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Question Number 117632 by syamil last updated on 12/Oct/20

$$Solutionfrom2xydy=(x^2-y^2)dx$$

Answered by 1549442205PVT last updated on 13/Oct/20

$$2xydy=(x^2-y^2)dx$$$$\iff ({\mathrm{x}}^2-{\mathrm{y}}^2)\mathrm{dx}-2\mathrm{xydy}=0$$$$\mathrm{Put}\mathrm{P}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2-{\mathrm{y}}^2,\mathrm{Q}(\mathrm{x},\mathrm{y})=-2\mathrm{xy}$$$$\frac{\partial \mathrm{P}}{\partial \mathrm{y}}=-2\mathrm{y},\frac{\partial \mathrm{Q}}{\partial \mathrm{x}}=-2\mathrm{y}.\frac{\partial \mathrm{P}}{\partial \mathrm{y}}=\frac{\partial \mathrm{Q}}{\partial \mathrm{x}}=-2\mathrm{y}\Rightarrow$$$$\mathrm{This}\mathrm{is}\mathrm{exact}\mathrm{equation}.\mathrm{Hence},\mathrm{we}\mathrm{have}$$$${\int }_0^{\mathrm{x}}({\mathrm{x}}^2-{\mathrm{y}}^2)\mathrm{dx}-{\int }_0^{\mathrm{y}}2\mathrm{xydy}=\mathrm{C}$$$$\iff (\frac{{\mathrm{x}}^3}{3}-{\mathrm{y}}^2\mathrm{x}){\mid }_0^{\mathrm{x}}-{\mathrm{xy}}^2{\mid }_0^{\mathrm{y}}=\mathrm{C}$$$$\iff \frac{{\mathrm{x}}^3}{3}-{\mathrm{y}}^2\mathrm{x}-{\mathrm{xy}}^2=\mathrm{C}$$

Answered by bobhans last updated on 13/Oct/20

$$\mathrm{letting}\mathrm{y}=\lambda \mathrm{x}\Rightarrow \mathrm{dy}=\lambda \mathrm{dx}+\mathrm{xd}\lambda$$$$\Rightarrow {2\mathrm{x}}^2\lambda (\lambda \mathrm{dx}+\mathrm{xd}\lambda )={\mathrm{x}}^2(1-{\lambda }^2)\mathrm{dx}$$$$\Rightarrow 2{\lambda }^2\mathrm{dx}+\mathrm{xd}\lambda =(1-{\lambda }^2)\mathrm{dx}$$$$\Rightarrow \mathrm{xd}\lambda =(1-3{\lambda }^2)\mathrm{dx}$$$$\Rightarrow \frac{\mathrm{d}\lambda }{1-3{\lambda }^2}=\frac{\mathrm{dx}}{\mathrm{x}}\Rightarrow \int \frac{\mathrm{d}\lambda }{(1+\lambda \sqrt{3})(1-\lambda \sqrt{3})}=\int \frac{\mathrm{dx}}{\mathrm{x}}$$$$\int (\frac{1}{1-\lambda \sqrt{3}}+\frac{1}{1+\lambda \sqrt{3}})\mathrm{d}\lambda =\int \frac{2\mathrm{dx}}{\mathrm{x}}$$$$-\frac{1}{\sqrt{3}}\ln (1-\lambda \sqrt{3})+\frac{1}{\sqrt{3}}\ln (1+\lambda \sqrt{3})=\ln \mid {\mathrm{Cx}}^2\mid$$$$\Rightarrow \ln \mid \frac{1+\lambda \sqrt{3}}{1-\lambda \sqrt{3}}\mid =\sqrt{3}\ln \mid {\mathrm{Cx}}^2\mid$$$$\ln \mid 1-3{\lambda }^2\mid =\ln \mid {\mathrm{Cx}}^2{\mid }^{\sqrt{3}}$$$$\Rightarrow 1-3\left(\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}\right)={\left({\mathrm{Cx}}^2\right)}^{\sqrt{3}}={\mathrm{Kx}}^{2\sqrt{3}}$$$$\Rightarrow {\mathrm{y}}^2=\frac{{\mathrm{x}}^2-{\mathrm{Kx}}^{2+2\sqrt{3}}}{3}$$

Answered by TANMAY PANACEA last updated on 13/Oct/20

$$2xydy+y^{2}dx=x^{2}dx$$$${xdy}^2+y^{2}dx=x^{2}dx$$$$d\left({xy}^2\right)=\frac{1}{3}{dx}^3$$$${xy}^2=\frac{x^3}{3}+c$$$$ormethod$$$$\frac{dy}{dx}=\frac{x^2-y^2}{2xy}=\frac{1-\frac{y^2}{x^2}}{2\frac{y}{x}}$$$$v=\frac{y}{x}\to \frac{dy}{dx}=x\frac{dv}{dx}+v$$$$v+x\frac{dv}{dx}=\frac{1-v^2}{2v}$$$$x\frac{dv}{dx}=\frac{1-v^2}{2v}-v$$$$\frac{2v}{1-3v^2}dv=\frac{dx}{x}$$$$\frac{2}{3}\int \frac{vdv}{\frac{1}{3}-v^2}=\int \frac{dx}{x}$$$$\frac{1}{3}\int \frac{d(\frac{1}{3}-v^2)}{\frac{1}{3}-v^2}=-\int \frac{dx}{x}$$$$\frac{1}{3}ln(\frac{1}{3}-v^2)=-lnx+lnc$$$$ln{(\frac{1}{3}-\frac{y^2}{x^2})}^{\frac{1}{3}}+lnx=lnc$$$$x{(\frac{1}{3}-\frac{y^2}{x^2})}^{\frac{1}{3}}=c$$$$x^3(\frac{1}{3}-\frac{y^2}{x^2})=c^3={\mathit{C}}_1$$$$\frac{x^3}{3}-{xy}^2=C_1$$

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