The solution of the differential equation dydx=x+y2x+2y−2 is

dydx=(x+y)2(x+2)(y−2)=[(x+2)+(y−2)]2(x+2)(y−2) put X=x+2 and Y=y+2dYdX=(X+Y)2XYput Y=vXv+XdvdX=v2+2v+1v=v+2v+1v⇒Xdvdx=2v+1v⇒v2v+1dv=dxX⇒∫1−11+2vdv=2∫dxXv−12ln⁡|1+2v|=2ln⁡X+ln⁡c2v-ln1+2v=4ln X+2lnc2v=ln⁡c2X4(1+2v)2YX=ln⁡c2X41+2YXe2YX=c2X3(X+2Y)⇒e2(y−2)x+2=c2(x+2)3(x+2+2y−4)=c2(x+2)3(x+2y-2)

The solution of the differential equation dydx=x+y2x+2y−2 is

dydx=(x+y)2(x+2)(y−2)=[(x+2)+(y−2)]2(x+2)(y−2) put X=x+2 and Y=y+2dYdX=(X+Y)2XYput Y=vXv+XdvdX=v2+2v+1v=v+2v+1v⇒Xdvdx=2v+1v⇒v2v+1dv=dxX⇒∫1−11+2vdv=2∫dxXv−12ln⁡|1+2v|=2ln⁡X+ln⁡c2v-ln1+2v=4ln X+2lnc2v=ln⁡c2X4(1+2v)2YX=ln⁡c2X41+2YXe2YX=c2X3(X+2Y)⇒e2(y−2)x+2=c2(x+2)3(x+2+2y−4)=c2(x+2)3(x+2y-2)

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The solution of the differential equation $\frac{d y}{d x} = \frac{{(x + y)}^{2}}{(x + 2) (y - 2)}$ is

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$e^{\frac{2 (y - 2)}{x + 2}} = = c^{2} (x + 2)^{3} (x + 2 y - 2)$ .

$e^{\frac{2 (y - 2)}{x + 2}} = c^{2} (x + 2)^{4} (x + 2 y - 2)$ ..

$e^{\frac{2 (y - 2)}{x + 2}} = c^{2} (x + 2)^{3} (x + 2 y + 2)$ …

$e^{\frac{2 (y - 2)}{x + 2}} = c^{2} (x + 2)^{3} (x - 2 y + 2)$

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} = \frac{(x + y)^{2}}{(x + 2) (y - 2)} = \frac{[(x + 2) + (y - 2)]^{2}}{(x + 2) (y - 2)} \\ put X = x + 2 and Y = y + 2 \\ \frac{d Y}{d X} = \frac{(X + Y)^{2}}{X Y} \\ p u t Y = v X \\ v + X \frac{d v}{d X} = \frac{v^{2} + 2 v + 1}{v} = v + \frac{2 v + 1}{v} \\ \Rightarrow X \frac{d v}{d x} = \frac{2 v + 1}{v} \\ \Rightarrow \frac{v}{2 v + 1} d v = \frac{d x}{X} \\ \Rightarrow \int (1 - \frac{1}{1 + 2 v}) d v = 2 \int \frac{d x}{X} \end{array}$

$\begin{array}{l} v - \frac{1}{2} \ln | 1 + 2 v | = 2 \ln X + \ln c \\ 2 v - \ln |1 + 2 v| = 4 \ln X + 2 \ln c \\ 2 v = \ln |c^{2} X^{4} (1 + 2 v)| \\ \frac{2 Y}{X} = \ln |c^{2} X^{4} (1 + \frac{2 Y}{X})| \\ e^{\frac{2 Y}{X}} = |c^{2} X^{3} (X + 2 Y)| \\ \Rightarrow e^{\frac{2 (y - 2)}{x + 2}} = c^{2} (x + 2)^{3} (x + 2 + 2 y - 4) = c^{2} (x + 2)^{3} (x + 2 y - 2) \end{array}$