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(A) Given differential equation is $\frac{dx}{dy} + 2yx = 2y$. Rearranging the terms,we get $\frac{dx}{dy} = 2y(1-x)$. Separating the variables,we hav

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$(x-1) = e^{-y^2}$

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(A) Given differential equation is $\frac{dx}{dy} + 2yx = 2y$. Rearranging the terms,we get $\frac{dx}{dy} = 2y(1-x)$. Separating the variables,we have $\int \frac{dx}{1-x} = \int 2y \, dy$. Integrating both sides,we get $-\ln|1-x| = y^2 + C$,which can be written as $\ln|1-x| = -y^2 - C$. Taking the exponential of both sides,$|1-x| = e^{-y^2 - C} = Ae^{-y^2}$,where $A = e^{-C}$. This implies $1-x = \pm Ae^{-y^2}$,or $x-1 = Ke^{-y^2}$ where $K = \mp A$. Since the curve passes through the point $(2,0)$,substitute $x=2$ and $y=0$: $2-1 = Ke^{-(0)^2} \Rightarrow 1 = K(1) \Rightarrow K=1$. Thus,the solution is $(x-1) = e^{-y^2}$.

The solution of the differential equation $\frac{dx}{dy} + 2yx = 2y$ which passes through the point $(2,0)$ is

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