# Doesn't the initial value problem have exactly one solution, since the determinant is $\neq 0$?

I am looking at the following theorem:

If $a_1, a_2 \in \mathbb{R}$ and $x_0,y_0,y_1 \in \mathbb{R}$

then the initial value problem $\left\{\begin{matrix} y''+a_1y'+a_2y=0, x \in \mathbb{R}\\ y(x_0)=y_0\\ y'(x_0)=y_1 \end{matrix}\right.$

has (at least ) one solution.

It is as follows:

We are looking for a solution of the form $\phi(x)=e^{rx}$, where $r \in \mathbb{R}$ or $\mathbb{C}$.

If $\phi$ is a solution of the initial value problem, then it has to hold:

$$\phi''+a_1 \phi'+a_2 \phi=0 (A)$$

and $\phi(x_0)=y_0, \phi'(x_0)=y_1$.

$\phi$ is a solution of the differential equation $A$ iff:

$$(r^2+a_1 r+a_2)e^{rx}=0, x \in \mathbb{R} \ \text{ iff }\\ r^2+a_1r+a_2=0 \to \text{the equation has always a solution in } \mathbb{C}. \text{So, there exists a }\\ r \in \mathbb{C} \text{ so that } \phi(x)=e^{rx} \text{is a solution of } A.$$

The equation $r^2+a_1r+a_2=0$ has:

• two real roots $r_1, r_2$ with $r_1 \neq r_2$
• a double root $r$
• two complex roots: $r_1$ and $r_2=\overline{r_1}$

Then we have the solution of $A$:

• $\phi_1(x)=e^{r_1 x}, \phi_2(x)=e^{r_2x}$
• $\phi_1(x)=e^{rx}, \phi_2(x)=xe^{rx}$
• $\phi_1(x)=e^{r_1x}, \phi_2(x)=e^{\overline{r_1}x}$

We are looking for a solution of the initial value problem of the form

$$\phi(x)=c_1 \phi_1(x)+c_2 \phi_2(x)$$

Then $\phi$ satifies the differential equation $A$ $\forall c_1, c_2 \in \mathbb{R}$ (or $\mathbb{C}$) and so it suffices that $\phi$ satisfies the following:

$$c_1 \phi_1(x_0)+c_2 \phi_2(x_0)=y_0\\ c_1 \phi_1'(x_0)+c_2 \phi_2'(x_0)=y_1$$

The above system has a unique solution as for $c_1, c_2$ if:

$$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix} \neq 0$$

• If $r_1, r_2$ real roots with $r_1 \neq r_2$ then

$$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix}=(r_2-r_1)e^{(r_1+r_2)x_0} \neq 0$$

• If $r$ is a double root, then:

$$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix}=e^{2rx} \neq 0$$

• If $r_1, r_2$ complex solutions, then:

$$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix}=(\overline{r_1}-r_1)e^{(r_1+\overline{r_1})x_0} \neq 0$$

Since in each case, the determinant $$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix}$$ is $\neq 0$ doesn't this imply that the initial value problem $\left\{\begin{matrix} y''+a_1y'+a_2y=0, x \in \mathbb{R}\\ y(x_0)=y_0\\ y'(x_0)=y_1 \end{matrix}\right.$ has exactly one solution?