Уравнение

[Гост]

[tex](2x + y)y'= x + 2y[/tex]

[KOPMOPAH]

[tex](2x + y)y'= x + 2y[/tex]

This is a first-order **homogeneous differential equation**.

To solve it, we can first rewrite the equation in the standard form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$:

$$\frac{dy}{dx} = \frac{x + 2y}{2x + y}$$

Now, divide the numerator and the denominator by $x$:

$$\frac{dy}{dx} = \frac{\frac{x}{x} + 2\frac{y}{x}}{2\frac{x}{x} + \frac{y}{x}} = \frac{1 + 2\frac{y}{x}}{2 + \frac{y}{x}}$$

## Substitution

Use the substitution for homogeneous equations:

Let $\mathbf{v = \frac{y}{x}}$, which means $\mathbf{y = vx}$.

Differentiating $\mathbf{y = vx}$ with respect to $x$ using the product rule gives:

$$\frac{dy}{dx} = v \cdot \frac{d(x)}{dx} + x \cdot \frac{dv}{dx}$$
$$\mathbf{\frac{dy}{dx} = v + x\frac{dv}{dx}}$$

## Separating Variables

Substitute $\frac{dy}{dx}$ and $\frac{y}{x}$ into the differential equation:

$$v + x\frac{dv}{dx} = \frac{1 + 2v}{2 + v}$$

Now, isolate $x\frac{dv}{dx}$:

$$x\frac{dv}{dx} = \frac{1 + 2v}{2 + v} - v$$
$$x\frac{dv}{dx} = \frac{1 + 2v - v(2 + v)}{2 + v}$$
$$x\frac{dv}{dx} = \frac{1 + 2v - 2v - v^2}{2 + v}$$
$$x\frac{dv}{dx} = \frac{1 - v^2}{2 + v}$$

Separate the variables $v$ and $x$:

$$\frac{2 + v}{1 - v^2} dv = \frac{1}{x} dx$$

## Integration

Integrate both sides:

$$\int \frac{2 + v}{1 - v^2} dv = \int \frac{1}{x} dx$$

### Right-Hand Side (RHS)

The RHS is straightforward:

$$\int \frac{1}{x} dx = \ln|x| + C$$

### Left-Hand Side (LHS)

For the LHS, we use **partial fraction decomposition**. Factor the denominator: $1 - v^2 = (1 - v)(1 + v)$.

$$\frac{2 + v}{1 - v^2} = \frac{A}{1 - v} + \frac{B}{1 + v}$$
$$2 + v = A(1 + v) + B(1 - v)$$

To find $A$ and $B$:
1. Set $v = 1$:
$$2 + 1 = A(1 + 1) + B(1 - 1) \implies 3 = 2A \implies \mathbf{A = \frac{3}{2}}$$
2. Set $v = -1$:
$$2 + (-1) = A(1 - 1) + B(1 - (-1)) \implies 1 = 2B \implies \mathbf{B = \frac{1}{2}}$$

So, the integral is:

$$\int \left(\frac{3/2}{1 - v} + \frac{1/2}{1 + v}\right) dv = \frac{3}{2} \int \frac{1}{1 - v} dv + \frac{1}{2} \int \frac{1}{1 + v} dv$$
$$= \frac{3}{2} (-\ln|1 - v|) + \frac{1}{2} (\ln|1 + v|)$$
$$\mathbf{= -\frac{3}{2} \ln|1 - v| + \frac{1}{2} \ln|1 + v|}$$

## General Solution

Equating LHS and RHS, and combining the constant of integration ($C$):

$$-\frac{3}{2} \ln|1 - v| + \frac{1}{2} \ln|1 + v| = \ln|x| + C$$

Multiply by 2 to clear fractions:

$$-3 \ln|1 - v| + \ln|1 + v| = 2 \ln|x| + 2C$$

Use logarithm properties $(\ln a^b = b \ln a)$ and $(\ln a - \ln b = \ln \frac{a}{b})$:

$$\ln|1 + v| - \ln|1 - v|^3 = \ln|x^2| + 2C$$
$$\ln\left|\frac{1 + v}{(1 - v)^3}\right| = \ln(x^2) + 2C$$

Let $2C = \ln K$, where $K > 0$ is a new constant:

$$\ln\left|\frac{1 + v}{(1 - v)^3}\right| = \ln(x^2) + \ln K$$
$$\ln\left|\frac{1 + v}{(1 - v)^3}\right| = \ln(K x^2)$$

Exponentiate both sides:

$$\left|\frac{1 + v}{(1 - v)^3}\right| = K x^2$$

Since $K$ is an arbitrary positive constant, we can absorb the absolute value and replace $\pm K$ with a new arbitrary constant $\mathbf{C_1}$ (where $C_1 \neq 0$):

$$\frac{1 + v}{(1 - v)^3} = C_1 x^2$$

## Final Solution

Substitute back $v = \frac{y}{x}$:

$$\frac{1 + \frac{y}{x}}{\left(1 - \frac{y}{x}\right)^3} = C_1 x^2$$

Simplify the fractions:

$$\frac{\frac{x + y}{x}}{\left(\frac{x - y}{x}\right)^3} = C_1 x^2$$

$$\frac{x + y}{x} \cdot \frac{x^3}{(x - y)^3} = C_1 x^2$$

$$\frac{x + y}{1} \cdot \frac{x^2}{(x - y)^3} = C_1 x^2$$

If $x \neq 0$, we can divide by $x^2$:

$$\frac{x + y}{(x - y)^3} = C_1$$

The **implicit general solution** is:

$$\mathbf{x + y = C_1 (x - y)^3}$$

(Note: $y=x$ is a singular solution lost during division by $1-v^2$ that may or may not be covered by $C_1$. In this case, $x+y = C_1 (x-y)^3$ yields $2x = 0$ for $y=x$, so $y=x$ is not a part of the general solution, but it is a solution of the original ODE).

Тъкмо да си опресниш английския и да търсиш по-кратко решение.