Find the general integral of the equation (2x+y+1)…

Find the general integral of the equation $(2x+y+1)\,dx-(4x+2y-3)\,dy=0$.

Explanation of Method

Since $4x+2y=2(2x+y)$, the linear parts of the numerator and denominator are proportional, so a translation cannot make it homogeneous. Instead substitute $u=2x+y$ to collapse both linear parts into one variable.

Step 1: Substitute $u=2x+y$

Here $y'=\dfrac{2x+y+1}{4x+2y-3}=\dfrac{u+1}{2u-3}$. Since $u'=2+y'$,

\[\frac{du}{dx}=2+\frac{u+1}{2u-3}=\frac{5u-5}{2u-3}.\]

Step 2: Separate and integrate

Using $\dfrac{2u-3}{u-1}=2-\dfrac{1}{u-1}$,

\[\frac{2u-3}{5(u-1)}\,du=dx \quad\Longrightarrow\quad 2u-\ln|u-1|=5x+K.\]

Step 3: Back-substitute $u=2x+y$

\[2(2x+y)-\ln|2x+y-1|=5x+K \quad\Longrightarrow\quad \boxed{\,2y-x-\ln|2x+y-1|=C.}\]

Result

Final answer:

\[\boxed{\,2y-x-\ln|2x+y-1|=C,\qquad x+\tfrac{y}{2}\neq\tfrac12.}\]