The solution of (2y - x) frac{dy}{dx} = 1 is | vedclass

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(A) Given the differential equation: $(2y - x) \frac{dy}{dx} = 1$. Rearranging the equation,we get: $\frac{dx}{dy} = 2y - x$. This is a linear differe

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$x = 2(y - 1) + ce^{-y}$,where $c$ is the constant of integration

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(A) Given the differential equation: $(2y - x) \frac{dy}{dx} = 1$. Rearranging the equation,we get: $\frac{dx}{dy} = 2y - x$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = 1$ and $Q(y) = 2y$. The integrating factor $(IF)$ is given by $e^{\int P(y) dy} = e^{\int 1 dy} = e^y$. Multiplying both sides by the $IF$: $e^y \frac{dx}{dy} + x e^y = 2y e^y$. This simplifies to: $\frac{d}{dy}(x e^y) = 2y e^y$. Integrating both sides with respect to $y$: $x e^y = \int 2y e^y dy$. Using integration by parts: $\int 2y e^y dy = 2(y e^y - e^y) + c = 2e^y(y - 1) + c$. Thus,$x e^y = 2e^y(y - 1) + c$. Dividing by $e^y$,we get $x = 2(y - 1) + ce^{-y}$.

The solution of $(2y - x) \frac{dy}{dx} = 1$ is

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