Equations reducible to homogeneous form Solve the following differential equation:
$$(2x+y-3)dy=(x+2y-3)dx$$
I have tried this & I found:
$$(y-x)^3=c(y+x+2)$$
But in my book the answer is:
$$(y-x)^3=c(y+x-2)$$
Please tell which one is correct..

$$  (x+2y-3)dx=(2x+y-3)dy$$
The two constants of $-3$ are keeping the equation from being homogeneous. We can eliminate the constants by an appropriately chosen substitution: 
$$X=x+a,\quad Y=y+b$$and let's see if we can choose suitable values for $a$ and $b$.
$$ (X - a + 2Y - 2b - 3)dX = (2X-2a + Y - b - 3)dY $$
It's easy to find what the "suitable" values should be, since we just want them to cancel out the constants! So we solve the following system of equations:
$$\begin{align*}
3 &= -a - 2b\\
3 &= -2a - b
\end{align*}$$
I'll let you verify that the correct values are $a=-1$ and $b=-1$. This leaves us with
$$(X + 2Y)dX = (2X+Y)dY$$
Now let $Y=VX$, and hence $dY = VdX + XdV$.
$$\begin{align*}
X(1 + 2V)dX &= X(2+V)(VdX+XdV) \\
X(1-V^2)dX &= X^2(2+V)dV\\
\dfrac{1}{X}dX &= \dfrac{2+V}{1-V^2}dV\\
\log\left|X\right| + C_1 &= \dfrac{1}{2}(-3 \log\left|1-V\right|+\log\left|1+V\right|)\\
C_2X^2 &= \dfrac{1+V}{(1-V)^3}\\
C_2X^2(1-V)^3 &= 1+V\\
C_2X^2(1-\dfrac{Y}{X})^3 &= 1+\dfrac{Y}{X}\\
C_2\dfrac{(X-Y)^3}{X}&= 1+\dfrac{Y}{X}\\
C_2(X-Y)^3 &= X + Y\\
C_2(x-y)^3 &= x + y - 2
\end{align*}$$
So your book has the correct solution.
