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Question Number 50430 by Abdo msup. last updated on 16/Dec/18
solve (x^2 +3)y^′ +(x^3 −1)y =x^2
$${solve}\:\left({x}^{\mathrm{2}} \:+\mathrm{3}\right){y}^{'} \:+\left({x}^{\mathrm{3}} −\mathrm{1}\right){y}\:={x}^{\mathrm{2}} \\ $$
Commented by Abdo msup. last updated on 17/Dec/18
(he) (x^2 +3)y^′ +(x^3 −1)y =0 ⇒ (x^2 +3)y^′ =−(x^3 −1)y ⇒(y^′ /y) =−((x^3 −1)/(x^2 +3)) ⇒ ln∣y∣ = ∫ ((1−x^3 )/(x^2 +3)) dx+c =∫ ((1−x(x^2 +3−3))/(x^2 +3))dx +c = ∫ ((1+3x)/(x^2 +3))dx −∫xdx +c = (3/2) ∫ ((2x)/(x^2 +3))dx + ∫ (dx/(x^2 +3)) −(x^2 /2) +c =(3/2)ln(x^2 +3)−(x^2 /2) + ∫ (dx/(x^2 +3)) but ∫ (dx/(x^2 +3)) =_(x=(√3)u) ∫ (((√3)du)/(3(1+u^2 ))) =(1/( (√3))) arctan((x/( (√3))))⇒ ln∣y∣ =(3/2)ln(x^2 +3)−(x^2 /2) +(1/( (√3))) arctan((x/( (√3)))) +c⇒ y = K e^(−(x^2 /2)) (x^2 +3)^(3/2) .e^((1/( (√3)))arctan((x/( (√3))))) ...be continued...
$$\left({he}\right)\:\:\left({x}^{\mathrm{2}} \:+\mathrm{3}\right){y}^{'} \:+\left({x}^{\mathrm{3}} −\mathrm{1}\right){y}\:=\mathrm{0}\:\Rightarrow \\ $$$$\left({x}^{\mathrm{2}} \:+\mathrm{3}\right){y}^{'} =−\left({x}^{\mathrm{3}} −\mathrm{1}\right){y}\:\Rightarrow\frac{{y}^{'} }{{y}}\:=−\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow \\ $$$${ln}\mid{y}\mid\:=\:\int\:\:\frac{\mathrm{1}−{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{3}}\:{dx}+{c}\:=\int\:\:\:\frac{\mathrm{1}−{x}\left({x}^{\mathrm{2}} \:+\mathrm{3}−\mathrm{3}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:+{c} \\ $$$$=\:\int\:\:\:\frac{\mathrm{1}+\mathrm{3}{x}}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:−\int{xdx}\:+{c} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:+\:\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{3}}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{c} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{3}}\:{but} \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{3}}\:=_{{x}=\sqrt{\mathrm{3}}{u}} \:\:\int\:\:\:\:\:\frac{\sqrt{\mathrm{3}}{du}}{\mathrm{3}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{{x}}{\:\sqrt{\mathrm{3}}}\right)\Rightarrow \\ $$$${ln}\mid{y}\mid\:=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{{x}}{\:\sqrt{\mathrm{3}}}\right)\:+{c}\Rightarrow \\ $$$${y}\:=\:{K}\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:.{e}^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{{x}}{\:\sqrt{\mathrm{3}}}\right)} \:\:…{be}\:{continued}… \\ $$$$ \\ $$

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