Menu Close

dy-dx-y-x-y-3-e-x-x-2-




Question Number 127783 by bemath last updated on 02/Jan/21
             (dy/dx) −(y/x) = ((y^3 .e^x )/x^2 )
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:−\frac{\mathrm{y}}{\mathrm{x}}\:=\:\frac{\mathrm{y}^{\mathrm{3}} .\mathrm{e}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} }\: \\ $$
Answered by liberty last updated on 02/Jan/21
 Bernoulli diff equation .   let v = y^(−2)  ⇒(dv/dx) = −2y^(−3)  (dy/dx) or (dy/dx) = −(1/2)y^3  (dv/dy)  then we find :−(1/2)y^3  (dv/dx)−(y/x) = ((y^3 .e^x )/x^2 )   ⇒(dv/dx)+(2/x)v = (e^x /x^2 ) . Put integrating factor    μ = e^(∫ (2/x) dx)  = x^2  . multiply both sides by μ  ⇒ x^2  (dv/dx) + 2vx = e^x  ; (d/dx)(x^2 v) = e^x   ⇒∫ d(x^2 v) = ∫ e^x  dx ; x^2 v = e^x +C  we find solution : x^2 ((1/y^2 )) = e^x +C   (y^2 /x^2 ) = (1/(e^x +C)) ⇒ y^2  = (x^2 /(e^x +C))
$$\:\mathrm{Bernoulli}\:\mathrm{diff}\:\mathrm{equation}\:. \\ $$$$\:\mathrm{let}\:\mathrm{v}\:=\:\mathrm{y}^{−\mathrm{2}} \:\Rightarrow\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:−\mathrm{2y}^{−\mathrm{3}} \:\frac{\mathrm{dy}}{\mathrm{dx}}\:\mathrm{or}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}^{\mathrm{3}} \:\frac{\mathrm{dv}}{\mathrm{dy}} \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{find}\::−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}^{\mathrm{3}} \:\frac{\mathrm{dv}}{\mathrm{dx}}−\frac{\mathrm{y}}{\mathrm{x}}\:=\:\frac{\mathrm{y}^{\mathrm{3}} .\mathrm{e}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\Rightarrow\frac{\mathrm{dv}}{\mathrm{dx}}+\frac{\mathrm{2}}{\mathrm{x}}\mathrm{v}\:=\:\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} }\:.\:\mathrm{Put}\:\mathrm{integrating}\:\mathrm{factor}\: \\ $$$$\:\mu\:=\:\mathrm{e}^{\int\:\frac{\mathrm{2}}{\mathrm{x}}\:\mathrm{dx}} \:=\:\mathrm{x}^{\mathrm{2}} \:.\:\mathrm{multiply}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{by}\:\mu \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} \:\frac{\mathrm{dv}}{\mathrm{dx}}\:+\:\mathrm{2vx}\:=\:\mathrm{e}^{\mathrm{x}} \:;\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{2}} \mathrm{v}\right)\:=\:\mathrm{e}^{\mathrm{x}} \\ $$$$\Rightarrow\int\:\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} \mathrm{v}\right)\:=\:\int\:\mathrm{e}^{\mathrm{x}} \:\mathrm{dx}\:;\:\mathrm{x}^{\mathrm{2}} \mathrm{v}\:=\:\mathrm{e}^{\mathrm{x}} +\mathrm{C} \\ $$$$\mathrm{we}\:\mathrm{find}\:\mathrm{solution}\::\:\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\right)\:=\:\mathrm{e}^{\mathrm{x}} +\mathrm{C}\: \\ $$$$\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{x}} +\mathrm{C}}\:\Rightarrow\:\mathrm{y}^{\mathrm{2}} \:=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{e}^{\mathrm{x}} +\mathrm{C}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *