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Question Number 183532 by MikeH last updated on 26/Dec/22
Answered by CElcedricjunior last updated on 26/Dec/22
3xy′−y=lnx+1 y(1)=−2 (h):3xy′−y=0=>y=e^(∫(1/(3x))dx) =ke^((1/3)ln∣x∣) =ke^((1/3)lnx) pour x∈R_+ ^∗ y=k(x)^(1/3) faisons varier k y(x)=k(x)(x)^(1/3) =>y′(x)=k′(x)(x)^(1/3) +((k(x))/(3(x^2 )^(1/3) )) =>3xy′−y=lnx+1 =>3xy′−y=3xk′(x)(x)^(1/3) +k(x)(x)^(1/3) −k(x)(x)^(1/3) =3xk′(x)(x)^(1/3) <=>k′(x)=((lnx+1)/(3(x^4 )^(1/3) )) <=>k(x)=∫((lnx+1)/(3(x^4 )^(1/3) ))dx=∫((lnx)/(3x(x)^(1/3) ))dx+(1/3)∫x^(−(4/3)) dx =(1/3)((1/(−(4/3)+1))x^(−(4/3)+1) )+(1/3)∫((lnx)/x)×(1/( (x)^(1/3) ))dx posons { ((u=lnx)),((v′=x^(−(4/3)) )) :}=> { ((u′=(1/x))),((v=−(3/( (x)^(1/3) )))) :} =−(1/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+∫x^(−(4/3)) dx =−(4/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+c c∈R....cedric y(x)=k(x)(x)^(1/3) y′(x)=(−(4/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+c)(x)^(1/3) .....junior 3xy′−y=lnx+1 y(1)=−2 (h):3xy′−y=0=>y=e^(∫(1/(3x))dx) =ke^((1/3)ln∣x∣) =ke^((1/3)lnx) pour x∈R_+ ^∗ y=k(x)^(1/3) faisons varier k y(x)=k(x)(x)^(1/3) =>y′(x)=k′(x)(x)^(1/3) +((k(x))/(3(x^2 )^(1/3) )) =>3xy′−y=lnx+1 =>3xy′−y=3xk′(x)(x)^(1/3) +k(x)(x)^(1/3) −k(x)(x)^(1/3) =3xk′(x)(x)^(1/3) <=>k′(x)=((lnx+1)/(3(x^4 )^(1/3) )) <=>k(x)=∫((lnx+1)/(3(x^4 )^(1/3) ))dx=∫((lnx)/(3x(x)^(1/3) ))dx+(1/3)∫x^(−(4/3)) dx =(1/3)((1/(−(4/3)+1))x^(−(4/3)+1) )+(1/3)∫((lnx)/x)×(1/( (x)^(1/3) ))dx posons { ((u=lnx)),((v′=x^(−(4/3)) )) :}=> { ((u′=(1/x))),((v=−(3/( (x)^(1/3) )))) :} =−(1/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+∫x^(−(4/3)) dx =−(4/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+c c∈R....cedric y(x)=k(x)(x)^(1/3) y′(x)=(−(4/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+c)(x)^(1/3) .....junior y(x)=−4−lnx+c(x)^(1/3) y(1)=−2=>−4−ln(1)+c=−2 =>c=2 y(x)=−4−lnx+2((x ))^(1/3) ======================== ..............le celebre cedric junior........ ====================== y(x)=−4−lnx+c(x)^(1/3) y(1)=−2=>−4−ln(1)+c=−2 =>c=2 y(x)=−4−lnx+2((x ))^(1/3) ======================== ..............le celebre cedric junior........ ======================
$$\mathrm{3}\boldsymbol{{xy}}'−\boldsymbol{{y}}=\boldsymbol{{lnx}}+\mathrm{1}\:\:\:\:\boldsymbol{{y}}\left(\mathrm{1}\right)=−\mathrm{2} \\ $$$$\left(\boldsymbol{{h}}\right):\mathrm{3}\boldsymbol{{xy}}'−\boldsymbol{{y}}=\mathrm{0}=>\boldsymbol{{y}}=\boldsymbol{{e}}^{\int\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{x}}}\boldsymbol{{dx}}} =\boldsymbol{{ke}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{ln}}\mid\boldsymbol{{x}}\mid} =\boldsymbol{{ke}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{lnx}}} \boldsymbol{{pour}}\:\boldsymbol{{x}}\in\mathbb{R}_{+} ^{\ast} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{k}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{faisons}}\:\boldsymbol{{varier}}\:\boldsymbol{{k}} \\ $$$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}\:=>\boldsymbol{{y}}'\left(\boldsymbol{{x}}\right)=\boldsymbol{{k}}'\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}\:+\frac{\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)}{\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}^{\mathrm{2}} }} \\ $$$$=>\mathrm{3}\boldsymbol{{xy}}'−\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{lnx}}+\mathrm{1} \\ $$$$=>\mathrm{3}\boldsymbol{\mathrm{xy}}'−\boldsymbol{{y}}=\mathrm{3}\boldsymbol{{xk}}'\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}+\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}−\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}\: \\ $$$$\:\:\:\:\:\:=\mathrm{3}\boldsymbol{{xk}}'\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}} \\ $$$$<=>\boldsymbol{{k}}'\left(\boldsymbol{{x}}\right)=\frac{\boldsymbol{{lnx}}+\mathrm{1}}{\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}^{\mathrm{4}} }} \\ $$$$<=>\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)=\int\frac{\boldsymbol{{lnx}}+\mathrm{1}}{\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}^{\mathrm{4}} }}\boldsymbol{{dx}}=\int\frac{\boldsymbol{{lnx}}}{\mathrm{3}\boldsymbol{{x}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}\boldsymbol{{dx}}+\frac{\mathrm{1}}{\mathrm{3}}\int\boldsymbol{{x}}^{−\frac{\mathrm{4}}{\mathrm{3}}} \boldsymbol{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{−\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{1}}\boldsymbol{{x}}^{−\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{1}} \right)+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\boldsymbol{\mathrm{lnx}}}{\boldsymbol{\mathrm{x}}}×\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}}}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{{posons}}\:\begin{cases}{\boldsymbol{{u}}=\boldsymbol{{lnx}}}\\{\boldsymbol{{v}}'=\boldsymbol{{x}}^{−\frac{\mathrm{4}}{\mathrm{3}}} }\end{cases}=>\begin{cases}{\boldsymbol{{u}}'=\frac{\mathrm{1}}{{x}}}\\{\boldsymbol{{v}}=−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}}\end{cases} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}−\frac{\boldsymbol{{lnx}}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}+\int\boldsymbol{{x}}^{−\frac{\mathrm{4}}{\mathrm{3}}} \boldsymbol{{dx}} \\ $$$$=−\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}−\frac{\boldsymbol{{lnx}}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}+\boldsymbol{{c}}\:\:\:\:\:\:\boldsymbol{{c}}\in\mathbb{R}….{cedric} \\ $$$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{y}}'\left(\boldsymbol{{x}}\right)=\left(−\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}−\frac{\boldsymbol{{lnx}}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}+\boldsymbol{{c}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}\:\:…..{junior} \\ $$$$\mathrm{3}\boldsymbol{{xy}}'−\boldsymbol{{y}}=\boldsymbol{{lnx}}+\mathrm{1}\:\:\:\:\boldsymbol{{y}}\left(\mathrm{1}\right)=−\mathrm{2} \\ $$$$\left(\boldsymbol{{h}}\right):\mathrm{3}\boldsymbol{{xy}}'−\boldsymbol{{y}}=\mathrm{0}=>\boldsymbol{{y}}=\boldsymbol{{e}}^{\int\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{x}}}\boldsymbol{{dx}}} =\boldsymbol{{ke}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{ln}}\mid\boldsymbol{{x}}\mid} =\boldsymbol{{ke}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{lnx}}} \boldsymbol{{pour}}\:\boldsymbol{{x}}\in\mathbb{R}_{+} ^{\ast} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{k}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{faisons}}\:\boldsymbol{{varier}}\:\boldsymbol{{k}} \\ $$$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}\:=>\boldsymbol{{y}}'\left(\boldsymbol{{x}}\right)=\boldsymbol{{k}}'\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}\:+\frac{\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)}{\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}^{\mathrm{2}} }} \\ $$$$=>\mathrm{3}\boldsymbol{{xy}}'−\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{lnx}}+\mathrm{1} \\ $$$$=>\mathrm{3}\boldsymbol{\mathrm{xy}}'−\boldsymbol{{y}}=\mathrm{3}\boldsymbol{{xk}}'\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}+\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}−\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}\: \\ $$$$\:\:\:\:\:\:=\mathrm{3}\boldsymbol{{xk}}'\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}} \\ $$$$<=>\boldsymbol{{k}}'\left(\boldsymbol{{x}}\right)=\frac{\boldsymbol{{lnx}}+\mathrm{1}}{\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}^{\mathrm{4}} }} \\ $$$$<=>\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)=\int\frac{\boldsymbol{{lnx}}+\mathrm{1}}{\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}^{\mathrm{4}} }}\boldsymbol{{dx}}=\int\frac{\boldsymbol{{lnx}}}{\mathrm{3}\boldsymbol{{x}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}\boldsymbol{{dx}}+\frac{\mathrm{1}}{\mathrm{3}}\int\boldsymbol{{x}}^{−\frac{\mathrm{4}}{\mathrm{3}}} \boldsymbol{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{−\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{1}}\boldsymbol{{x}}^{−\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{1}} \right)+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\boldsymbol{\mathrm{lnx}}}{\boldsymbol{\mathrm{x}}}×\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}}}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{{posons}}\:\begin{cases}{\boldsymbol{{u}}=\boldsymbol{{lnx}}}\\{\boldsymbol{{v}}'=\boldsymbol{{x}}^{−\frac{\mathrm{4}}{\mathrm{3}}} }\end{cases}=>\begin{cases}{\boldsymbol{{u}}'=\frac{\mathrm{1}}{{x}}}\\{\boldsymbol{{v}}=−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}}\end{cases} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}−\frac{\boldsymbol{{lnx}}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}+\int\boldsymbol{{x}}^{−\frac{\mathrm{4}}{\mathrm{3}}} \boldsymbol{{dx}} \\ $$$$=−\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}−\frac{\boldsymbol{{lnx}}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}+\boldsymbol{{c}}\:\:\:\:\:\:\boldsymbol{{c}}\in\mathbb{R}….{cedric} \\ $$$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{k}}\left(\boldsymbol{{x}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{y}}'\left(\boldsymbol{{x}}\right)=\left(−\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}−\frac{\boldsymbol{{lnx}}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}}+\boldsymbol{{c}}\right)\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}\:\:…..{junior} \\ $$$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=−\mathrm{4}−\boldsymbol{{lnx}}+\boldsymbol{{c}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{y}}\left(\mathrm{1}\right)=−\mathrm{2}=>−\mathrm{4}−\boldsymbol{{ln}}\left(\mathrm{1}\right)+\boldsymbol{{c}}=−\mathrm{2} \\ $$$$=>\boldsymbol{{c}}=\mathrm{2} \\ $$$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=−\mathrm{4}−\boldsymbol{{lnx}}+\mathrm{2}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}\:} \\ $$$$======================== \\ $$$$…………..{le}\:{celebre}\:{cedric}\:{junior}…….. \\ $$$$====================== \\ $$$$ \\ $$$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=−\mathrm{4}−\boldsymbol{{lnx}}+\boldsymbol{{c}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{y}}\left(\mathrm{1}\right)=−\mathrm{2}=>−\mathrm{4}−\boldsymbol{{ln}}\left(\mathrm{1}\right)+\boldsymbol{{c}}=−\mathrm{2} \\ $$$$=>\boldsymbol{{c}}=\mathrm{2} \\ $$$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=−\mathrm{4}−\boldsymbol{{lnx}}+\mathrm{2}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}\:} \\ $$$$======================== \\ $$$$…………..{le}\:{celebre}\:{cedric}\:{junior}…….. \\ $$$$====================== \\ $$$$ \\ $$
Commented by mr W last updated on 26/Dec/22
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Commented by MikeH last updated on 26/Dec/22
merci
$$\mathrm{merci}\: \\ $$
Answered by qaz last updated on 27/Dec/22
3xy′−y=lnx+1 ⇒y′−(1/(3x))y=((lnx)/(3x))+(1/(3x)) ⇒y=e^(∫(dx/(3x))) (C+∫(((lnx)/(3x))+(1/(3x)))e^(−∫(dx/(3x))) dx) =(x)^(1/3) (C+∫(((lnx)/(3x))+(1/(3x)))(dx/( (x)^(1/3) ))) =(x)^(1/3) (C−((lnx+4)/( (x)^(1/3) ))) =C(x)^(1/3) −lnx−4 y(1)=−2 ⇒C=2 ⇒y=2(x)^(1/2) −lnx−4
$$\mathrm{3}{xy}'−{y}={lnx}+\mathrm{1} \\ $$$$\Rightarrow{y}'−\frac{\mathrm{1}}{\mathrm{3}{x}}{y}=\frac{{lnx}}{\mathrm{3}{x}}+\frac{\mathrm{1}}{\mathrm{3}{x}} \\ $$$$\Rightarrow{y}={e}^{\int\frac{{dx}}{\mathrm{3}{x}}} \left({C}+\int\left(\frac{{lnx}}{\mathrm{3}{x}}+\frac{\mathrm{1}}{\mathrm{3}{x}}\right){e}^{−\int\frac{{dx}}{\mathrm{3}{x}}} {dx}\right) \\ $$$$=\sqrt[{\mathrm{3}}]{{x}}\left({C}+\int\left(\frac{{lnx}}{\mathrm{3}{x}}+\frac{\mathrm{1}}{\mathrm{3}{x}}\right)\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{x}}}\right) \\ $$$$=\sqrt[{\mathrm{3}}]{{x}}\left({C}−\frac{{lnx}+\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{{x}}}\right) \\ $$$$={C}\sqrt[{\mathrm{3}}]{{x}}−{lnx}−\mathrm{4} \\ $$$${y}\left(\mathrm{1}\right)=−\mathrm{2}\:\:\:\:\:\Rightarrow{C}=\mathrm{2} \\ $$$$\Rightarrow{y}=\mathrm{2}\sqrt[{\mathrm{2}}]{{x}}−{lnx}−\mathrm{4} \\ $$
Commented by universe last updated on 27/Dec/22
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