4.20. Obtain the general solution of the equation

$$
\frac{d y}{d x}=\frac{x+3 y-5}{4 x-y-7}
$$
Solution. We form the determinant $\Delta=\left|\begin{array}{cc}1 & 3 \\ 4 & -1\end{array}\right|=-13 \neq 0$.

We have the 2nd case. Introduce new arguments $\alpha=x-h$ and function $\beta=y-k$, where $h$ and $k$ are found from the system of linear equations

$$
\left\{\begin{array}{l}
h+3 k-5=0 \\
4 h-k-7=0
\end{array}\right.
$$

The system has the solution $h=2, k=1$. The substitution is $\alpha=x-2, \beta=y-1$. We obtain the homogeneous equation

$$
\frac{d \beta}{d \alpha}=\frac{\alpha+3 \beta}{4 \alpha-\beta}, \text { or } \frac{d \beta}{d \alpha}=\frac{1+3 \beta / \alpha}{4-\beta / \alpha} \text {. }
$$

We introduce a new function $u(\alpha)=\frac{\beta(\alpha)}{\alpha}, \frac{d \beta}{d \alpha}=u+\alpha \frac{d \beta}{d \alpha}$. The equation becomes $u+\alpha \frac{d u}{d \alpha}=\frac{1+3 u}{4-u}$. Separating variables: $\frac{(4-u) d u}{u^{2}-u+1}=\frac{d \alpha}{\alpha}$. Integrating:

$$
\int \frac{(4-u) d u}{u^{2}-u+1}=\int \frac{d \alpha}{\alpha}, \int \frac{(4-u) d u}{(u-0.5)^{2}+0.75}=\ln |\alpha|+\ln C
$$

We use the Euler substitution:

$$
u-0.5=\sqrt{0.75} z, u=\sqrt{0.75} z+0.5, d u=\sqrt{0.75} d z
$$

$$
\begin{aligned}
& \int \frac{(4-\sqrt{0.75} z+0.5) \sqrt{0.75} d z}{0.75 z^{2}+0.75}=\ln |\alpha|+\ln C \\
& \frac{4.5 \sqrt{0.75}}{0.75} \int \frac{d z}{z^{2}+1}+\int \frac{z d z}{z^{2}+1}=\ln (C|\alpha|) \\
& \frac{4.5 \sqrt{0.75}}{0.75} \operatorname{arctg} z+0.5 \ln \left|z^{2}+1\right|=\ln (C|\alpha|)
\end{aligned}
$$

Returning to the old variables:

$$
\begin{aligned}
& \frac{4.5 \sqrt{0.75}}{0.75} \operatorname{arctg} \frac{u-0.5}{\sqrt{0.75}}+0.5 \ln \left|\frac{(u-0.5)^{2}}{0.75}+1\right|=\ln (C|\alpha|) \\
& \frac{4.5 \sqrt{0.75}}{0.75} \operatorname{arctg} \frac{(\beta / \alpha-0.5)}{\sqrt{0.75}}+0.5 \ln \left|\frac{(\beta / \alpha-0.5)^{2}}{0.75}+1\right|=\ln (C|\alpha|) \\
& \frac{4.5 \sqrt{0.75}}{0.75} \operatorname{arctg} \frac{((y-1) /(x-2))-0.5}{\sqrt{0.75}}+ \\
& \quad+0.5 \ln \left|\frac{((y-1) \beta /(x-2)-) 0.5)^{2}}{0.75}+1\right|=\ln (C|x-2|)
\end{aligned}
$$