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dy-dx-2xe-y-y-1-0-




Question Number 169892 by weltr last updated on 11/May/22
(dy/dx) = 2xe^(−y)  ,  y(1) = 0
$$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{xe}^{−{y}} \:,\:\:{y}\left(\mathrm{1}\right)\:=\:\mathrm{0} \\ $$
Answered by mahdipoor last updated on 11/May/22
⇒(dy/e^(−y) )=e^y dy=2xdx⇒e^y =x^2 +c  ⇒x=1 , y=0 ⇒ c=0  ⇒⇒e^y =x^2
$$\Rightarrow\frac{{dy}}{{e}^{−{y}} }={e}^{{y}} {dy}=\mathrm{2}{xdx}\Rightarrow{e}^{{y}} ={x}^{\mathrm{2}} +{c} \\ $$$$\Rightarrow{x}=\mathrm{1}\:,\:{y}=\mathrm{0}\:\Rightarrow\:{c}=\mathrm{0} \\ $$$$\Rightarrow\Rightarrow{e}^{{y}} ={x}^{\mathrm{2}} \\ $$

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