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Question Number 67380 by mathmax by abdo last updated on 26/Aug/19

solve the (d.e.) ( x^2 −x+1 )y^′ −(2x+3)y =x^2 e^x

$${solve}\:{the}\:\left({d}.{e}.\right)\:\:\:\:\left(\:{x}^{\mathrm{2}} −{x}+\mathrm{1}\:\:\:\:\:\:\right){y}^{'} −\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:={x}^{\mathrm{2}} \:{e}^{{x}} \\ $$

Commented by mathmax by abdo last updated on 28/Aug/19

(he) →(x^2 −x+1)y^′ −(2x+3)y =0 ⇒(x^2 −x+1)y^′ =(2x+3)y ⇒ (y^′ /y) =((2x+3)/(x^2 −x+1)) ⇒ln∣y∣ =∫ ((2x+3)/(x^2 −x+1))dx +c but ∫ ((2x+3)/(x^2 −x +1)) =∫ ((2x−1 +4)/(x^2 −x+1))dx =∫ ((2x−1)/(x^2 −x+1)) dx +4 ∫ (dx/(x^2 −x +1)) ∫ ((2x−1)/(x^2 −x+1))dx =ln(x^2 −x+1) +c_1 ∫ (dx/(x^2 −x+1)) =∫ (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)t) (4/3) ∫ (1/(t^2 +1))((√3)/2)dt =(2/(√3))arctan(((2x−1)/(√3))) +c_2 ⇒ ln∣y∣ =ln(x^2 −x+1)+(8/(√3)) arctan(((2x−1)/(√3))) +c ⇒ y(x) =K(x^2 −x+1) e^((8/(√3)) arctan(((2x−1)/(√3)))) let use mvc method y^′ =K^′ (x^2 −x+1) e^((8/((√3) )) arctan(((2x−1)/((√3) ))) ) +K{ (2x−1)e^((8/(√3))arctan(((2x−1)/(√3)))) +(x^2 −x +1)×(8/(√3)) (2/((√3)(1+(((2x−1)/(√3)))^2 ))) }}e^((8/(√3))arctan(((2x−1)/(√3)))) ={K^′ (x^2 −x+1)+(2x−1)K+ ((16)/3) K(x^2 −x +1)×(1/(1+(((2x−1)^2 )/3)))} ×e^((8/(√3))arctan(((2x−1)/(√3)))) ....be continued....

$$\left({he}\right)\:\rightarrow\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right){y}^{'} −\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:=\mathrm{0}\:\Rightarrow\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right){y}^{'} =\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:\Rightarrow \\ $$$$\frac{{y}^{'} }{{y}}\:=\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\Rightarrow{ln}\mid{y}\mid\:=\int\:\:\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}\:+{c}\:\:{but} \\ $$$$\int\:\:\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}\:=\int\:\frac{\mathrm{2}{x}−\mathrm{1}\:+\mathrm{4}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}\:=\int\:\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx}\:+\mathrm{4}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\ $$$$\int\:\:\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}\:={ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:=\int\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \frac{\mathrm{4}}{\mathrm{3}}\:\:\:\int\:\:\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${ln}\mid{y}\mid\:={ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:+{c}\:\Rightarrow \\ $$$${y}\left({x}\right)\:={K}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:{e}^{\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} \:\:\:{let}\:{use}\:{mvc}\:{method} \\ $$$${y}^{'} \:={K}^{'} \left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:{e}^{\frac{\mathrm{8}}{\sqrt{\mathrm{3}}\:}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}\:}\right)\:} \:+{K}\left\{\:\left(\mathrm{2}{x}−\mathrm{1}\right){e}^{\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} \right. \\ $$$$\left.+\left.\left({x}^{\mathrm{2}} −{x}\:+\mathrm{1}\right)×\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}\:\:\:\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}\left(\mathrm{1}+\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \right)}\:\right\}\right\}{e}^{\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} \\ $$$$=\left\{{K}^{'} \left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\left(\mathrm{2}{x}−\mathrm{1}\right){K}+\:\frac{\mathrm{16}}{\mathrm{3}}\:{K}\left({x}^{\mathrm{2}} −{x}\:+\mathrm{1}\right)×\frac{\mathrm{1}}{\mathrm{1}+\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}}}\right\} \\ $$$$×{e}^{\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} ....{be}\:{continued}.... \\ $$

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