Example 5. Find the general solution of the differential equation

$$
d r-r d \varphi=0
$$
Solution. In this equation, the unknown function is denoted by the letter $r$, and its argument by the letter $\varphi$. Separating the variables, we get

$$
\frac{d r}{r}=d \varphi
$$

from which

$$
\ln r=\varphi+C_{1} .
$$

From the last equation, we find

$$
r=e^{\varphi+C_{1}}=C e^{\varphi}
$$

where $C=e^{C_{1}}$.

Therefore, the general solution is given by the formula

$$
r=C e^{\varphi}
$$

## Problems

Integrate the differential equations, find the specified particular solutions, and plot them:

1. $y^{\prime}=4 x^{3} ; y=0$ when $x=0$.
2. $\left(x^{2}+4\right) y^{\prime}-2 x y=0, y=5$ when $x=1$.
3. $x y^{\prime}=\frac{y}{\ln x}, y=1$ when $x=e$.
4. $y^{\prime}=\frac{x}{\sqrt{1-x^{2}}}, y=1$ when $x=0$.
5. $y^{\prime}=y^{2}, y=1$ when $x=-1$.
6. $y^{\prime} \operatorname{tg} x-y=1, y=1$ when $x=\frac{\pi}{2}$.

Integrate the differential equations with separable variables:
7. $(x+3) d y-(y+3) d x=0$.
8. $y^{\prime}=\frac{4}{x^{2}-4}$.
9. $\left(x y^{2}+x\right) d x+\left(y+x^{2} y\right) d y=0$.
10. $y^{\prime}=y^{2} \cos x$.
11. $\sin x d x+\frac{d y}{\sqrt{y}}=0$.
12. $d r-r \operatorname{ctg} \varphi d \varphi=0$.

## Answers

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

## § 9.2. Homogeneous Differential Equations of the First Order

A differential equation

$$
P(x, y) d x+Q(x, y) d y=0
$$

is called homogeneous if the coefficients $P(x, y)$ and $Q(x, y)$ are homogeneous functions of the same degree $n$, i.e., functions for which for any $k$ the identities

$$
P(k x, k y) \equiv k^{n} P(x, y), Q(k x, k y) \equiv k^{n} Q(x, y)
$$

hold.

Equation (9.11) can be brought to the form

$$
y^{\prime}=\varphi\left(\frac{y}{x}\right)
$$

Using the substitution

$$
y=u x
$$

where $u$ is a new unknown function, the homogeneous equation is transformed into an equation with separable variables.

The equation

$$
y^{\prime}=f\left(\frac{a_{1} x+b_{1} y+c_{1}}{a_{2} x+b_{2} y+c_{2}}\right)
$$

for which

$$
\Delta=\left|\begin{array}{ll}
a_{1} & b_{1} \\
a_{2} & b_{2}
\end{array}\right| \neq 0
$$

can be transformed by

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

where the constants $h$ and $k$ are found from the system of equations

$$
\left.\begin{array}{r}
a_{1} h+b_{1} k+c_{1}=0 \\
a_{2} h+b_{2} k+c_{2}=0
\end{array}\right\}
$$

When $\Delta=0$, the transformation $a_{1} x+b_{1} y=t$ reduces equation (9.15) to an equation with separable variables.