Category: Differential Equation

Question Number 116614 by bemath last updated on 05/Oct/20 $$({\mathrm{x}}^2\cos {}^2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-\mathrm{y})\mathrm{dx}+\mathrm{x}\mathrm{dy}=0$$$$\mathrm{where}\mathrm{y}\left(1\right)=\frac{\pi }{4}$$ Commented by TANMAY PANACEA last updated on 05/Oct/20 $${i}\:{think}\:{it}\:{should}\:{be}\:{cos}^{\mathrm{2}}…

Question Number 182036 by Maisam last updated on 03/Dec/22 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 116457 by bobhans last updated on 04/Oct/20 $$\mathrm{Use}\mathrm{Laplace}\mathrm{transform}\mathrm{to}\mathrm{solve}\mathrm{the}$$$$\mathrm{initial}\mathrm{value}\mathrm{problem}{\mathrm{ty}}^{″}+(4\mathrm{t}-2){\mathrm{y}}^{\prime }+(13\mathrm{t}-4)\mathrm{y}=0$$$$\mathrm{where}\mathrm{y}\left(0\right)=0\mathrm{and}{\mathrm{y}}^{\prime }\left(0\right)=0$$ Answered by Olaf last updated on 04/Oct/20 $$\mathrm{L}\left({y}''\right)\:=\:\mathrm{p}^{\mathrm{2}} \mathrm{L}\left(\mathrm{p}\right)−\mathrm{p}{y}\left(\mathrm{0}\right)−{y}'\left(\mathrm{0}\right)…

Question Number 116329 by bemath last updated on 03/Oct/20 $$\mathrm{solve}\mathrm{the}\mathrm{diff}\mathrm{equation}$$$$\left(1\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}+3\mathrm{y}-5}{\mathrm{x}-\mathrm{y}-1}$$$$\left(2\right)(3\mathrm{y}-7\mathrm{x}-3)\mathrm{dx}+(7\mathrm{y}-3\mathrm{x}-7)\mathrm{dy}=0$$ Answered by mr W last updated on 03/Oct/20 $$\left(\mathrm{1}\right)\:\:\frac{{dy}}{{dx}}=\frac{{x}+\mathrm{3}{y}−\mathrm{5}}{{x}−{y}−\mathrm{1}}…

Question Number 181799 by amin96 last updated on 30/Nov/22 $$\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{e}\mathit{d}\mathit{i}\mathit{f}\mathit{g}\mathit{e}\mathit{r}\mathit{e}\mathit{n}\mathit{t}\mathit{i}\mathit{a}\mathit{l}\mathit{e}\mathit{q}\mathit{u}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}$$$$\mathit{y}\sqrt{1+{\left({\mathit{y}}^{\prime }\right)}^2}={\mathit{y}}^{\prime }$$ Answered by mr W last updated on 30/Nov/22 $${y}^{\mathrm{2}} \left(\mathrm{1}+\left({y}'\right)^{\mathrm{2}}…