Example 4. Integrate the differential equation

$$
(4 y-3 x-5) y^{\prime}+7 x-3 y+2=0
$$
Solution. This is an equation of the form (9.15)

$$
y^{\prime}=\frac{-7 x+3 y-2}{-3 x+4 y-5}
$$

here $\Delta=\left|\begin{array}{ll}-7 & 3 \\ -3 & 4\end{array}\right|=-28+9=-19 \neq 0$.

We introduce new variables by the formulas (9.17):

$$
\left.\begin{array}{l}
x=u+h ; \\
y=v+k,
\end{array}\right\}
$$

where $h, k$ must satisfy the equations (9.18):

$$
\left.\begin{array}{l}
-7 h+3 k-2=0 \\
-3 h+4 k-5=0
\end{array}\right\}
$$

Solving these equations, we find $h=\frac{7}{19}, k=\frac{29}{19}$. Thus, the formulas (9.17) take the form:

$$
\left.\begin{array}{l}
x=u+\frac{7}{19} \\
y=v+\frac{29}{19}
\end{array}\right\}
$$

from which

$$
d x=d u, d y=d v
$$

Substituting this into the original equation, we get

$$
y^{\prime}=\frac{d v}{d u}=\frac{-7 u+3 v-\frac{49}{19}+\frac{87}{19}-2}{-3 u+4 v-\frac{21}{19}+\frac{116}{19}-5}=\frac{-7 u+3 v}{-3 u+4 v}
$$

or

$$
\frac{d v}{d u}=\frac{-7+3\left(\frac{v}{u}\right)}{-3+4\left(\frac{v}{u}\right)}
$$

We make the substitution

$$
\frac{v}{u}=z,
$$

from which

$$
v=u z, \frac{d v}{d u}=\frac{d z}{d u} \cdot u+z \cdot 1
$$

therefore, the equation (A) takes the form

$$
\frac{d z}{d u} \cdot u+z=\frac{-7+3 z}{-3+4 z}
$$

Transforming the last equation

$$
\frac{d z}{d u} \cdot u=\frac{3 z-7}{4 z-3}-z=\frac{3 z-7-4 z^{2}+3 z}{4 z-3}=\frac{-\left(4 z^{2}-6 z+7\right)}{4 z-3}
$$

Separating the variables, we get

$$
\frac{(4 z-3) d z}{4 z^{2}-6 z+7}=-\frac{d u}{u}
$$

or

$$
\frac{(4 z-3) d z}{2\left(2 z^{2}-3 z+\frac{7}{2}\right)}=-\frac{d u}{u}
$$

Using the formula

$$
\int \frac{f^{\prime}(z) d z}{f(z)}=\ln f(z)
$$

from the last equation we find

$$
\frac{1}{2} \ln \left(2 z^{2}-3 z+\frac{7}{2}\right)=-\ln u+\frac{1}{2} \ln C_{1}
$$

Multiplying by 2:

$$
\begin{aligned}
& \ln \left(2 z^{2}-3 z+\frac{7}{2}\right)=\ln C_{1}-2 \ln u \\
& \ln \left(2 z^{2}-3 z+\frac{7}{2}\right)+\ln u^{2}=\ln C_{1} \\
& \ln \left[u^{2}\left(2 z^{2}-3 z+\frac{7}{2}\right)\right]=\ln C_{1}
\end{aligned}
$$

Substituting here the expression for $z=\frac{v}{u}$, we get

$$
\ln \left(2 v^{2}-3 u v+\frac{7}{2} u^{2}\right)=\ln C_{1}
$$

Transitioning to the variables $x, y$ by the formulas:

$$
\begin{aligned}
& \left.\begin{array}{l}
u=x-\frac{7}{19} \\
v=y-\frac{29}{19}
\end{array}\right\} \\
& \ln \left[2\left(y-\frac{29}{19}\right)^{2}-3\left(x-\frac{7}{19}\right)\left(y-\frac{29}{19}\right)+\frac{7}{2}\left(x-\frac{7}{19}\right)^{2}\right]=\ln C_{1}
\end{aligned}
$$

$$
\begin{gathered}
2 y^{2}-\frac{4 \cdot 29}{19} y+2\left(\frac{29}{19}\right)^{2}-3 x y+\frac{21}{19} y+\frac{87}{19} x-\frac{21 \cdot 29}{19^{2}}+ \\
+\frac{7}{2} x^{2}-\frac{49}{19} x+\frac{7 \cdot 49}{2 \cdot 19^{2}}=C_{1}
\end{gathered}
$$

Therefore, the general integral has the form

$$
2 y^{2}-3 x y+\frac{7}{2} x^{2}+2 x-5 y=C
$$

## Problems

Integrate homogeneous differential equations:

1. $2 x y y^{\prime}=y^{2}-4 x^{2}$.
2. $2 x^{2} y^{\prime}-4 x y-y^{2}=0$.
3. $\left(x^{2}+y^{2}\right) d x-2 x y d y=0$.
4. $y^{2}-4 x y+4 x^{2} y^{\prime}=0$.
5. $\left(2 x y+y^{2}\right) d x+\left(2 x y+x^{2}\right) d y=0$.
6. $x y^{\prime}-y+x \operatorname{tg} \frac{y}{x}=0$.
7. $x d y-y d x=y d y$.
8. $(x+y) d x-(x-y) d y=0$.
9. $\left(x^{2}+y^{2}\right) d x-2 x^{2} d y=0$.
10. $\left(x^{2}+y^{2}+x y\right) d x=x^{2} d y$.

Integrate equations reducible to homogeneous:
11. $3 y-7 x+7=(3 x-7 y-3) \frac{d y}{d x}$.
12. $(2 x+8) d x+(3 y-5 x-11) d y=0$.

## Answers

$$
\begin{aligned}
& \quad \text { 1. } y^{2}+4 x^{2}-8 C x=0 . \text { 2. } 2 x^{2}+x y-C y=0 . \text { 3. } x^{2}-y^{2}=C x . \text { 4. } y= \\
& =\frac{4 x}{\ln x+C} \cdot \text { 5. } x y(x+y)=C .6 . x \sin \frac{y}{x}=C . \text { 7. } \ln y+\frac{x}{y}=C . \text { 8. } \operatorname{arctg} \frac{y}{x}- \\
& -\ln \sqrt{x^{2}+y^{2}}=C . \text { 9. } 2 x=(x-y)(\ln x+C) . \text { 10. } y=x \operatorname{tg}(\ln C x) \text {. 11. }(x+ \\
& +y-1)^{5}(x-y-1)^{2}=C .12 . C(3 y-2 x+1)^{3}(y-x-1)^{2}=0 .
\end{aligned}
$$

## § 9.3. Linear differential equations of the first order. Bernoulli equation

A linear differential equation of the first order is an equation of the first degree with respect to $y$ and $y^{\prime}$, i.e., an equation of the form

$$
\varphi_{1}(x) y^{\prime}+\varphi_{2}(x) y+\varphi_{3}(x)=0\left(\varphi_{1}(x) \neq 0\right)
$$

or

$$
y^{\prime}+p(x) y=q(x)
$$

The solution of a linear equation is sought in the form

$$
y=u(x) v(x)
$$

Substituting the expressions for $y$ and $y^{\prime}$ into equation (9.20) reduces it to the form

$$
v \frac{d u}{d x}+\left[\frac{d v}{d x}+p(x) v\right] u=q(x)
$$

As $v$, one chooses one of the functions satisfying the equation

$$
\frac{d v}{d x}+p(x) v=0
$$

then the function $u$ is determined from the equation

$$
v \frac{d u}{d x}=q(x)
$$

If the function $q(x) \equiv 0$, then equation (9.20) takes the form

$$
y^{\prime}+p(x) y=0
$$

and is called a linear homogeneous equation, its general solution is expressed by the formula

$$
y=C e^{-\int p(x) d x}
$$

To solve the non-homogeneous linear