Question Number 175210 by mnjuly1970 last updated on 23/Aug/22 $$ \\ $$$$\:\:\:{y}'{x}\:+\:{y}\:=\:{y}^{\:\mathrm{2}} {ln}\left({x}\right) \\ $$$$\:\:\:\:{u}={y}^{\:−\mathrm{1}} \:\Rightarrow\:{u}'\:=−{y}'{y}^{\:−\mathrm{2}} \\ $$$$\:\:\:\:\:−{y}'{y}^{\:−\mathrm{2}} {x}\:−{y}^{−\mathrm{1}} =\:−{ln}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:{u}'{x}\:−{u}\:=\:−{ln}\left({x}\right) \\ $$$$\:\:\:\:{u}'−\frac{\mathrm{1}}{{x}}\:{u}\:=\frac{−{ln}\left({x}\right)}{{x}}\: \\…
Question Number 44120 by peter frank last updated on 21/Sep/18 Answered by ajfour last updated on 21/Sep/18 $$\frac{{dy}}{{dx}}=\frac{{dy}/{d}\theta}{{dx}/{d}\theta}=\frac{{a}\mathrm{sin}\:\theta}{{a}+{a}\mathrm{cos}\:\theta}\:=\frac{\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\left[\frac{{d}}{{d}\theta}\left(\frac{{dy}}{{dx}}\right)\right]/\left(\frac{{dx}}{{d}\theta}\right) \\ $$$$\:\:\:=\frac{\mathrm{cos}\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)−\mathrm{sin}\:\theta\left(−\mathrm{sin}\:\theta\right)}{\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} ×{a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}…
Question Number 44091 by peter frank last updated on 21/Sep/18 Answered by tanmay.chaudhury50@gmail.com last updated on 22/Sep/18 $$ \\ $$$${ln}\left({y}+\mathrm{1}\right)\left\{\frac{\mathrm{2}\left({y}+\mathrm{1}\right)−{y}}{{y}\left({y}+\mathrm{1}\right)}\right\} \\ $$$${ln}\left({y}+\mathrm{1}\right)\left[\frac{\mathrm{2}}{{y}}−\frac{\mathrm{1}}{{y}+\mathrm{1}}\right] \\ $$$$ \\…
Question Number 175053 by mnjuly1970 last updated on 17/Aug/22 Commented by infinityaction last updated on 18/Aug/22 $$\sqrt{\mathrm{2}}{tanh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$ Commented by Frix last updated…
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Question Number 43854 by peter frank last updated on 16/Sep/18 $${use}\:{the}\:{first}\:{principle}\:{y}=\mathrm{ln}\:\sqrt{\mathrm{cos}\:{x}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 16/Sep/18 $${y}+\bigtriangleup{y}={ln}\sqrt{{cos}\left({x}+\bigtriangleup{x}\right)}\:\: \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left({cos}\left({x}+\bigtriangleup{x}\right)\right\}−\frac{\mathrm{1}}{\mathrm{2}}\left\{{lncosx}\right\}\right. \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}\right\}…
Question Number 174838 by mnjuly1970 last updated on 12/Aug/22 $$ \\ $$$$\:\:\:\:\:{calculate} \\ $$$$ \\ $$$$\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:{x}^{\:\mathrm{2}} }{\left(\:{sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\:\mathrm{2}} }{dx}=? \\ $$$$ \\ $$ Answered…
Question Number 174766 by mnjuly1970 last updated on 10/Aug/22 $$ \\ $$$$\:\:\:\:{Q}:\:\:{by}\:{using}\:{fourier}\:{series}, \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\Omega}=\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:\boldsymbol{{n}}−\mathrm{1}} }{\left(\mathrm{2}\boldsymbol{{n}}−\mathrm{1}\:\right)^{\:\mathrm{3}} }\:=\:\frac{\boldsymbol{\pi}^{\:\mathrm{3}} }{\mathrm{32}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−−− \\ $$…