Question Number 133296 by rs4089 last updated on 21/Feb/21 Answered by SEKRET last updated on 21/Feb/21 $$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}''+\mathrm{3}\boldsymbol{\mathrm{xy}}'+\boldsymbol{\mathrm{y}}=\:\frac{\mathrm{1}}{\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} } \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}''+\mathrm{3}\boldsymbol{\mathrm{xy}}'+\boldsymbol{\mathrm{y}}=\mathrm{0}\:\:\:\:\boldsymbol{\mathrm{y}}=\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}} \:\:\:\:\boldsymbol{\mathrm{y}}'=\boldsymbol{\mathrm{mx}}^{\boldsymbol{\mathrm{m}}−\mathrm{1}} \:\:\:\:\:\boldsymbol{\mathrm{y}}''=\boldsymbol{\mathrm{m}}\left(\boldsymbol{\mathrm{m}}−\mathrm{1}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}−\mathrm{2}} \\…
Question Number 67761 by ugwu Kingsley last updated on 31/Aug/19 $${solve}\:{by}\:{laplace}\:{transform}\:{method} \\ $$$$ \\ $$$$\overset{\bullet\bullet\:} {{x}}\:+{w}_{\mathrm{0}} ^{\mathrm{2}} {x}={coswt}\:\:\:{x}\left(\mathrm{0}\right)={x}_{\mathrm{0}} \:\overset{\bullet} {{x}}\left(\mathrm{0}\right)={v}_{\mathrm{0}} \:\:\:\:\:{w}^{\mathrm{2}} \neq\:{w}_{\mathrm{0}} ^{\mathrm{2}} \\ $$…
Question Number 67759 by ugwu Kingsley last updated on 31/Aug/19 $${using}\:{variation}\:{of}\:{parameters}\:{method} \\ $$$$ \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} {y}''−\left({x}+\mathrm{2}\right){y}'=\mathrm{2}{x}+\mathrm{4} \\ $$$$ \\ $$$$ \\ $$$${x}^{\mathrm{2}} {y}''+\mathrm{2}{xy}'−\mathrm{2}{y}={x}^{\mathrm{2}} {lnx}+\mathrm{3}{x} \\…
Question Number 67719 by aliesam last updated on 30/Aug/19 Answered by mind is power last updated on 30/Aug/19 $$\Leftrightarrow\frac{{d}\left({y}+{x}\right)}{{dx}}+{x}\left({x}+{y}\right)={x}^{\mathrm{3}} \left({x}+{y}\right)^{\mathrm{5}} \\ $$$${let}\:{z}={y}+{x} \\ $$$$\Rightarrow\frac{{dz}}{{dx}}+{xz}={x}^{\mathrm{3}} {z}^{\mathrm{5}}…
Question Number 2160 by Yozzis last updated on 05/Nov/15 $${Solve}\:{the}\:{d}.{e} \\ $$$${sin}\left(\frac{{d}^{\mathrm{3}} {y}}{{dt}^{\mathrm{3}} }\right)+\mathrm{3}{t}^{\mathrm{2}} {y}=\mathrm{6}{t}. \\ $$ Commented by Yozzi last updated on 07/Nov/15 $${I}\:{was}\:{helping}\:{a}\:{friend}\:{with}\:{his}\:…
Question Number 2161 by Yozzis last updated on 05/Nov/15 $${Solve}\:{the}\:{d}.{e}\: \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }\sqrt{\frac{{dy}}{{dt}}}−\frac{{t}}{{y}}=\mathrm{0}. \\ $$ Answered by prakash jain last updated on 07/Nov/15 $$\mathrm{Solving}\:\mathrm{as}\:\mathrm{a}\:\mathrm{series}\:\mathrm{expansion}…
Question Number 2159 by Yozzis last updated on 05/Nov/15 $${Suppose}\:{that}\:{y}_{{n}\:} \:{satisfies}\:{the}\:{equations}\: \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}_{{n}} }{{dx}^{\mathrm{2}} }−{x}\frac{{dy}_{{n}} }{{dx}}+{n}^{\mathrm{2}} {y}=\mathrm{0},\:{y}_{{n}} \left(\mathrm{1}\right)=\mathrm{1} \\ $$$${y}_{{n}} \left({x}\right)=\left(−\mathrm{1}\right)^{{n}} {y}_{{n}} \left(−{x}\right).…
Question Number 2054 by Yozzi last updated on 01/Nov/15 $${Find}\:{all}\:{real}\:{solutions}\:{y}\:{to}\:{the}\:{equation} \\ $$$$\:\:\:\:{sin}\left(\frac{{dy}}{{dx}}\right)+{sin}\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)=\mathrm{0}\:. \\ $$$${For}\:{each}\:{solution}\:{determine}\:{the}\:{value}\:{of} \\ $$$${Q}={max}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}\left(\frac{{d}^{{r}} {y}}{{dx}^{{r}} }\right)\right),\:{giving}\:{the}\:{value}\left({s}\right) \\ $$$${of}\:{x}\:{for}\:{which}\:{Q}\:{arises}. \\…
Question Number 2052 by Yozzi last updated on 01/Nov/15 $${Find}\:{the}\:{solution}\:{of}\:{the}\:{d}.{e} \\ $$$$\:\left({sinhx}\right)\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} +\mathrm{2}\frac{{dy}}{{dx}}−{sinhx}=\mathrm{0} \\ $$$${which}\:{satisfies}\:{y}=\mathrm{0}\:{at}\:{x}=\mathrm{0}. \\ $$ Commented by prakash jain last updated on 01/Nov/15…
Question Number 2051 by Yozzi last updated on 01/Nov/15 $${Solve}\:{the}\:{d}.{e}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \frac{{dy}}{{dx}}+{xy}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{1} \\ $$$${by}\:{letting}\:{y}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{v}}\:{where} \\ $$$${v}\:{is}\:{a}\:{function}\:{of}\:{x}.\: \\ $$ Answered by 123456 last…