Menu Close

Question-133296

Tinku Tara 0 Comments
Pin




Question Number 133296 by rs4089 last updated on 21/Feb/21
Answered by SEKRET last updated on 21/Feb/21
x^2 y′′+3xy′+y= (1/((1−x)^2 )) x^2 y′′+3xy′+y=0 y= x^m y′=mx^(m−1) y′′=m(m−1)x^(m−2) x^2 ∙(m(m−1))x^(m−2) +3x∙m∙x^(m−1) +x^m =0 m^2 +2m+1=0 m_1 = −1 m_2 = −1 x≠0 y_1 = (C_1 /x) y_2 =((C_2 ln(x))/x) y=y_1 +y_2 y=(C_1 /x)+((C_2 ln(x))/x) Q=((1/x))∙(((lnx)/x))^′ − (((ln(x))/x))∙((1/x))′= (1/x^3 ) y′′+(3/x) y′+(1/x^2 ) y = (1/(x^2 (1−x)^2 )) u_1 =∫((−ln(x))/((1−x)^2 ))dx= ln((x/(1−x)))+((ln(x))/(1−x)) u_2 =∫ (1/((x−1)^2 ))dx= (1/(1−x)) y_u = (1/x)ln((x/(1−x))) y=(1/x)ln((x/(1−x)))+(c_1 /x)+((c_2 ln(x))/x)
$$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}''+\mathrm{3}\boldsymbol{\mathrm{xy}}'+\boldsymbol{\mathrm{y}}=\:\frac{\mathrm{1}}{\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} } \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}''+\mathrm{3}\boldsymbol{\mathrm{xy}}'+\boldsymbol{\mathrm{y}}=\mathrm{0}\:\:\:\:\boldsymbol{\mathrm{y}}=\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}} \:\:\:\:\boldsymbol{\mathrm{y}}'=\boldsymbol{\mathrm{mx}}^{\boldsymbol{\mathrm{m}}−\mathrm{1}} \:\:\:\:\:\boldsymbol{\mathrm{y}}''=\boldsymbol{\mathrm{m}}\left(\boldsymbol{\mathrm{m}}−\mathrm{1}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}−\mathrm{2}} \\ $$$$\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \centerdot\left(\boldsymbol{\mathrm{m}}\left(\boldsymbol{\mathrm{m}}−\mathrm{1}\right)\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}−\mathrm{2}} +\mathrm{3}\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{m}}\centerdot\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}−\mathrm{1}} +\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{m}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{m}}+\mathrm{1}=\mathrm{0}\:\:\:\:\:\boldsymbol{\mathrm{m}}_{\mathrm{1}} =\:−\mathrm{1}\:\:\:\:\boldsymbol{\mathrm{m}}_{\mathrm{2}} =\:−\mathrm{1}\:\:\boldsymbol{\mathrm{x}}\neq\mathrm{0} \\ $$$$\:\boldsymbol{\mathrm{y}}_{\mathrm{1}} =\:\frac{\boldsymbol{\mathrm{C}}_{\mathrm{1}} }{\boldsymbol{\mathrm{x}}}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}_{\mathrm{2}} =\frac{\boldsymbol{\mathrm{C}}_{\mathrm{2}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\:\:\:\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{y}}_{\mathrm{1}} +\boldsymbol{\mathrm{y}}_{\mathrm{2}} \:\:\:\:\:\:\boldsymbol{\mathrm{y}}=\frac{\boldsymbol{\mathrm{C}}_{\mathrm{1}} }{\boldsymbol{\mathrm{x}}}+\frac{\boldsymbol{\mathrm{C}}_{\mathrm{2}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}} \\ $$$$\boldsymbol{\mathrm{Q}}=\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\centerdot\left(\frac{\boldsymbol{\mathrm{lnx}}}{\boldsymbol{\mathrm{x}}}\right)\:^{'} \:−\:\left(\frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\right)\centerdot\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)'=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} } \\ $$$$\:\boldsymbol{\mathrm{y}}''+\frac{\mathrm{3}}{\boldsymbol{\mathrm{x}}}\:\boldsymbol{\mathrm{y}}'+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:\boldsymbol{\mathrm{y}}\:=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} } \\ $$$$\:\boldsymbol{\mathrm{u}}_{\mathrm{1}} =\int\frac{−\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\:\boldsymbol{\mathrm{ln}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}}\right)+\frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}−\boldsymbol{\mathrm{x}}} \\ $$$$\:\boldsymbol{\mathrm{u}}_{\mathrm{2}} =\int\:\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\:\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{\mathrm{x}}} \\ $$$$\:\boldsymbol{\mathrm{y}}_{\boldsymbol{\mathrm{u}}} =\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{ln}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}}\right) \\ $$$$\:\boldsymbol{\mathrm{y}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{ln}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}}\right)+\frac{\boldsymbol{\mathrm{c}}_{\mathrm{1}} }{\boldsymbol{\mathrm{x}}}+\frac{\boldsymbol{\mathrm{c}}_{\mathrm{2}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}} \\ $$

Pin

Leave a Reply

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