Example 3. Solve the equation $y^{\prime}=\frac{2 x+2 y+5}{x+y-1}$.
Solution. The system $\left\{\begin{array}{l}2 x+2 y+5=0 \\ x+y-1=0\end{array}\right.$ is unsolvable, so the equation cannot be reduced to a homogeneous one. Notice that the right-hand side is a function of $x+y$. The equation can be reduced to one with separable variables (see §2). Let $x+y=z$. Then $2 x+2 y+5=2 z+5$. After differentiating the previous equality, we get: $1+y^{\prime}=z^{\prime}, y^{\prime}=z^{\prime}-1$. We arrive at the equation $z^{\prime}-1=\frac{2 z+5}{z-1}$, from which $z^{\prime}=\frac{3 z+4}{z-1}$, or $\left(z^{\prime}=\frac{d z}{d x}\right) \int \frac{z-1}{3 z+4} d z=\int d x$. We integrate and return to the variables $x$ and $y$. We get $\frac{1}{3}\left(z-\frac{7}{3} \ln |3 z+4|\right)=x+C$.

Answer. $y-2 x-\frac{7}{3} \ln |3 x+3 y+4|=3 C$.

## Exercises

1. $x^{2}+y^{2}-2 x y y^{\prime}=0$. 2. $x^{2} y^{\prime}=y^{2}+x y$.
2. $y^{\prime}=\frac{1-3 x-3 y}{1+x+y}$. 4. $y^{\prime}=\frac{y^{2}}{x^{2}}-\frac{y}{x} ; x=-1, y=1$.
3. $(x+y) d x+(x+2 y) d y=0$.
4. $(2 x-y+4) d y+(x-2 y+5) d x=0$.
5. $x y^{\prime} \cos \frac{y}{x}=y \cos \frac{y}{x}-x$.
6. $\left(4 x^{2}+3 x y+y^{2}\right) d x+\left(4 y^{2}+3 x y+x^{2}\right) d y=0$.
7. $x y^{\prime}=\sqrt{x^{2}-y^{2}}+y ; \quad x=1, \quad y=1$.
8. $y^{3} d y+x\left(2 x^{2}+3 y^{2}\right) d x=0 ; x=0, y=1$. 11. $y^{\prime}=\frac{2 x+y-3}{3 x-5 y+2}$.

## Answers

1. $x^{2}-y^{2}=C x$. 2. $x=y(C-\ln |x|)$. 3. $3 x+y+2 \ln |x+y-1|=C$. 4. $y=\frac{2 x}{1-C x^{2}} ; \quad y=\frac{2 x}{1-3 x^{2}} \cdot$ 5. $x^{2}+2 x y+2 y^{2}=C \cdot 6 \cdot(x+y-1)^{3}=$ $=C(x-y-3) . \quad 7 . \quad \sin \frac{y}{x}+\ln x=C .8 . \quad\left(x^{2}+y^{2}\right)^{3}(x+y)^{2}=C$.

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-084.jpg?height=76&width=1114&top_left_y=393&top_left_x=83)
${\sqrt{x^{2}+y^{2}}}^{x}=y^{2}+2 x^{2}$. 11. $\frac{1}{2} \ln \left|5(y-1)^{2}-2(y-1)(x-1)+2(x-1)^{2}\right|-$ $-\frac{2}{3} \operatorname{arctg} \frac{5 y-x-4}{3(x-1)}=C$.

## § 4. Linear Differential Equations of the First Order and Bernoulli Equations

$1^{\circ}$. A linear differential equation of the first order is an equation that is linear in the unknown function and its derivative. It has the form

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

An equation of the form

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

is called a Bernoulli equation with exponent $\alpha$. When $\alpha=0$, the Bernoulli equation becomes linear.

A linear equation and a Bernoulli equation can be solved using a unified method, specifically by the substitution

$$
y=u \cdot v, \quad y^{\prime}=u^{\prime} v+v^{\prime} u
$$

Here, $u$ and $v$ are two unknown functions that must satisfy the equation

$$
u^{\prime} v+v^{\prime} u+p(x) \cdot u v=q(x) \cdot u^{\alpha} \cdot v^{\alpha}
$$

An additional condition is imposed on the new unknown functions $u$ and $v$ (in addition to the fact that they must satisfy the previous equation):

$$
v^{\prime} u+p(x) u v=0, \quad \text { i.e. } \quad v^{\prime}+p(x) v=0
$$

If this equality holds, then the equation automatically implies

$$
u^{\prime} v=q(x) \cdot u^{\alpha} \cdot v^{\alpha}
$$

The previous equation is with separable variables. Solving it and substituting the obtained $v$ into the last equation, we determine the function $u$. Thus, since $y=u v$, the general solution of the original equation is found (in explicit form).

Note. From the above, it follows that the substitution $y=u v$ reduces a linear equation or a Bernoulli equation to an equivalent system of equations:

$$
y^{\prime}+p(x) y=q(x) y^{\alpha} \Leftrightarrow\left\{\begin{array}{c}
v^{\prime}+p(x) v=0 \\
u^{\prime} \cdot v=q(x) u^{\alpha} \cdot v^{\alpha}
\end{array}\right.
$$

$2^{\circ}$. A linear equation can also be solved by another method.

1) If $q(x) \equiv 0$, then the equation $y^{\prime}+p(x) \cdot y=0$ is called a linear homogeneous equation. It is an equation with separable variables and has the general solution

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

2) The general solution of a non-homogeneous equation (i.e., when $q(x) \not \equiv 0$) can be found using the method of variation of the arbitrary constant, which involves seeking it in the form

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

This expression, along with its derivative, is substituted into the original equation, resulting in a new differential equation for the unknown function $C(x)$. Finding this function and substituting it into the last equality, we obtain the solution.

Below, we will only use the first method for solving linear equations and Bernoulli equations.

## Examples with Solutions