The general solution of the differential equa | vedclass

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(A) Given the differential equation $\frac{dy}{dx} = -\frac{x+y+1}{x-3y+5}$.Let $x = X+h$ and $y = Y+k$. We choose $h, k$ such that $h+k+1 = 0$ and $h

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$3(y-1)^2 - 2(x+2)(y-1) - (x+2)^2 = c$

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(A) Given the differential equation $\frac{dy}{dx} = -\frac{x+y+1}{x-3y+5}$.Let $x = X+h$ and $y = Y+k$. We choose $h, k$ such that $h+k+1 = 0$ and $h-3k+5 = 0$.Solving these,we get $h = -2$ and $k = 1$. So,$x = X-2$ and $y = Y+1$.The equation becomes $\frac{dY}{dX} = -\frac{X+Y}{X-3Y}$.This is a homogeneous equation. Let $Y = vX$,then $\frac{dY}{dX} = v + X\frac{dv}{dX}$.$v + X\frac{dv}{dX} = -\frac{1+v}{1-3v} = \frac{1+v}{3v-1}$.$X\frac{dv}{dX} = \frac{1+v}{3v-1} - v = \frac{1+v-3v^2+v}{3v-1} = \frac{-3v^2+2v+1}{3v-1}$.Separating variables: $\int \frac{3v-1}{-3v^2+2v+1} dv = \int \frac{1}{X} dX$.Integrating,we get $-\frac{1}{2} \ln| -3v^2+2v+1 | = \ln|X| + C$.This simplifies to $-3v^2+2v+1 = \frac{c}{X^2}$.Substituting $v = \frac{Y}{X} = \frac{y-1}{x+2}$:$-3(\frac{y-1}{x+2})^2 + 2(\frac{y-1}{x+2}) + 1 = \frac{c}{(x+2)^2}$.$-3(y-1)^2 + 2(y-1)(x+2) + (x+2)^2 = c$.Rearranging gives $3(y-1)^2 - 2(x+2)(y-1) - (x+2)^2 = c$.

The general solution of the differential equation $\frac{dy}{dx} + \frac{x+y+1}{x-3y+5} = 0$ is

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