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Question Number 60695 by maxmathsup by imad last updated on 24/May/19

solve (√(3+x^2 ))y^(′′) −(2x+1)y^′ =x^2 e^(−x^2 )

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

Commented by maxmathsup by imad last updated on 27/May/19

let y^′ =z so (e)⇔(√(3+x^2 ))z^′ −(2x+1)z =x^2 e^(−x^2 ) (he) →(√(3+x^2 ))z^′ −(2x+1)z =0 ⇒(√(3+x^2 ))z^′ =(2x+1)z ⇒(z^′ /z) =((2x+1)/(√(x^2 +3))) ⇒ ln∣z∣ =∫ ((2x+1)/(√(x^2 +3))) dx+c =2 ∫ (x/(√(x^2 +3))) + ∫ (dx/(√(x^2 +3))) +c =2(√(x^2 +3)) +∫ (dx/(√(x^2 +3))) dx ∫ (dx/(√(x^2 +3))) dx =_(x=(√3)u) ∫ (((√3)du)/((√3)(√(1+u^2 )))) =ln(u +(√(1+u^2 ))) +c_0 =ln((x/(√3)) +(√(1+(x^2 /3) ))) +c_0 ⇒ ln∣z∣ =2(√(x^2 +3)) +ln((x/(√3)) +(√(1+(x^2 /3)))) +c ⇒ z(x) =k ((x/(√3)) +(√(1+(x^2 /3)))) e^(2(√(x^2 +3))) mvc method give z^′ (x) =K^′ ((x/(√3)) +(√(1+(x^2 /3))))e^(2(√(1+(x^2 /3)))) +K{( (1/(√3)) +(((2x)/(√3))/(2(√(1+(x^2 /3))))))e^(2(√(1+(x^2 /3)))) +((x/(√3)) +(√(1+(x^2 /3)))) (2 ((2x)/3)) (1/(√(1+(x^2 /3)))) e^(2(√(1+(x^2 /3)))) } =K^′ (((x+(√(3+x^2 )))/(√3))) e^((2/(√3))(√(3+x^2 ))) +K { (1/(√3)) + (1/(√(3+x^2 )))} e^((2/(√3))(√(3+x^2 ))) +((4x)/3)( ((x+(√(3+x^2 )))/(√3))) (1/(√(3+x^2 ))) e^((2/(√3))(√(3+x^2 ))) ....be continued....

$${let}\:{y}^{'} \:={z}\:\:{so}\:\left({e}\right)\Leftrightarrow\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }{z}^{'} \:−\left(\mathrm{2}{x}+\mathrm{1}\right){z}\:={x}^{\mathrm{2}} \:{e}^{−{x}^{\mathrm{2}} } \\ $$$$\left({he}\right)\:\rightarrow\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }{z}^{'} \:−\left(\mathrm{2}{x}+\mathrm{1}\right){z}\:=\mathrm{0}\:\Rightarrow\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }{z}^{'} \:=\left(\mathrm{2}{x}+\mathrm{1}\right){z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}}\:\Rightarrow \\ $$$${ln}\mid{z}\mid\:=\int\:\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}}\:{dx}+{c}\:\:=\mathrm{2}\:\int\:\:\frac{{x}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}}\:+\:\int\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}}\:+{c} \\ $$$$=\mathrm{2}\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}\:\:+\int\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}}\:{dx} \\ $$$$\int\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}}\:{dx}\:=_{{x}=\sqrt{\mathrm{3}}{u}} \:\:\:\:\int\:\:\:\frac{\sqrt{\mathrm{3}}{du}}{\sqrt{\mathrm{3}}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:={ln}\left({u}\:+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)\:+{c}_{\mathrm{0}} \\ $$$$={ln}\left(\frac{{x}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\:}\right)\:+{c}_{\mathrm{0}} \:\Rightarrow \\ $$$${ln}\mid{z}\mid\:=\mathrm{2}\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}\:+{ln}\left(\frac{{x}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}\right)\:+{c}\:\Rightarrow \\ $$$${z}\left({x}\right)\:={k}\:\left(\frac{{x}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}\right)\:{e}^{\mathrm{2}\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}} \:\:\:\:\:\:{mvc}\:\:{method}\:{give}\: \\ $$$${z}^{'} \left({x}\right)\:={K}^{'} \left(\frac{{x}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}\right){e}^{\mathrm{2}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}} \:\:\:+{K}\left\{\left(\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\frac{\frac{\mathrm{2}{x}}{\sqrt{\mathrm{3}}}}{\mathrm{2}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}}\right){e}^{\mathrm{2}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}} \right. \\ $$$$\left.+\left(\frac{{x}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}\right)\:\left(\mathrm{2}\:\frac{\mathrm{2}{x}}{\mathrm{3}}\right)\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}}\:{e}^{\mathrm{2}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}} \right\} \\ $$$$={K}^{'} \left(\frac{{x}+\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }}{\sqrt{\mathrm{3}}}\right)\:{e}^{\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }} \:+{K}\:\left\{\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }}\right\}\:{e}^{\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }} \\ $$$$+\frac{\mathrm{4}{x}}{\mathrm{3}}\left(\:\frac{{x}+\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }}{\sqrt{\mathrm{3}}}\right)\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }}\:{e}^{\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }} \:\:\:\:\:\:....{be}\:{continued}.... \\ $$

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