Example 4. Solve the equation $(x+y+1) d x+(2 x+2 y-1) d y=0$.






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Solution. The system of linear algebraic equations

$$
\left\{\begin{aligned}
x+y+1 & =0 \\
2 x+2 y-1 & =0
\end{aligned}\right.
$$

is inconsistent. In this case, the method used in the previous example is not applicable. To integrate the equation, we apply the substitution \(x+y=z\), \(d y=d z-d x\). The equation becomes

$$
(2-z) d x+(2 z-1) d z=0
$$

Separating the variables, we get

$$
d x-\frac{2 z-1}{z-2} d z=0, \quad \text { hence } \quad x-2 z-3 \ln |z-2|=C
$$

Returning to the variables \(x, y\), we obtain the general integral of the given equation \(x+2 y+3 \ln |x+y-2|=C\).

## Problems for Independent Solution

Solve the following equations:

108. \(x+y-2+(1-x) y^{\prime}=0\).
109. \((3 y-7 x+7) d x-(3 x-7 y-3) d y=0\).
110. \((x+y-2) d x+(x-y+4) d y=0\).
111. \((x+y) d x+(x-y-2) d y=0\).
112. \(2 x+3 y-5+(3 x+2 y-5) y^{\prime}=0\).
113. \(8 x+4 y+1+(4 x+2 y+1) y^{\prime}=0\).
114. \((x-2 y-1) d x+(3 x-6 y+2) d y=0\).
115. \((x+y) d x+(x+y-1) d y=0\).

B. Sometimes the equation can be reduced to a homogeneous one by the substitution \(y=\boldsymbol{z}^{\alpha}\). This is the case when all terms in the equation have the same dimension, if the variable \(x\) is assigned a dimension of 1, the variable \(y\) - a dimension of \(\alpha\), and the derivative \(\frac{d y}{d x}\) - a dimension of \(\alpha-1\).