Let alpha |x| = |y| e^{xy-beta},where alpha, | vedclass

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(A) The given differential equation is $x dy - y dx + xy(x dy + y dx) = 0$.Dividing by $xy$,we get $\frac{dy}{y} - \frac{dx}{x} + (x dy + y dx) = 0$.I

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$4$

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(A) The given differential equation is $x dy - y dx + xy(x dy + y dx) = 0$.Dividing by $xy$,we get $\frac{dy}{y} - \frac{dx}{x} + (x dy + y dx) = 0$.Integrating both sides,we get $\int \frac{dy}{y} - \int \frac{dx}{x} + \int d(xy) = \int 0$.This simplifies to $\ln|y| - \ln|x| + xy = C$,which is $\ln|\frac{y}{x}| + xy = C$.Given $y(1) = 2$,we substitute $x=1$ and $y=2$ into the equation:$\ln|\frac{2}{1}| + (1)(2) = C \implies C = \ln 2 + 2$.Substituting $C$ back into the general solution:$\ln|\frac{y}{x}| + xy = \ln 2 + 2$.Rearranging the terms: $\ln|\frac{y}{x}| - \ln 2 = 2 - xy$.$\ln|\frac{y}{2x}| = -(xy - 2)$.Taking the exponential of both sides: $|\frac{y}{2x}| = e^{-(xy - 2)}$.$|y| = 2|x| e^{-(xy - 2)}$.Multiplying by $e^{xy-2}$: $|y| e^{xy-2} = 2|x|$.Comparing this with the given form $\alpha |x| = |y| e^{xy-\beta}$,we get $\alpha = 2$ and $\beta = 2$.Therefore,$\alpha + \beta = 2 + 2 = 4$.

Let $\alpha |x| = |y| e^{xy-\beta}$,where $\alpha, \beta \in \mathbb{N}$,be the solution of the differential equation $x dy - y dx + xy(x dy + y dx) = 0$ with the initial condition $y(1) = 2$. Then $\alpha + \beta$ is equal to:

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