The solution of the differential equation,2x^ | vedclass

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

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

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(A) Given the differential equation: $2x^2y \frac{dy}{dx} = 	an(x^2y^2) - 2xy^2$. Rearranging the terms,we get: $2x^2y \frac{dy}{dx} + 2xy^2 = 	an(x^2y^2)$. Notice that the left side is the derivative of $x^2y^2$ with respect to $x$: $\frac{d}{dx}(x^2y^2) = 2x^2y \frac{dy}{dx} + 2xy^2$. Let $z = x^2y^2$. Then the equation becomes $\frac{dz}{dx} = 	an z$. Separating the variables: $\int \cot z \, dz = \int dx$. Integrating both sides: $\ln(\sin z) = x + C$. Substituting $z = x^2y^2$: $\ln(\sin(x^2y^2)) = x + C$. Given $y(1) = \sqrt{\frac{\pi}{2}}$,at $x=1$,$z = (1)^2 \cdot (\sqrt{\frac{\pi}{2}})^2 = \frac{\pi}{2}$. Substituting these values: $\ln(\sin(\frac{\pi}{2})) = 1 + C \Rightarrow \ln(1) = 1 + C \Rightarrow 0 = 1 + C \Rightarrow C = -1$. Thus,$\ln(\sin(x^2y^2)) = x - 1$. Taking the exponential of both sides: $\sin(x^2y^2) = e^{x-1}$.

The solution of the differential equation,$2x^2y \frac{dy}{dx} = \tan(x^2y^2) - 2xy^2$ given $y(1) = \sqrt{\frac{\pi}{2}}$ is

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