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Question Number 93707 by i jagooll last updated on 14/May/20

y y′′ = (y′)^2

$$\mathrm{y}\:\mathrm{y}''\:=\:\left(\mathrm{y}'\right)^{\mathrm{2}} \: \\ $$

Commented by i jagooll last updated on 14/May/20

cool man

Commented by john santu last updated on 14/May/20

let v(y) = (dy/dx) ⇒(d^2 y/dx^2 ) = (d/dx)(v(y))= ((d(v(y)))/dy).(dy/dx) v(y)((d(v(y)))/dy) = v(y)^2 ⇒solve yv(y)((d(v(y)))/dy)=v(y)^2 factor −v(y){−y((d(v(y)))/dy)+v(y)}=0 v(y){−y((d(v(y)))/dy)+v(y)}=0 { ((v(y)=0)),((((d(v(y)))/dy)=((v(y))/y))) :} for ((d(v(y)))/dy)=((v(y))/y) ((d(v(y)))/(v(y))) = (dy/y) ⇒∫ ((d(v(y)))/(v(y))) = ∫ (dy/y) ln(v(y)) = ln(y) + c_1 v(y) = Cy ⇒ (dy/dx) = Cy (dy/y) = Cx ⇒ ln y = (1/2)Cx^2 +c_2 y = e^((1/2)Cx^2 +c_2 ) . for v(y) = 0 ⇒ (dy/dx) =0 y = c_(3 ) ⇒ { ((y=e^((1/2)Cx^2 +c_2 ) )),((y=c_3 )) :}

$$\mathrm{let}\:\mathrm{v}\left(\mathrm{y}\right)\:=\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\Rightarrow\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{v}\left(\mathrm{y}\right)\right)=\:\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{y}\right)\right)}{\mathrm{dy}}.\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\mathrm{v}\left(\mathrm{y}\right)\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{y}\right)\right)}{\mathrm{dy}}\:=\:\mathrm{v}\left(\mathrm{y}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{solve}\:\mathrm{yv}\left(\mathrm{y}\right)\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{y}\right)\right)}{\mathrm{dy}}=\mathrm{v}\left(\mathrm{y}\right)^{\mathrm{2}} \\ $$$$\mathrm{factor}\:−\mathrm{v}\left(\mathrm{y}\right)\left\{−\mathrm{y}\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{y}\right)\right)}{\mathrm{dy}}+\mathrm{v}\left(\mathrm{y}\right)\right\}=\mathrm{0} \\ $$$$\mathrm{v}\left(\mathrm{y}\right)\left\{−\mathrm{y}\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{y}\right)\right)}{\mathrm{dy}}+\mathrm{v}\left(\mathrm{y}\right)\right\}=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{v}\left(\mathrm{y}\right)=\mathrm{0}}\\{\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{y}\right)\right)}{\mathrm{dy}}=\frac{\mathrm{v}\left(\mathrm{y}\right)}{\mathrm{y}}}\end{cases} \\ $$$$\mathrm{for}\:\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{y}\right)\right)}{\mathrm{dy}}=\frac{\mathrm{v}\left(\mathrm{y}\right)}{\mathrm{y}} \\ $$$$\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{y}\right)\right)}{\mathrm{v}\left(\mathrm{y}\right)}\:=\:\frac{\mathrm{dy}}{\mathrm{y}}\:\Rightarrow\int\:\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{y}\right)\right)}{\mathrm{v}\left(\mathrm{y}\right)}\:=\:\int\:\frac{\mathrm{dy}}{\mathrm{y}} \\ $$$$\mathrm{ln}\left(\mathrm{v}\left(\mathrm{y}\right)\right)\:=\:\mathrm{ln}\left(\mathrm{y}\right)\:+\:\mathrm{c}_{\mathrm{1}} \\ $$$$\mathrm{v}\left(\mathrm{y}\right)\:=\:\mathrm{Cy}\:\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{Cy} \\ $$$$\frac{\mathrm{dy}}{\mathrm{y}}\:=\:\mathrm{Cx}\:\Rightarrow\:\mathrm{ln}\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Cx}^{\mathrm{2}} +\mathrm{c}_{\mathrm{2}} \\ $$$$\mathrm{y}\:=\:\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Cx}^{\mathrm{2}} +\mathrm{c}_{\mathrm{2}} } \:.\: \\ $$$$\mathrm{for}\:\mathrm{v}\left(\mathrm{y}\right)\:=\:\mathrm{0}\:\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\mathrm{0} \\ $$$$\mathrm{y}\:=\:\mathrm{c}_{\mathrm{3}\:} \Rightarrow\begin{cases}{\mathrm{y}=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Cx}^{\mathrm{2}} +\mathrm{c}_{\mathrm{2}} } }\\{\mathrm{y}=\mathrm{c}_{\mathrm{3}} \:}\end{cases} \\ $$$$ \\ $$

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