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Question Number 91353 by jagoll last updated on 30/Apr/20

(dy/dx) = e^(3x−y^2 )

$$\frac{{dy}}{{dx}}\:=\:{e}^{\mathrm{3}{x}−{y}^{\mathrm{2}} } \: \\ $$

Answered by niroj last updated on 30/Apr/20

  (dy/dx) = e^(3x) .e^(−y^2 )     (1/e^(−y^2 ) )dy= e^(3x) dx    ∫e^y^2  dy...= ∫e^(3x) dx......     (π/2)error f(iy)= (e^(3x) /3)+C

$$\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{e}^{\mathrm{3x}} .\mathrm{e}^{−\mathrm{y}^{\mathrm{2}} } \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{e}^{−\mathrm{y}^{\mathrm{2}} } }\mathrm{dy}=\:\mathrm{e}^{\mathrm{3x}} \mathrm{dx} \\ $$$$\:\:\int\mathrm{e}^{\mathrm{y}^{\mathrm{2}} } \mathrm{dy}...=\:\int\mathrm{e}^{\mathrm{3x}} \mathrm{dx}...... \\ $$$$\:\:\:\frac{\pi}{\mathrm{2}}\mathrm{error}\:\mathrm{f}\left(\mathrm{iy}\right)=\:\frac{\mathrm{e}^{\mathrm{3x}} }{\mathrm{3}}+\mathrm{C} \\ $$

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