Example. Find the integral curves of the differential equation

$$
y^{\prime}=\frac{x^{2}+2 x y-5 y^{2}}{2 x^{2}-6 x y}
$$
Solution.

1. Transform the given equation to the form

$$
y^{\prime}=\frac{1+2 \frac{y}{x}-5\left(\frac{y}{x}\right)^{2}}{2-6 \frac{y}{x}}
$$

(we divided the numerator and denominator of the right-hand side of the given equation by $x^{2}$).

2. Make the substitution $y=x z(x)$, where $z(x)$ is the new unknown function. Then $y^{\prime}=z+x z^{\prime}$ and the equation reduces to the form

$$
z^{\prime} x+z=\frac{1+2 z-5 z^{2}}{2-6 z}
$$

i.e., to an equation with separable variables.

As a result of simple transformations, we obtain

$$
\frac{d z}{d x} x=\frac{1+z^{2}}{2-6 z}
$$

3. Separate the variables $\left(1+z^{2} \neq 0\right.$ and $\left.x \neq 0\right)$:

$$
\frac{2-6 z}{1+z^{2}} d z=\frac{d x}{x}
$$

4. Integrating, we get

$$
2 \operatorname{arctg} z-3 \ln \left(1+z^{2}\right)=\ln |x|+C
$$

Substituting $z$ with $y / x$, we get

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

Answer. The integral curves are defined by the equation

$$
2 \operatorname{arctg} \frac{y}{x}-\ln \frac{\left(x^{2}+y^{2}\right)^{3}}{|x|^{5}}=C
$$

for all possible values of $C$.

Conditions of the Problems. Find the integral curves of the differential equations.

1. $y^{\prime}=e^{y / x}+\frac{y}{x}$.
2. $y^{\prime}=\frac{y}{x}-1$.
3. $x^{2} y^{\prime}=x y+y^{2} e^{-x / y}$.
4. $x \cos \frac{y}{x} d y+\left(x-y \cos \frac{y}{x}\right) d x=0$.
5. $\left(x^{2}+2 x y\right) d x+x y d y=0$.
6. $x y^{\prime} \ln \left(\frac{y}{x}\right)=x+y \ln \left(\frac{y}{x}\right)$.
7. $y d x+(2 \sqrt{x y}-x) d y=0$.
8. $\left(4 y^{2}+x^{2}\right) y^{\prime}=x y$.
9. $x y^{\prime} \sin \left(\frac{y}{x}\right)+x=y \sin \left(\frac{y}{x}\right)$.
10. $x y+y^{2}=\left(2 x^{2}+x y\right) y^{\prime}$.

Answers.

1. $e^{-y / x}+\ln |x|=C$.
2. $x e^{y / x}=C$.
3. $e^{x / y}+\ln |x|=C$.
4. $\ln |x|+\sin \frac{y}{x}=C$.
5. $\ln |x+y|+\frac{x}{x+y}=C$
6. $\ln x-\frac{y}{x}\left(\ln \frac{y}{x}-1\right)=C$.
7. $\sqrt{\frac{y}{x}}+\ln |y|=C$.
8. $\ln |y|=\frac{x^{2}}{8 y^{2}}+C$.
9. $C x=e^{\cos (y / x)}$.
10. $y^{2}=C x e^{-y / x}$.

## 11.4. Linear Equations of the First Order

Problem Statement. Solve the Cauchy problem for the equation

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

with the initial condition

$$
y\left(x_{0}\right)=y_{0}
$$

Plan of Solution.

1st Method.

1. Write down the corresponding homogeneous linear equation:

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

This is a separable variables equation.

2. Separating the variables and integrating, we obtain the general solution of the homogeneous equation (2)

$$
y=C \exp \left\{-\int p(x) d x\right\}
$$

3. Apply the method of variation of the arbitrary constant:

a) seek the solution of the non-homogeneous equation (1) in the form (3), considering $C$ as an unknown function of $x$, i.e., assuming $C=C(x)$;

b) substitute $y$ and $y^{\prime}$, determined from relation (3) where $C=C(x)$, into equation (1). From the resulting differential equation, determine the function $C(x)$.

4. Thus, the general solution of the non-homogeneous equation (1) is obtained in the form

$$
y=C(x) \exp \left\{-\int p(x) d x\right\}
$$

Here, $C(x)$ contains an arbitrary constant $C_{0}$.

5. Using the initial conditions ( $1^{\prime}$ ), find the value of $C_{0}$ and obtain the solution of the given Cauchy problem.

Write the answer in the form $y=\varphi(x)$.

2nd Method.

1. Seek the solution of equation (1) in the form

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

where $u$ and $v$ are unknown functions of $x$.

2. Equation (1) takes the form

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

3. Since we now have two unknown functions, and the equation they must satisfy is only one, we can take another equation arbitrarily, as long as it does not narrow the set of solutions $y$.

Let one of the functions (for example, $u(x)$) satisfy the equation

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

Then equation (5) becomes

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

Solve equation (6) (with separable variables) to find $u$, not identically zero, to avoid narrowing the set of solutions $y$.

4. Substitute $u(x)$ into equation (7) and solve it for $v$.
5. Write the general solution of the equation in the form $y(x)=u(x) v(x)$.
6. Using the initial conditions ( $1^{\prime}$ ), obtain the solution of the given Cauchy problem.

Write the answer in the form $y=\varphi(x)$.