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(B) Given differential equation is $x dy + 2y dx = 0$. Rearranging the terms,we get $x dy = -2y dx$. Separating the variables,we have $\frac{dy}{y} =

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$x^2y = 4$

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(B) Given differential equation is $x dy + 2y dx = 0$. Rearranging the terms,we get $x dy = -2y dx$. Separating the variables,we have $\frac{dy}{y} = -2 \frac{dx}{x}$. Integrating both sides,$\int \frac{dy}{y} = -2 \int \frac{dx}{x}$. This gives $\ln|y| = -2 \ln|x| + C$. Using properties of logarithms,$\ln|y| + 2 \ln|x| = C$,which simplifies to $\ln|y| + \ln|x^2| = C$. Thus,$\ln|yx^2| = C$,which implies $yx^2 = e^C = k$. Given the initial condition $x = 2$ and $y = 1$,substitute these values into $x^2y = k$: $(2)^2(1) = k \Rightarrow 4(1) = k \Rightarrow k = 4$. Therefore,the particular solution is $x^2y = 4$.

The particular solution of the differential equation $x dy + 2y dx = 0$,when $x = 2$ and $y = 1$ is

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