Example 2. Form a differential equation for which the functions $y_{1}(x)=e^{x^{2}}, y_{2}(x)=e^{-x^{2}}$ form a fundamental system of solutions.
Solution. Let us form an equation of the form (42):

$$
\left|\begin{array}{ccc}
e^{x^{2}} & e^{-x^{2}} & y \\
2 x e^{x^{2}} & -2 x e^{-x^{2}} & y^{\prime} \\
\left(2+4 x^{2}\right) e^{x^{2}} & \left(4 x^{2}-2\right) e^{-x^{2}} & y^{\prime \prime}
\end{array}\right|=0, \quad \text { or }\left|\begin{array}{ccc}
1 & 1 & y \\
2 x & -2 x & y^{\prime} \\
2+4 x^{2} & 4 x^{2}-2 & y^{\prime \prime}
\end{array}\right|=0 .
$$

Expanding the last determinant by the elements of the 3rd column, we will have

$$
x y^{\prime \prime}-y^{\prime}-4 x^{3} y=0
$$

In this example, the Wronskian determinant $W\left[y_{1}, y_{2}\right]=-4 x$ is zero at $x=0$. This does not contradict the general theory, according to which the Wronskian determinant of the fundamental system of solutions of a linear homogeneous differential equation

$$
y^{(n)}+p_{1}(x) y^{(n-1)}+\ldots+p_{n}(x) y=0
$$

with continuous coefficients on the interval $[a, b]$ does not vanish at any point $x$ of the interval $[a, b]$. Writing equation (44) in the form

$$
y^{\prime \prime}-\frac{1}{x} y^{\prime}-4 x^{2} y=0
$$

we see that the coefficient of $y^{\prime}$ is discontinuous at $x=0$, so the continuity of the coefficients of equation (45) is violated at the point $x=0$.

## Problems for Independent Solution

Form the differential equations for which the given systems of functions form fundamental systems of solutions:

675. $y_{1}(x)=\operatorname{sh} x, \quad y_{2}(x)=\operatorname{ch} x$.
676. $y_{1}(x)=x, \quad y_{2}(x)=e^{x}$.
677. $y_{1}(x)=e^{x}, \quad y_{2}(x)=e^{x^{2} / 2}$.
678. $y_{1}(x)=1, \quad y_{2}(x)=x, y_{3}(x)=x^{2}$.
679. $y_{1}(x)=x, \quad y_{2}(x)=\sin x, \quad y_{3}(x)=\cos x$.

$7^{\circ}$. Different problems. Let $y_{1}, y_{2}, \ldots, y_{n}$ be a fundamental system of a linear homogeneous equation

$$
y^{(n)}+p_{1}(x) y^{(n-1)}+\ldots+p_{n}(x) y=0
$$

Then the Ostrogradsky-Liouville formula holds

$$
W(x)=W\left(x_{0}\right) e^{-\int_{s_{0}}^{t} p_{1}(t) d t}
$$

where $W(x)=W\left[y_{1}, y_{2}, \ldots, y_{n}\right]$ is the Wronskian determinant, and $x_{0}$ is any value of $x$ from the interval $[a, b]$, on which the coefficients $p_{1}(x), p_{2}(x), \ldots, p_{n}(x)$ of the equation are continuous.