Question Number 46694 by MJS last updated on 30/Oct/18
$$\left({y}''{y}−\left({y}'\right)^{\mathrm{2}} \right)\mathrm{e}^{\frac{{y}'}{{y}}} ={y}^{\mathrm{2}} \\ $$$$\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{at}\:\mathrm{all}… \\ $$
Answered by ajfour last updated on 30/Oct/18
![ln [y′′y−(y′)^2 ]+((y′)/y)=2ln y ⇒ ln [((y′′)/y)−(((y′)/y))^2 ]=−((y′)/y) let ((y′)/y)=t ⇒ (((y′′y−y′)^2 )/y^2 )=(dt/dx) ⇒ ln (dt/dx) = −t or dt = e^(−t) dx ⇒ ∫e^t dt = ∫dx ⇒ e^t = x+c_1 e^((y′)/y) = x+c_1 ⇒ ∫ (dy/y) = ∫ln (x+c_1 )dx ln y = xln (x+c_1 )−∫((xdx)/(x+c_1 )) ln y = xln (x+c_1 )−x+c_1 ln (x+c_1 )+c_2 ________________________ .](https://www.tinkutara.com/question/Q46698.png)
$$\mathrm{ln}\:\left[{y}''{y}−\left({y}'\right)^{\mathrm{2}} \right]+\frac{{y}'}{{y}}=\mathrm{2ln}\:{y} \\ $$$$\Rightarrow\:\mathrm{ln}\:\left[\frac{{y}''}{{y}}−\left(\frac{{y}'}{{y}}\right)^{\mathrm{2}} \right]=−\frac{{y}'}{{y}} \\ $$$${let}\:\:\frac{{y}'}{{y}}={t}\:\:\:\:\Rightarrow\:\:\frac{\left({y}''{y}−{y}'\right)^{\mathrm{2}} }{{y}^{\mathrm{2}} }=\frac{{dt}}{{dx}} \\ $$$$\Rightarrow\:\:\mathrm{ln}\:\frac{{dt}}{{dx}}\:=\:−{t} \\ $$$${or}\:\:\:\:{dt}\:=\:{e}^{−{t}} {dx} \\ $$$$\Rightarrow\:\:\:\int{e}^{{t}} {dt}\:=\:\int{dx} \\ $$$$\Rightarrow\:\:\:\:{e}^{{t}} \:=\:{x}+{c}_{\mathrm{1}} \\ $$$$\:\:\:{e}^{\frac{{y}'}{{y}}} \:=\:{x}+{c}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:\:\int\:\frac{{dy}}{{y}}\:=\:\int\mathrm{ln}\:\left({x}+{c}_{\mathrm{1}} \right){dx} \\ $$$$\:\:\:\:\mathrm{ln}\:{y}\:=\:{x}\mathrm{ln}\:\left({x}+{c}_{\mathrm{1}} \right)−\int\frac{{xdx}}{{x}+{c}_{\mathrm{1}} } \\ $$$$\:\:\mathrm{ln}\:{y}\:=\:{x}\mathrm{ln}\:\left({x}+{c}_{\mathrm{1}} \right)−{x}+{c}_{\mathrm{1}} \mathrm{ln}\:\left({x}+{c}_{\mathrm{1}} \right)+{c}_{\mathrm{2}} \: \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\:. \\ $$
Commented by MJS last updated on 30/Oct/18
$$\mathrm{thank}\:\mathrm{you}! \\ $$