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yy-y-2-0-




Question Number 126460 by BHOOPENDRA last updated on 20/Dec/20
yy′′−(y′)^2 =0
$${yy}''−\left({y}'\right)^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by BHOOPENDRA last updated on 20/Dec/20
thanks alot sir
$${thanks}\:{alot}\:{sir} \\ $$
Answered by mr W last updated on 20/Dec/20
p=y′  y′′=(dp/dx)=p(dp/dy)  yp(dp/dy)−p^2 =0  ⇒p=0 ⇒y=C  ⇒y(dp/dy)=p ⇒(dp/p)=(dy/y)⇒ln p=ln Cy   ⇒p=Cy ⇒(dy/dx)=Cy⇒(dy/y)=Cdx  ⇒ln y=Cx+C_2   ⇒y=C_1 e^(Cx)
$${p}={y}' \\ $$$${y}''=\frac{{dp}}{{dx}}={p}\frac{{dp}}{{dy}} \\ $$$${yp}\frac{{dp}}{{dy}}−{p}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{p}=\mathrm{0}\:\Rightarrow{y}={C} \\ $$$$\Rightarrow{y}\frac{{dp}}{{dy}}={p}\:\Rightarrow\frac{{dp}}{{p}}=\frac{{dy}}{{y}}\Rightarrow\mathrm{ln}\:{p}=\mathrm{ln}\:{Cy}\: \\ $$$$\Rightarrow{p}={Cy}\:\Rightarrow\frac{{dy}}{{dx}}={Cy}\Rightarrow\frac{{dy}}{{y}}={Cdx} \\ $$$$\Rightarrow\mathrm{ln}\:{y}={Cx}+{C}_{\mathrm{2}} \\ $$$$\Rightarrow{y}={C}_{\mathrm{1}} {e}^{{Cx}} \\ $$
Commented by Coronavirus last updated on 20/Dec/20
mehode interessante
$${mehode}\:{interessante} \\ $$

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