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Category: Differential Equation

Solve-the-following-system-of-differential-equations-for-functions-x-t-and-y-t-m-d-2-x-dt-2-kx-x-2-y-2-2-m-d-2-y-dt-2-ky-x-2-y-2-2-

Question Number 4625 by Yozzis last updated on 14/Feb/16 $${Solve}\:{the}\:{following}\:{system}\:{of}\:{differential}\: \\ $$$${equations}\:{for}\:{functions}\:{x}\left({t}\right)\:{and}\:{y}\left({t}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }=\frac{{kx}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }=\frac{{ky}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}}…

Give-an-example-of-differential-equation-which-has-no-solutions-

Question Number 3656 by prakash jain last updated on 17/Dec/15 $$\mathrm{Give}\:\mathrm{an}\:\mathrm{example}\:\mathrm{of}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{which} \\ $$$$\mathrm{has}\:\mathrm{no}\:\mathrm{solutions}. \\ $$ Answered by Filup last updated on 18/Dec/15 $${Weierstrass}\:{function} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{fuction}\:\mathrm{that}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{at}\:\mathrm{all}…

sin-4-xdx-

Question Number 134694 by Engr_Jidda last updated on 06/Mar/21 $$\int{sin}^{\mathrm{4}} {xdx} \\ $$ Answered by john_santu last updated on 06/Mar/21 $$\mathcal{J}\:=\:\int\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\right)^{\mathrm{2}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\right)^{\mathrm{2}} \:{dx}…

Question-68946

Question Number 68946 by ramirez105 last updated on 17/Sep/19 Commented by kaivan.ahmadi last updated on 17/Sep/19 $$\frac{\partial{u}}{\partial{x}}=\left(\mathrm{2}{xy}−{tany}\right)\Rightarrow{u}\left({x},{y}\right)={x}^{\mathrm{2}} {y}−{xtany}+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}={x}^{\mathrm{2}} −{xsec}^{\mathrm{2}} {y}={x}^{\mathrm{2}} −{xsec}^{\mathrm{2}} {y}+\frac{{dh}}{{dy}}\Rightarrow\frac{{dh}}{{dy}}=\mathrm{0}\Rightarrow{h}={c}' \\…