For the differential equation x y frac{dy}{dx | vedclass

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(A) The given differential equation is $xy \frac{dy}{dx} = (x+2)(y+2)$.Separating the variables,we get:$\left( \frac{y}{y+2} \right) dy = \left( \frac

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$y - x + 2 = \log \left[ x^2 (y+2)^2 \right]$

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(A) The given differential equation is $xy \frac{dy}{dx} = (x+2)(y+2)$.Separating the variables,we get:$\left( \frac{y}{y+2} \right) dy = \left( \frac{x+2}{x} \right) dx$$\Rightarrow \left( 1 - \frac{2}{y+2} \right) dy = \left( 1 + \frac{2}{x} \right) dx$Integrating both sides:$\int \left( 1 - \frac{2}{y+2} \right) dy = \int \left( 1 + \frac{2}{x} \right) dx$$y - 2 \log |y+2| = x + 2 \log |x| + C$$y - x - C = 2 \log |x| + 2 \log |y+2|$$y - x - C = \log \left[ x^2 (y+2)^2 \right] \quad \dots(1)$Since the curve passes through $(1, -1)$:$-1 - 1 - C = \log \left[ (1)^2 (-1+2)^2 \right]$$-2 - C = \log(1) = 0 \Rightarrow C = -2$Substituting $C = -2$ into equation $(1)$:$y - x + 2 = \log \left[ x^2 (y+2)^2 \right]$.

For the differential equation $x y \frac{dy}{dx} = (x+2)(y+2)$,find the solution curve passing through the point $(1, -1)$.

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