Word Count : 773

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For the differential equation: [ (6xy - 3y^2 + 2y),dx + 2(x - y),dy = 0 ] let (M = 6xy - 3y^2 + 2y) and (N = 2(x - y)). Check if it is exact: [ \frac{\partial M}{\partial y} = 6x - 6y + 2, \quad \frac{\partial N}{\partial x} = 2 ] Since (\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}), it is not exact. Consider an integrating factor of the form (\mu = \mu(y)). Check (\frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N)): [ \mu' M + \mu \frac{\partial M}{\partial y} = \mu \frac{\partial N}{\partial x} \implies \frac{\mu'}{\mu} = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{2 - (6x - 6y + 2)}{6xy - 3y^2 + 2y} = \frac{-6x + 6y}{6xy - 3y^2 + 2y} = \frac{6(y-x)}{6xy - 3y^2 + 2y} ] Factor 3 in denominator: [ \frac{\mu'}{\mu} = \frac{2(y-x)}{2y(3x - y + 1)} \text{ is not solely a function of } y ] Try (\mu = \mu(x)). Then: [ \frac{\mu'}{\mu} = \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{(6x - 6y + 2) - 2}{2(x-y)} = \frac{6x - 6y}{2(x-y)} = 3 ] Hence (\mu(x) = e^{\int 3 dx} = e^{3x}). ___ ______ ______ ____ ___ _______.
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