{{dx}}\right){e}^{{u}} }{{e}^{{u}} }\right)\:=\:{u} \\ $$$$\frac{{d}^{\mathrm{2}} {u}}{{dx}^{\mathrm{2}} }\:=\:{u} \\ $$$$\frac{{d}^{\mathrm{2}} {u}}{{dx}^{\mathrm{2}} }\:−\:{u}\:=\:\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{1}\:=\:\mathrm{0} \\ $$$${r}\:=\:\pm\mathrm{1} \\ $$$${u}\:=\:\mathrm{A}{e}^{{x}} +\mathrm{B}{e}^{−{x}} \\ $$$${y}\:=\:{e}^{{u}} \:=\:{e}^{\mathrm{A}{e}^{{x}} +\mathrm{B}{e}^{−{x}} } \\ $$$$ \\ $$$$ \\ $$

Answered by mnjuly1970 last updated on 04/Oct/20

(dy/dx) =p⇒ (d^2 y/dx^2 )= (dp/dx)=p(dp/dy) p(dp/dy) −y^(−1) p^2 =yln(y) p^2 =u (bernoulli)⇒2p(dp/dy) =(du/dy) (du/dy) −2(u/y) =2yln(y) μ =e^(∫((−2)/y)dy ) =(1/y^2 ) ( I.F) u=y^2 (∫(1/y^2 )2yln(y)dy +k)=y^2 (lny)^2 +ky^2 p=y(√((lny)^2 +k)) ∫(dy/(y(√((lny)^2 +k)))) =x+ c_1 t=ln(y) ⇒dt =(1/y) dy ∫(dt/( (√(t^2 +k)))) = x +c_(1 ) ⇒t=(√k) sinh(r) dt=(√k) cosh(r)dr (1/( (√k)))∫ (√k) ((cosh(r))/(cosh(r)))dr =x+c_1 r=x+c_1 ⇒sinh^(−1) ((t/( (√k))))=x+c_1 ((t/( (√k))))=sinh(x+c_1 ) t=(√k) sinh(x+c_1 )⇒y=e^(√k) ^(sinh(x+c_1 )) m.n.197o

$$\frac{{dy}}{{dx}}\:={p}\Rightarrow\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\:\frac{{dp}}{{dx}}={p}\frac{{dp}}{{dy}} \\ $$$${p}\frac{{dp}}{{dy}}\:−{y}^{−\mathrm{1}} {p}^{\mathrm{2}} ={yln}\left({y}\right) \\ $$$${p}^{\mathrm{2}} ={u}\:\:\left({bernoulli}\right)\Rightarrow\mathrm{2}{p}\frac{{dp}}{{dy}}\:=\frac{{du}}{{dy}} \\ $$$$\frac{{du}}{{dy}}\:−\mathrm{2}\frac{{u}}{{y}}\:=\mathrm{2}{yln}\left({y}\right) \\ $$$$\mu\:={e}^{\int\frac{−\mathrm{2}}{{y}}{dy}\:\:} =\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:\left(\:{I}.{F}\right) \\ $$$${u}={y}^{\mathrm{2}} \left(\int\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\mathrm{2}{yln}\left({y}\right){dy}\:\:+{k}\right)={y}^{\mathrm{2}} \left({lny}\right)^{\mathrm{2}} +{ky}^{\mathrm{2}} \\ $$$${p}={y}\sqrt{\left({lny}\right)^{\mathrm{2}} +{k}} \\ $$$$\int\frac{{dy}}{{y}\sqrt{\left({lny}\right)^{\mathrm{2}} +{k}}}\:={x}+\:{c}_{\mathrm{1}} \\ $$$${t}={ln}\left({y}\right)\:\Rightarrow{dt}\:=\frac{\mathrm{1}}{{y}}\:{dy} \\ $$$$\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} +{k}}}\:=\:{x}\:+{c}_{\mathrm{1}\:} \:\:\Rightarrow{t}=\sqrt{{k}}\:{sinh}\left({r}\right) \\ $$$${dt}=\sqrt{{k}}\:{cosh}\left({r}\right){dr} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{k}}}\int\:\:\sqrt{{k}}\:\frac{{cosh}\left({r}\right)}{{cosh}\left({r}\right)}{dr}\:={x}+{c}_{\mathrm{1}} \\ $$$$\:{r}={x}+{c}_{\mathrm{1}} \Rightarrow{sinh}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{{k}}}\right)={x}+{c}_{\mathrm{1}} \\ $$$$\left(\frac{{t}}{\:\sqrt{{k}}}\right)={sinh}\left({x}+{c}_{\mathrm{1}} \right) \\ $$$${t}=\sqrt{{k}}\:{sinh}\left({x}+{c}_{\mathrm{1}} \right)\Rightarrow{y}={e}^{\sqrt{{k}}} \:^{{sinh}\left({x}+{c}_{\mathrm{1}} \right)} \: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:{m}.{n}.\mathrm{197}{o} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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