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Question Number 84477 by jagoll last updated on 13/Mar/20

$$(\mathrm{ycos}\mathrm{x}+{2\mathrm{xe}}^{\mathrm{y}})\mathrm{dx}+(\sin \mathrm{x}+{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{y}}-1)\mathrm{dy}=0$$

Answered by john santu last updated on 13/Mar/20

$$\frac{\partial \mathrm{M}}{\partial \mathrm{y}}=\cos \mathrm{x}+{2\mathrm{xe}}^{\mathrm{y}}$$$$\frac{\partial \mathrm{N}}{\partial \mathrm{x}}=\cos \mathrm{x}+{2\mathrm{xe}}^{\mathrm{y}}\Rightarrow \frac{\partial \mathrm{M}}{\partial \mathrm{y}}=\frac{\partial \mathrm{N}}{\partial \mathrm{x}}$$$$\mathrm{diff}\mathrm{equation}\mathrm{exact}$$$$\mathrm{F}(\mathrm{x},\mathrm{y})=\int (\mathrm{ycos}\mathrm{x}+{2\mathrm{xe}}^{\mathrm{y}})\mathrm{dx}+\mathrm{g}\left(\mathrm{y}\right)$$$$=\mathrm{ysin}\mathrm{x}+{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{y}}+\mathrm{g}\left(\mathrm{y}\right)$$$$(\ast ){\mathrm{g}}^{\prime }\left(\mathrm{y}\right)=\mathrm{N}(\mathrm{x},\mathrm{y})-\frac{\partial }{\partial \mathrm{y}}[\mathrm{ysin}\mathrm{x}+{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{y}}]$$$${\mathrm{g}}^{\prime }\left(\mathrm{y}\right)=\sin \mathrm{x}+{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{y}}-1-[\sin \mathrm{x}+{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{y}}]$$$$\mathrm{g}\left(\mathrm{y}\right)=\int -\mathrm{dy}=-\mathrm{y}$$$$\therefore \mathrm{F}(\mathrm{x},\mathrm{y})=\mathrm{C}$$$$\therefore \mathrm{ysin}\mathrm{x}+{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{y}}-\mathrm{y}=\mathrm{C}$$

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

$$ycosxdx+sinxdy+2{xe}^{y}dx+x^2{e}^{y}dy-dy=0$$$$yd\left(sinx\right)+sinxd\left(y\right)+e^{y}d\left(x^2\right)+x^{2}d\left(e^y\right)-dy=0$$$$d(y.sinx)+d(x^2.e^y)-dy=dC$$$$ysinx+x^2{e}^y-y=C$$

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