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Question Number 83159 by jagoll last updated on 28/Feb/20

$$3\mathrm{x}(\mathrm{xy}-2)\mathrm{dx}+({\mathrm{x}}^3+2\mathrm{y})\mathrm{dy}=0$$$$\mathrm{find}\mathrm{the}\mathrm{solution}$$

Commented by niroj last updated on 28/Feb/20

$$({3\mathrm{x}}^2\mathrm{y}-6\mathrm{x})\mathrm{dx}+({\mathrm{x}}^3+2\mathrm{y})\mathrm{dy}=0$$$$\mathrm{let},$$$$\mathrm{M}={3\mathrm{x}}^2\mathrm{y}-6\mathrm{x}\mathrm{\&}\mathrm{N}={\mathrm{x}}^3+2\mathrm{y}$$$$\frac{\mathrm{dM}}{\mathrm{dy}}={3\mathrm{x}}^2,\frac{\mathrm{dN}}{\mathrm{dx}}={3\mathrm{x}}^2$$$$\therefore \frac{\mathrm{dM}}{\mathrm{dy}}=\frac{\mathrm{dN}}{\mathrm{dx}}$$$$\mathrm{we}\mathrm{know}$$$$\mathrm{p}=\int \mathrm{Mdx}=\int ({3\mathrm{x}}^2\mathrm{y}-6\mathrm{x})\mathrm{dx}$$$$\mathrm{p}=\frac{{3\mathrm{x}}^3}{3}\mathrm{y}-\frac{{6\mathrm{x}}^2}{2}={\mathrm{x}}^3\mathrm{y}-{3\mathrm{x}}^2$$$$\mathrm{again},$$$$\mathrm{N}-\frac{\mathrm{dp}}{\mathrm{dy}}=({\mathrm{x}}^3+2\mathrm{y})-({\mathrm{x}}^3-0)$$$$={\mathrm{x}}^3+2\mathrm{y}-{\mathrm{x}}^3=2\mathrm{y}$$$$\mathrm{and}$$$$\mathrm{p}+\int (\mathrm{N}-\frac{\mathrm{dp}}{\mathrm{dy}})\mathrm{dy}=\mathrm{C}$$$${\mathrm{x}}^3\mathrm{y}-{3\mathrm{x}}^2+\int 2\mathrm{y}\mathrm{dy}=\mathrm{C}$$$${\mathrm{x}}^3\mathrm{y}-{3\mathrm{x}}^2+\frac{{2\mathrm{y}}^2}{2}=\mathrm{C}$$$${\mathrm{x}}^3\mathrm{y}-{3\mathrm{x}}^2+{\mathrm{y}}^2=\mathrm{C}//.$$$$$$$$$$

Commented by jagoll last updated on 28/Feb/20

$$\mathrm{type}\mathrm{exact}$$$$\frac{\partial \mathrm{M}}{\partial \mathrm{y}}=\frac{\partial \mathrm{N}}{\partial \mathrm{x}}$$$$\mathrm{F}(\mathrm{x},\mathrm{y})=\int ({3\mathrm{x}}^2\mathrm{y}-6\mathrm{x})\mathrm{dx}+\mathrm{g}\left(\mathrm{y}\right)$$$$\mathrm{F}(\mathrm{x},\mathrm{y})={\mathrm{x}}^3\mathrm{y}-{3\mathrm{x}}^2+\mathrm{g}\left(\mathrm{y}\right)$$$$$$

Commented by jagoll last updated on 28/Feb/20

$${\mathrm{g}}^{\prime }\left(\mathrm{y}\right)={\mathrm{x}}^3+2\mathrm{y}-\frac{\partial }{\partial \mathrm{y}}({\mathrm{x}}^3\mathrm{y}-{3\mathrm{x}}^2)$$$${\mathrm{g}}^{\prime }\left(\mathrm{y}\right)={\mathrm{x}}^3+2\mathrm{y}-{\mathrm{x}}^3=2\mathrm{y}$$$$\mathrm{g}\left(\mathrm{y}\right)={\mathrm{y}}^2$$$$\mathrm{F}(\mathrm{x},\mathrm{y})={\mathrm{x}}^3\mathrm{y}-{3\mathrm{x}}^2+{\mathrm{y}}^2=\mathrm{C}$$

Commented by peter frank last updated on 28/Feb/20

$$thankyouboth$$

Commented by niroj last updated on 28/Feb/20

$$\mathrm{you}\mathrm{must}\mathrm{welcome}\mathrm{dear}.$$

Answered by TANMAY PANACEA last updated on 28/Feb/20

$$3x^{2}ydx-6xdx+x^{3}dy+2ydy=0$$$$yd\left(x^3\right)+x^{3}d\left(y\right)+d\left(y^2\right)-3d\left(x^2\right)=0$$$$d(x^3.y)+d\left(y^2\right)-3d\left(x^2\right)=dc$$$$intregating...$$$$x^{3}y+y^2-3x^2=c$$

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