Example 2. Verify that the function $y=C e^{x}$ is the general solution of the equation $y^{\prime}-y=0$ and find the particular solution that satisfies the initial condition $\left.y\right|_{x=1}=-1$.

![](https://cdn.mathpix.com/cropped/2024_05_22_a0612bb33256580e6dcdg-008.jpg?height=468&width=391&top_left_y=308&top_left_x=125)

Fig. 5
Solution. We have $y=C e^{x}, y^{\prime}=C e^{x}$. Substituting the expressions for $y$ and $y^{\prime}$ into the given equation, we get $C e^{x}-C e^{x} \equiv 0$, i.e., the function $y=C e^{x}$ satisfies the given equation for any value of the constant $C$.

Let us set an arbitrary initial condition $\left.y\right|_{x=x_{0}}=y_{0}$. Substituting $x_{0}$ and $y_{0}$ for $x$ and $y$ in the function $y=C e^{x}$, we get $y_{0}=C e^{x_{0}}$, from which $C=y_{0} e^{-x_{0}}$. The function $y=y_{0} e^{x-x_{0}}$ satisfies the initial condition. Indeed, setting $x=x_{0}$, we get $y=y_{0} e^{x_{0}-x_{0}}=y_{0}$. The function $y=C e^{x}$ is the general solution of the given equation.

For $x_{0}=1$ and $y_{0}=-1$, we obtain the particular solution $y=-e^{x-1}$.

From a geometric point of view, the general solution defines a family of integral curves, which are the graphs of exponential functions; the particular solution is the integral curve passing through the point $M_{0}(1,-1)$ (Fig. 5).

A relation of the form $\Phi(x, y, C)=0$, implicitly defining the general solution, is called the general integral of a first-order differential equation.

A relation obtained from the general integral for a specific value of the constant $C$ is called a particular integral of the differential equation.

The problem of solving or integrating a differential equation consists in finding the general solution or the general integral of the given differential equation. If an initial condition is additionally given, then it is required to select a particular solution or particular integral that satisfies the given initial condition.

Since from a geometric point of view the coordinates $x$ and $y$ are equal, we will consider the equation $\frac{d x}{d y}=\frac{1}{f(x, y)}$ along with the equation $\frac{d y}{d x}=f(x, y)$.

## Problems for Independent Solution

1. Find the coinciding solutions of the two differential equations:
a) $y^{\prime}=y^{2}+2 x-x^{4}$
b) $y^{\prime}=-y^{2}-y+2 x+x^{2}+x^{4}$.

In the following problems, using some sufficient condition for uniqueness, identify the regions in which the given equations have unique solutions.
2. $y^{\prime}=x^{2}+y^{2}$.
3. $y^{\prime}=\frac{x}{y}$.
4. $y^{\prime}=y+3 \sqrt[3]{y}$.
5. $y^{\prime}=\sqrt{x-y}$.
6. $y^{\prime}=\sqrt{x^{2}-y}-x$.
7. $y^{\prime}=\sqrt{1-y^{2}}$.
8. $y^{\prime}=\frac{y+1}{x-y}$.
9. $y^{\prime}=\sin y-\cos x$.
10. $y^{\prime}=1-\operatorname{ctg} y$.
11. $y^{\prime}=\sqrt[3]{3 x-y}-1$.

12. Show that for the equation $y^{\prime}=|y|^{1 / 2}$, uniqueness of the solution is violated at each point on the $O x$ axis.
13. Find the integral curve of the equation $y^{\prime}=\sin (x y)$ passing through the point $O(0,0)$.

In the following problems, show that the given functions are solutions of the specified differential equations:

14. $y=\frac{\sin x}{x}, x y^{\prime}+y=\cos x$.
15. $y=C e^{-2 x}+\frac{1}{3} e^{x}, y^{\prime}+2 y=e^{x}$.
16. $y=2+C \sqrt{1-x^{2}},\left(1-x^{2}\right) y^{\prime}+x y=2 x$.

## § 2. Method of Isoclines

The equation

$$
y^{\prime}=f(x, y)
$$

defines in each point $(x, y)$, where the function $f(x, y)$ exists, the value of $y^{\prime}$, i.e., the slope of the tangent to the integral curve at that point.

If a value of some quantity is defined in each point of a region $D$, it is said that a field of this quantity is defined in the region $D$. Thus, the differential equation (1) defines a field of directions.

The triplet of numbers $\left(x ; y ; y^{\prime}\right)$ defines the direction of a line passing through the point $(x, y)$. The collection of segments of these lines gives a geometric picture of the field of directions.

The problem of integrating the differential equation (1) can now be interpreted as follows: find a curve such that the tangent to it at each point has a direction coinciding with the direction of the field at that point.

The problem of constructing integral curves is often solved by introducing isoclines. An isocline is the geometric locus of points where the tangents to the sought integral curves have the same direction. The family of isoclines of the differential equation (1) is defined by the equation

$$
f(x, y)=k
$$

where $k$ is a parameter. By assigning close numerical values to the parameter $k$, we obtain a sufficiently dense network of isoclines, which can be used to approximately construct the integral curves of the differential equation (1).

Remark 1. The zero isocline $f(x, y)=0$ gives the equation of the lines on which the points of maximum and minimum of the integral curves may lie.

For greater accuracy in constructing integral curves, we also find the geometric locus of points of inflection. For this, we find $y^{\prime \prime}$ using the equation (1):

$$
y^{\prime \prime}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} y^{\prime}=\frac{\partial f}{\partial x}+f(x, y) \frac{\partial f}{\partial y}
$$

and set it to zero. The line defined by the equation

$$
\frac{\partial f}{\partial x}+f(x, y) \frac{\partial f}{\partial y}=0
$$

is the possible geometric locus of points of inflection.