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Question Number 74821 by sridhar nayak last updated on 01/Dec/19

Answered by mind is power last updated on 01/Dec/19

$$\iff ({6\mathrm{e}}^{\mathrm{y}}-2\mathrm{x})\mathrm{dy}-\mathrm{dx}=0...\mathrm{E}$$$$\mathrm{try}\mathrm{too}\mathrm{find}\mathrm{k}\left(\mathrm{y}\right)\mathrm{To}\mathrm{mak}\mathrm{it}\mathrm{exacte}$$$$\iff \mathrm{k}\left(\mathrm{y}\right)({6\mathrm{e}}^{\mathrm{y}}-2\mathrm{x})\mathrm{dy}-\mathrm{k}\left(\mathrm{y}\right)\mathrm{dx}=0$$$$\frac{\partial }{\partial \mathrm{x}}\left(\mathrm{k}\left(\mathrm{y}\right)({6\mathrm{e}}^{\mathrm{y}}-2\mathrm{x})\right)=\frac{\partial }{\mathrm{dy}}(-\mathrm{k}\left(\mathrm{y}\right))$$$$\Rightarrow -2\mathrm{k}=-{\mathrm{k}}^{\prime }\Rightarrow {\mathrm{k}}^{\prime }=2\mathrm{k}\Rightarrow \mathrm{k}={\mathrm{e}}^{2\mathrm{y}}$$$$\mathrm{E}\iff {\mathrm{e}}^{2\mathrm{y}}({6\mathrm{e}}^{\mathrm{y}}-2\mathrm{x})\mathrm{dy}-{\mathrm{e}}^{2\mathrm{y}}\mathrm{dx}=0$$$$\mathrm{let}\mathrm{F}(\mathrm{x},\mathrm{y})=\mathrm{c}$$$$\mathrm{such}\mathrm{dF}={\mathrm{e}}^{2\mathrm{y}}({6\mathrm{e}}^{\mathrm{y}}-2\mathrm{x})\mathrm{dy}-{\mathrm{e}}^{2\mathrm{y}}\mathrm{dx}=0$$$$\Rightarrow \frac{\partial \mathrm{F}}{\partial \mathrm{x}}=-{\mathrm{e}}^{2\mathrm{y}}\Rightarrow \mathrm{F}(\mathrm{x},\mathrm{y})=-{\mathrm{xe}}^{2\mathrm{y}}+\mathrm{h}\left(\mathrm{y}\right)$$$$\frac{\partial \mathrm{F}}{\partial \mathrm{y}}={6\mathrm{e}}^{3\mathrm{y}}-{2\mathrm{xe}}^{2\mathrm{y}}=-{2\mathrm{xe}}^{2\mathrm{y}}+{\mathrm{h}}^{\prime }\left(\mathrm{y}\right)$$$$\Rightarrow {\mathrm{h}}^{\prime }\left(\mathrm{y}\right)={6\mathrm{e}}^{3\mathrm{y}}\Rightarrow \mathrm{h}\left(\mathrm{y}\right)={2\mathrm{e}}^{3\mathrm{y}}+\mathrm{c}$$$$\mathrm{F}(\mathrm{x},\mathrm{y})=-{\mathrm{xe}}^{2\mathrm{y}}+{2\mathrm{e}}^{3\mathrm{y}}+\mathrm{c}=0$$$$$$$$$$

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