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(A) $x^{2} dy + (xy + y^{2}) dx = 0$$\Rightarrow x^{2} dy = -(xy + y^{2}) dx$$\Rightarrow \frac{dy}{dx} = \frac{-(xy + y^{2})}{x^{2}}$ ............. $

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$y + 2x = 3x^{2}y$

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(A) $x^{2} dy + (xy + y^{2}) dx = 0$$\Rightarrow x^{2} dy = -(xy + y^{2}) dx$$\Rightarrow \frac{dy}{dx} = \frac{-(xy + y^{2})}{x^{2}}$ ............. $(1)$Let $F(x, y) = \frac{-(xy + y^{2})}{x^{2}}$.$	herefore F(\lambda x, \lambda y) = \frac{-(\lambda x \cdot \lambda y + (\lambda y)^{2})}{(\lambda x)^{2}} = \frac{-\lambda^{2}(xy + y^{2})}{\lambda^{2}x^{2}} = \lambda^{0} F(x, y)$.Therefore,the given differential equation is a homogeneous equation. To solve it,we make the substitution $y = vx$.$\Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting the values of $y$ and $\frac{dy}{dx}$ in equation $(1)$,we get:$v + x \frac{dv}{dx} = \frac{-(x \cdot vx + (vx)^{2})}{x^{2}} = -(v + v^{2}) = -v - v^{2}$.$\Rightarrow x \frac{dv}{dx} = -v^{2} - 2v = -v(v + 2)$.$\Rightarrow \frac{dv}{v(v + 2)} = -\frac{dx}{x}$.Using partial fractions: $\frac{1}{v(v + 2)} = \frac{1}{2} \left( \frac{1}{v} - \frac{1}{v + 2} ight)$.$\Rightarrow \frac{1}{2} \int \left( \frac{1}{v} - \frac{1}{v + 2} ight) dv = -\int \frac{dx}{x}$.$\Rightarrow \frac{1}{2} [\ln|v| - \ln|v + 2|] = -\ln|x| + C_1$.$\Rightarrow \ln \left( \frac{v}{v + 2} ight) = -2\ln|x| + 2C_1 = \ln \left( \frac{C}{x^{2}} ight)$.$\Rightarrow \frac{v}{v + 2} = \frac{C}{x^{2}}$.Substituting $v = \frac{y}{x}$:$\Rightarrow \frac{y/x}{y/x + 2} = \frac{C}{x^{2}} \Rightarrow \frac{y}{y + 2x} = \frac{C}{x^{2}} \Rightarrow \frac{x^{2}y}{y + 2x} = C$.Given $y = 1$ at $x = 1$:$\frac{1^{2} \cdot 1}{1 + 2(1)} = C \Rightarrow C = \frac{1}{3}$.$\Rightarrow \frac{x^{2}y}{y + 2x} = \frac{1}{3} \Rightarrow 3x^{2}y = y + 2x$.

Find the particular solution satisfying the given condition:
$x^{2} dy + (xy + y^{2}) dx = 0$; $y = 1$ where $x = 1$.

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Find the particular solution satisfying the given condition:$x^{2} dy + (xy + y^{2}) dx = 0$; $y = 1$ where $x = 1$.