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Category: Differential Equation

Question-67939

Question Number 67939 by ramirez105 last updated on 02/Sep/19 Commented by mr W last updated on 02/Sep/19 $${sir},\:{you}\:{got}\:{y}^{\mathrm{2}} \left({y}+\mathrm{2}{x}\right)={C},\:{but}\:{this} \\ $$$${doesn}'{t}\:{satisfy}\:{the}\:{original}\:{equ}. \\ $$$${is}\: \\ $$$$\frac{{dy}}{{y}}\:=−\frac{{dx}}{\mathrm{2}{x}+\mathrm{3}{y}}\:\Rightarrow\int\:\frac{{dy}}{{y}}\:=−\int\frac{{dx}}{\mathrm{2}{x}+\mathrm{3}{y}}\:+{c}…

homogenous-differential-equation-please-answer-y-x-2-xy-2y-2-dx-x-3y-2-xy-x-2-2y-0-can-someone-answer-this-

Question Number 67900 by ramirez105 last updated on 02/Sep/19 $${homogenous}\:{differential}\:{equation}. \\ $$$${please}\:{answer}. \\ $$$${y}\left({x}^{\mathrm{2}} +{xy}−\mathrm{2}{y}^{\mathrm{2}} \right){dx}+{x}\left(\mathrm{3}{y}^{\mathrm{2}} −{xy}−{x}^{\mathrm{2}} \right)\mathrm{2}{y}=\mathrm{0} \\ $$$$ \\ $$$${can}\:{someone}\:{answer}\:{this}?? \\ $$ Terms…

verify-that-x-and-x-are-the-solution-of-the-homogeneous-equation-corresponding-to-1-x-y-2-xy-1-y-2-x-1-2-x-0-lt-x-lt-1-and-find-the-general-solution-

Question Number 133403 by Engr_Jidda last updated on 21/Feb/21 $${verify}\:{that}\:\varrho^{{x}} \:{and}\:{x}\:{are}\:{the}\:{solution} \\ $$$${of}\:{the}\:{homogeneous}\:{equation}\:{corresponding} \\ $$$${to}\:\left(\mathrm{1}−{x}\right){y}^{\mathrm{2}} +{xy}^{\mathrm{1}} −{y}=\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \varrho^{{x}\:} ,\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${and}\:{find}\:{the}\:{general}\:{solution}. \\ $$ Terms of…

Question-133296

Question Number 133296 by rs4089 last updated on 21/Feb/21 Answered by SEKRET last updated on 21/Feb/21 $$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}''+\mathrm{3}\boldsymbol{\mathrm{xy}}'+\boldsymbol{\mathrm{y}}=\:\frac{\mathrm{1}}{\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} } \\ $$$$\:\:\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}} \\…