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Category: Differential Equation

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}} \\…

solve-by-laplace-transform-method-x-w-0-2-x-coswt-x-0-x-0-x-0-v-0-w-2-w-0-2-

Question Number 67761 by ugwu Kingsley last updated on 31/Aug/19 $${solve}\:{by}\:{laplace}\:{transform}\:{method} \\ $$$$ \\ $$$$\overset{\bullet\bullet\:} {{x}}\:+{w}_{\mathrm{0}} ^{\mathrm{2}} {x}={coswt}\:\:\:{x}\left(\mathrm{0}\right)={x}_{\mathrm{0}} \:\overset{\bullet} {{x}}\left(\mathrm{0}\right)={v}_{\mathrm{0}} \:\:\:\:\:{w}^{\mathrm{2}} \neq\:{w}_{\mathrm{0}} ^{\mathrm{2}} \\ $$…

using-variation-of-parameters-method-x-2-2-y-x-2-y-2x-4-x-2-y-2xy-2y-x-2-lnx-3x-

Question Number 67759 by ugwu Kingsley last updated on 31/Aug/19 $${using}\:{variation}\:{of}\:{parameters}\:{method} \\ $$$$ \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} {y}''−\left({x}+\mathrm{2}\right){y}'=\mathrm{2}{x}+\mathrm{4} \\ $$$$ \\ $$$$ \\ $$$${x}^{\mathrm{2}} {y}''+\mathrm{2}{xy}'−\mathrm{2}{y}={x}^{\mathrm{2}} {lnx}+\mathrm{3}{x} \\…

Question-67719

Question Number 67719 by aliesam last updated on 30/Aug/19 Answered by mind is power last updated on 30/Aug/19 $$\Leftrightarrow\frac{{d}\left({y}+{x}\right)}{{dx}}+{x}\left({x}+{y}\right)={x}^{\mathrm{3}} \left({x}+{y}\right)^{\mathrm{5}} \\ $$$${let}\:{z}={y}+{x} \\ $$$$\Rightarrow\frac{{dz}}{{dx}}+{xz}={x}^{\mathrm{3}} {z}^{\mathrm{5}}…