# Why do the Existence and Uniqueness Theorem and The Principle of Superposition not contradict each other?

I have a question regarding the Existence and Uniqueness Theorem and The Principle of Superposition, which my book (Elementary Differential Equations and Boundary Value Problems) defines in the following ways:

Existence and Uniqueness Theorem

Consider the initial value problem $$y'' + p(t)y' + q(t)y = g(t), \qquad y(t_0) = y_0, \qquad y'(t_0) = y'_0$$ where $p$, $q$, and $g$ are continuous on an open interval $I$ that contains the point $t_0$. Then, there is exactly one solution $y = \phi(t)$ of this problem, and the solution exists throughout the interval $I$.

Principle of Superposition

If $y_1$ and $y_2$ are two solutions of the differential equation $y'' + p(t)y' + q(t)y = 0$, then the linear combination $c_1 y_1 + c_2 y_2$ is also a solution for any values of the constants $c_1$ and $c_2$.

From the existence and uniqueness theorem, we know there is only one equation $y = \phi(t)$ that satisfies the equation $y'' + p(t)y' + q(t)y = 0$. From the principle of superposition, we know that $y = c\phi(t)$ is also a solution. How is this possible? Is it because the existence and uniqueness theorem is for particular solutions, where as the principle of superposition is for general solutions?

From the existence and uniqueness theorem, we know that given $y_0$ and $y'_0$, there is only one solution $y = \phi(t)$ (depending on $y_0$ and $y'_0$) of the equation $y'' + p(t)y' + q(t)y = 0$ satisfying $y(t_0) = y_0$ and $y'(t_0) = y'_0$.
But when you make linear combinations of solutions you change their values $y_0$ and $y'_0$.
Existence and Uniqueness Theorem doesn't say that there is only one function $y = \phi(t)$ satisfying the equation $y'' + p(t)y' + q(t)y = 0$. In fact, the number of solutions for this equation is infinite.
Existence and Uniqueness Theorem says that there is only one function $y = \phi(t)$ satisfying simultaneously the equation $y'' + p(t)y' + q(t)y = 0$ and the initial conditions $y(t_0) = y_0$, $y'(t_0) = y'_0$.