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Category: Differential Equation

dy-dx-x-y-ln-x-y-

Question Number 128395 by bramlexs22 last updated on 06/Jan/21 $$\:\:\:\frac{{dy}}{{dx}}\:=\:\left(\frac{{x}}{{y}}\right)\:\mathrm{ln}\:\left(\frac{{x}}{{y}}\right) \\ $$ Answered by mr W last updated on 07/Jan/21 $${let}\:{x}={yu} \\ $$$$\frac{{dx}}{{dy}}={u}+{y}\frac{{du}}{{dy}} \\ $$$$\mathrm{1}=\left({u}+{y}\frac{{du}}{{dy}}\right){u}\mathrm{ln}\:{u}…

yy-b-x-a-ay-2-1-x-2-

Question Number 128106 by bemath last updated on 04/Jan/21 $$\:\mathrm{yy}'\:+\:\mathrm{b}\left(\mathrm{x}−\mathrm{a}\right)=\:\frac{\mathrm{ay}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\: \\ $$ Commented by mr W last updated on 05/Jan/21 $${once}\:{i}\:{asked}\:{this}\:{question}\:{due}\:{to} \\ $$$${Q}\mathrm{127997},\:{but}\:{then}\:{i}\:{could}\:{solve}\:{it}\:{myself}…

dy-dx-y-x-y-3-e-x-x-2-

Question Number 127783 by bemath last updated on 02/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:−\frac{\mathrm{y}}{\mathrm{x}}\:=\:\frac{\mathrm{y}^{\mathrm{3}} .\mathrm{e}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} }\: \\ $$ Answered by liberty last updated on 02/Jan/21 $$\:\mathrm{Bernoulli}\:\mathrm{diff}\:\mathrm{equation}\:. \\ $$$$\:\mathrm{let}\:\mathrm{v}\:=\:\mathrm{y}^{−\mathrm{2}}…

x-2-y-2-2x-dy-2y-dx-0-y-0-1-

Question Number 127780 by bemath last updated on 02/Jan/21 $$\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2x}\right)\:\mathrm{dy}−\mathrm{2y}\:\mathrm{dx}\:=\:\mathrm{0} \\ $$$$\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$ Answered by liberty last updated on 02/Jan/21 $$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…

Question-127589

Question Number 127589 by bramlexs22 last updated on 31/Dec/20 Answered by liberty last updated on 31/Dec/20 $$\left(\bullet\right)\:\mathrm{let}\:\mathrm{y}'=\mathrm{z}\:\Rightarrow\:\mathrm{xz}'\:+\:\mathrm{z}\:=\:\mathrm{3x}^{\mathrm{2}} −\mathrm{x}\: \\ $$$$\:\:\:\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left(\mathrm{xz}\right)\:=\:\mathrm{3x}^{\mathrm{2}} −\mathrm{x}\: \\ $$$$\:\:\:\mathrm{xz}\:=\:\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \:+\mathrm{C}_{\mathrm{1}}…