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Question Number 116054 by Study last updated on 30/Sep/20

$$(x^2+2xy+1)dx+(x^2+y^2-1)dy=0$$$$y=?$$

Commented by mohammad17 last updated on 30/Sep/20

$$M(x,y)=x^2+2xy+1\Rightarrow M_y=2x\to \left(1\right)$$$$N(x,y)=x^2+y^2-1\Rightarrow N_x=2x\to \left(2\right)$$$$from\left(1\right)and\left(2\right)theequationisexact$$$$$$$$\because M(x,y)dx+N(x,y)dy=0$$$$$$$$\therefore \int (x^2+2xy+1)dx+\int (x^2+y^2-1)dy=C$$$$$$$$\therefore \frac{1}{3}{x}^3+\frac{1}{3}{y}^3+2x^{2}y+x-y=C$$$$$$$$\ll m.o\gg$$

Answered by MWSuSon last updated on 30/Sep/20

$$\mathrm{M}={\mathrm{x}}^2+2\mathrm{xy}+1$$$$\mathrm{N}={\mathrm{x}}^2+{\mathrm{y}}^2-1$$$$\frac{\partial \mathrm{M}}{\partial \mathrm{y}}=\frac{\partial \mathrm{N}}{\partial \mathrm{x}}$$$$\int \mathrm{Mdx}+\int (\mathrm{N}-\mathrm{f}\left(\mathrm{y}\right))\mathrm{dy}=\mathrm{C}$$$$\int ({\mathrm{x}}^2+2\mathrm{xy}+1)\mathrm{dx}+\int ({\mathrm{y}}^2-1)\mathrm{dy}=\mathrm{C}$$$$\frac{{\mathrm{x}}^3}{3}+{\mathrm{x}}^2\mathrm{y}+\mathrm{x}+\frac{{\mathrm{y}}^3}{3}-\mathrm{y}=\mathrm{C}$$

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