The solution of the differential equation (2x | vedclass

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(D) Given differential equation: $(2x - 4y + 3) \frac{dy}{dx} + (x - 2y + 1) = 0$. Rearranging,we get $\frac{dy}{dx} = -\frac{x - 2y + 1}{2x - 4y + 3}

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$\log [4(x - 2y) + 5] = 4(x + 2y) + C$

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(D) Given differential equation: $(2x - 4y + 3) \frac{dy}{dx} + (x - 2y + 1) = 0$. Rearranging,we get $\frac{dy}{dx} = -\frac{x - 2y + 1}{2x - 4y + 3} = -\frac{(x - 2y) + 1}{2(x - 2y) + 3}$. Let $v = x - 2y$. Then $\frac{dv}{dx} = 1 - 2 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2} (1 - \frac{dv}{dx})$. Substituting these into the equation: $\frac{1}{2} (1 - \frac{dv}{dx}) = -\frac{v + 1}{2v + 3}$. $1 - \frac{dv}{dx} = -\frac{2v + 2}{2v + 3} \Rightarrow \frac{dv}{dx} = 1 + \frac{2v + 2}{2v + 3} = \frac{2v + 3 + 2v + 2}{2v + 3} = \frac{4v + 5}{2v + 3}$. Separating variables: $\frac{2v + 3}{4v + 5} dv = dx$. Multiply by $2$: $\frac{4v + 6}{4v + 5} dv = 2 dx$. $\int (1 + \frac{1}{4v + 5}) dv = \int 2 dx$. $v + \frac{1}{4} \log |4v + 5| = 2x + C$. Substitute $v = x - 2y$: $(x - 2y) + \frac{1}{4} \log |4(x - 2y) + 5| = 2x + C$. $\frac{1}{4} \log |4(x - 2y) + 5| = x + 2y + C$. $\log |4(x - 2y) + 5| = 4x + 8y + C'$.

The solution of the differential equation $(2x - 4y + 3) \frac{dy}{dx} + (x - 2y + 1) = 0$ is ($C$ is an arbitrary constant).

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