The solution of the differential equation (y^ | vedclass

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(A) Given differential equation is $(y^{2}+2x) \frac{dy}{dx}=y$. Rearranging the terms, we get $\frac{dx}{dy} = \frac{y^{2}+2x}{y} = y + \frac{2x}{y}$

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$x=y^{2}(1+\log_{e} y)$

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(A) Given differential equation is $(y^{2}+2x) \frac{dy}{dx}=y$. Rearranging the terms, we get $\frac{dx}{dy} = \frac{y^{2}+2x}{y} = y + \frac{2x}{y}$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = -\frac{2}{y}$ and $Q(y) = y$. The integrating factor $IF = e^{\int P(y) dy} = e^{\int -\frac{2}{y} dy} = e^{-2 \log_{e} y} = e^{\log_{e} y^{-2}} = y^{-2} = \frac{1}{y^{2}}$. The general solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + C$. Substituting the values, $x \cdot \frac{1}{y^{2}} = \int y \cdot \frac{1}{y^{2}} dy + C = \int \frac{1}{y} dy + C = \log_{e} y + C$. Thus, $x = y^{2}(\log_{e} y + C)$. Given that $x=1$ when $y=1$, we substitute these values: $1 = 1^{2}(\log_{e} 1 + C) \Rightarrow 1 = 1(0 + C) \Rightarrow C = 1$. Substituting $C=1$ into the general solution, we get $x = y^{2}(\log_{e} y + 1)$.

The solution of the differential equation $(y^{2}+2x) \frac{dy}{dx}=y$ satisfies $x=1, y=1$. Then the solution is

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