Example 6. Construct a differential equation if its general solution is known to be $y=\frac{1}{6} x^{4}+\frac{C}{x^{2}}$.
Solution. We will use the definition: we need to form an equation containing the argument $x$, the unknown function $y$, and its derivative $y'$. For this, we will differentiate the given function, and then from the two equations combined into a system, we will eliminate the constant $C$. So,

$$
\left\{\begin{array}{l}
y = \frac{1}{6} x^{4} + \frac{C}{x^{2}}, \\
y' = \frac{2 x^{3}}{3} - \frac{2 C}{x^{3}}
\end{array} \Rightarrow \left\{\begin{array}{l}
2 y = \frac{x^{4}}{3} + \frac{2 C}{x^{2}} \\
x y' = \frac{2 x^{4}}{3} - \frac{2 C}{x^{2}}
\end{array} \Rightarrow 2 y + x y' = x^{4}\right.\right.
$$

or

$$
y' + 2 \frac{y}{x} = x^{3}
$$

Thus, as the required differential equation, we can take one of the last equalities.

## Exercises

Check whether the given function is a general solution of the given differential equation.

1. $y = \frac{C}{x} - \frac{x}{2}, \quad y' + \frac{x + y}{x} = 0$.
2. $\left(x^{2} + y^{2}\right)^{3}(x + y)^{2} = C$, $\left(4 y^{2} + 3 x y + x^{2}\right) y' + 4 x^{2} + 3 x y + y^{2} = 0$.
3. $y \left(x^{2} + C x\right) = 1, \quad y' + \frac{y}{x} + x y^{2} = 0$.
4. $\left\{\begin{array}{l}x = \ln |t| + \arcsin t + C, \\ y = t - \sqrt{1 - t^{2}},\end{array} \quad y' - y - \sqrt{1 - y'^{2}} = 0\right.$.

Construct the direction field for the given differential equation and several integral curves based on this field.
5. $y' = \frac{1}{2} x$.
6. $y' = -\frac{x}{y}$.
7. $y' = 1 + y^{2}$.
8. $y' = x^{2} + \frac{y^{2}}{4}$.

Formulate a differential equation for the given family of integral curves.
9. $x y \left(C - \frac{1}{2} \ln^{2} x\right) = 1$.
10. $y = C x + C \ln C$.
11. $y = \frac{x}{x^{2} + C}$.
12. $y = \left(\frac{1}{2} x + \frac{1}{4} \sin 2 x + C\right) \frac{1}{\cos x}$.
13. $y = x^{4} \left(\frac{1}{2} \ln |x| + C\right)^{2}$.

## Answers

1. Yes. 2. Yes. 3. Yes. 4. Yes. 9. $x y' + y = x y^{2} \ln x$. 10. $x y' + y' \ln y' = y$. 11. $2 x y^{2} - y + x y' = 0$. 12. $y' = y \tan x + \cos x$. 13. $y' = \frac{4 y}{x} + x \sqrt{y}$.

## § 2. Equations with Separated and Separable Variables

1. An equation with separated variables is a differential equation of the form

$$
\varphi(x) d x = g(y) d y
$$

Considering the sought function $y = y(x)$, the equation can be written as an equality of two differentials

$$
\varphi(x) d x = g(y(x)) d y(x) \Leftrightarrow d(\Phi(x)) = d(G(y(x)))
$$

This is equivalent to the fact that the primitives $\Phi(x)$ and $G(y(x))$ of the left and right parts differ by an arbitrary constant:

$$
\int \varphi(x) d x = \int g(y) d y + C \Leftrightarrow \Phi(x) = G(y(x)) + C
$$

This equality is the general integral, or the general solution of the given equation.

$2^{\circ}$. An equation of the form $\varphi_{1}(x) g_{1}(y) d x + \varphi_{2}(x) g_{2}(y) d y = 0$ is called an equation with separable variables. It can be reduced to an equation with separated variables after term-by-term division by $\varphi_{2}(x) g_{1}(y) \neq 0$.

$3^{\circ}$. An equation of the form $y' = f(a x + b y)$ can be reduced to an equation with separated variables using the substitution $a x + b y = v$, where $v$ is a new unknown function.

## Examples with Solutions