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Author: Tinku Tara

Find-all-real-solutions-y-to-the-equation-sin-dy-dx-sin-d-2-y-dx-2-0-For-each-solution-determine-the-value-of-Q-max-r-1-n-sin-d-r-y-dx-r-giving-the-value-s-of-x-for-which-Q-ar

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}. \\…

Find-the-solution-of-the-d-e-sinhx-dy-dx-2-2-dy-dx-sinhx-0-which-satisfies-y-0-at-x-0-

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…

Solve-the-d-e-x-2-dy-dx-xy-x-2-y-2-1-by-letting-y-1-x-1-v-where-v-is-a-function-of-x-

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…

1-sin-3-x-dx-

Question Number 133117 by SOMEDAVONG last updated on 19/Feb/21 $$\int\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{3}} \mathrm{x}}\mathrm{dx}=? \\ $$ Answered by bemath last updated on 19/Feb/21 $$=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cot}\:\mathrm{x}\:\mathrm{csc}\:\mathrm{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}\mid\:+\:\mathrm{C} \\ $$ Terms of…