How to show that a solution of $L(y) = 0$ is real valed

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In the book of Int. to Ordinary DEs by Coddington, at page 109 in question 5.b, it is asked that

Let $$L(y) = y^{(n)} + a_1(x)y^{n-1} + ...+ a_n (x) y,$$ where $a_1, ..., a_n$ are continuous real-valued functions on the interval $I$.

**b-)**Let $\phi$ be a solution of $L(y) = 0$ satisfying $$\phi(x_0) = \alpha_1, \phi'(x_0) = \alpha_2,..., \phi^{n-1} (x_0) = \alpha_n,$$ where $x_0 \in I$, and $\alpha_1, ..., \alpha_n$ are real constants. Prove that $\phi$ is real-valued.

To show this, I need to take an arbitrary value from the the real line, and show that the image under $\phi$ is a real number, but I couldn't find a way to show that, so any help or hint is appreciated.

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Let $\phi =f+ig$ where $f$ and $g$ are real valued. Then $g$ is a solution of the differential given equation with the initial conditions $g(0)=0,g'(0)=0,...,g^{n-1} (0)=0$. [Just take imaginary parts in the equation $\phi^{j}(0)=\alpha_k$]. By an earlier theorem in Coddington's book this implies $g=0$ so $\phi =f$ is real valued.