The general solution of the differential equa | vedclass

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

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${x^2} + {y^2} - xy + x + y = c$

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(B) Given differential equation: $(2x - y + 1)dx + (2y - x + 1)dy = 0$.Rearranging,we get $\frac{dy}{dx} = -\frac{2x - y + 1}{2y - x + 1} = \frac{2x - y + 1}{x - 2y - 1}$.Let $x = X + h$ and $y = Y + k$. Substituting these,we get $\frac{dY}{dX} = \frac{2X - Y + 2h - k + 1}{X - 2Y + h - 2k - 1}$.For the equation to be homogeneous,we set $2h - k + 1 = 0$ and $h - 2k - 1 = 0$. Solving these gives $h = -1$ and $k = -1$.Thus,$\frac{dY}{dX} = \frac{2X - Y}{X - 2Y}$.Substitute $Y = vX$,so $\frac{dY}{dX} = v + X\frac{dv}{dX}$.$v + X\frac{dv}{dX} = \frac{2 - v}{1 - 2v} \implies X\frac{dv}{dX} = \frac{2 - v}{1 - 2v} - v = \frac{2 - v - v + 2v^2}{1 - 2v} = \frac{2(v^2 - v + 1)}{1 - 2v}$.Separating variables: $\frac{dX}{X} = \frac{1 - 2v}{2(v^2 - v + 1)} dv$.Integrating both sides: $\int \frac{dX}{X} = -\frac{1}{2} \int \frac{2v - 1}{v^2 - v + 1} dv$.$\ln|X| = -\frac{1}{2} \ln|v^2 - v + 1| + \ln|c_1| \implies \ln|X| + \frac{1}{2} \ln|v^2 - v + 1| = \ln|c_1|$.$X^2(v^2 - v + 1) = c$. Substituting $v = \frac{Y}{X}$,we get $X^2(\frac{Y^2}{X^2} - \frac{Y}{X} + 1) = c \implies Y^2 - XY + X^2 = c$.Substituting $X = x + 1$ and $Y = y + 1$: $(y + 1)^2 - (y + 1)(x + 1) + (x + 1)^2 = c$.Expanding: $(y^2 + 2y + 1) - (xy + x + y + 1) + (x^2 + 2x + 1) = c$.$x^2 + y^2 - xy + x + y = c'$ (where $c' = c - 1$).

The general solution of the differential equation $(2x - y + 1)dx + (2y - x + 1)dy = 0$ is

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