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Category: Differential Equation

Question-198104

Question Number 198104 by sonu753 last updated on 10/Oct/23 Answered by mr W last updated on 11/Oct/23 $$\frac{{dx}}{{dy}}−\frac{{x}}{{y}}=−{y} \\ $$$$\left[{d}.{e}.\:{of}\:{type}\:{y}'+{p}\left({x}\right){y}={q}\left({x}\right)\right] \\ $$$$\int{p}\left({y}\right){dy}=−\int\frac{{dy}}{{y}}=−\mathrm{ln}\:{y} \\ $$$${u}\left({y}\right)={e}^{−\mathrm{ln}\:{y}} =\frac{\mathrm{1}}{{y}}…

please-check-my-answer-x-2y-5-dx-2x-y-4-dy-0-X-x-a-amp-Y-y-b-X-a-2Y-2b-5-dX-2X-2a-Y-b-4-dY-0-a-2b-5-2a-b-4-a-1-b-2-X-2Y-dX-2X-Y-dY-0-Y-XV-dY-dX-V-X-dV-dx-X-2XV-2X-XV-V-X

Question Number 197389 by uchihayahia last updated on 16/Sep/23 $$ \\ $$$${please}\:{check}\:{my}\:{answer} \\ $$$$\:\left({x}−\mathrm{2}{y}+\mathrm{5}\right){dx}+\left(\mathrm{2}{x}−{y}+\mathrm{4}\right){dy}=\mathrm{0} \\ $$$$\:{X}={x}+{a}\:\&\:{Y}={y}+{b} \\ $$$$ \\ $$$$\:\left({X}−{a}−\mathrm{2}{Y}+\mathrm{2}{b}+\mathrm{5}\right){dX}+\left(\mathrm{2}{X}−\mathrm{2}{a}−{Y}+{b}+\mathrm{4}\right){dY}=\mathrm{0} \\ $$$$\:-{a}+\mathrm{2}{b}=-\mathrm{5} \\ $$$$\:-\mathrm{2}{a}+{b}=-\mathrm{4} \\…

Question-196934

Question Number 196934 by Amidip last updated on 03/Sep/23 Answered by aleks041103 last updated on 04/Sep/23 $${M}\left({x},{y}\right){dx}+{N}\left({x},{y}\right){dy}=\mathrm{0} \\ $$$${if}\:\:\partial_{{y}} {M}=\partial_{{x}} {N},\:{then}\:\exists{F}\left({x},{y}\right): \\ $$$${M}\left({x},{y}\right){dx}+{N}\left({x},{y}\right){dy}={dF}=\mathrm{0}\Rightarrow{F}={const}. \\ $$$${in}\:{our}\:{case}\:\partial_{{y}}…

Question-196458

Question Number 196458 by peter frank last updated on 25/Aug/23 Answered by Peace last updated on 25/Aug/23 $$\int\left({sin}\left({y}\right)+{ycos}\left({y}\right)\right){dh}=\int{x}\left(\mathrm{2}{ln}\left(\mathrm{x}\right)+\mathrm{1}\right)\mathrm{dx} \\ $$$$\Leftrightarrow+{ysin}\left({y}\right)+{c}={x}^{\mathrm{2}} {ln}\left({x}\right) \\ $$$${x}^{\mathrm{2}} {ln}\left({x}\right)−{ysin}\left({y}\right)+{c}=\mathrm{0} \\…

Solve-y-3y-y-2-1-y-2-

Question Number 196199 by Frix last updated on 19/Aug/23 $$\mathrm{Solve}: \\ $$$${y}'''=\frac{\mathrm{3}{y}'\left({y}''\right)^{\mathrm{2}} }{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} } \\ $$ Answered by aleks041103 last updated on 20/Aug/23 $$\left({ln}\left({y}''\right)\right)'=\frac{{y}'''}{{y}''}=\frac{\mathrm{3}{y}'{y}''}{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}}\:\frac{\left(\left({y}'\right)^{\mathrm{2}}…