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For the scalar linear ODE with periodic coefficients,

$$

x' = a(t)x,\quad\quad a(t + T) = a(t),

$$

show that the solution is of the form

$$

x(t) = x_0e^{\mu t}p(t),

$$

where $\mu$ and $x_0$ are constants, and $p(t)$ is a $T$-periodic function.

How can I show the solution is of the form mentioned? Can I just say by Floquet Theory, the solution is of the form $X(t) = \sum\limits_{n=1}^{n}c_nx_n(t)$ where $x_n(t) = e^{\mu_nt}p_n(t)$ but then how do I get the $x_0$ for $c_1$?

By Floquet Theory, we can define $X_k = X(t)v_k$.Suppose $Bv_k = \lambda_kv_k$.

$$

X_k(t + T) = X(t + T)v_k = X(t)Bv_k = X(t)\lambda_kv_k = \lambda_kX(t)v_k = \lambda_kX_k(t)

$$

Let $\underbrace{\lambda_k}_{\text{characteristic multipliers}} = \text{exp}\left[\overbrace{\rho_k}^{\text{characteristic exponents}}T\right]$.

Define $p_k(t) = \frac{X_k(t)}{e^{\rho_k t}}$.

Then

\begin{alignat*}{3}

p_k(t + T) & = & \frac{X_k(t + T)}{\text{exp}\left[\rho_k(t + T)\right]}\\

& = & \frac{\lambda_kX_k(t)}{\text{exp}\left[\rho_k(t + T)\right]}\\

& = & \frac{\lambda_kX_k(t)}{\lambda_k\text{exp}\left[\rho_kt\right]}\\

& = & \frac{X_k(t)}{\text{exp}\left[\rho_kt\right]}\\

& = & p_k(t)

\end{alignat*}

Thus $p(t)$ is T periodic.

Since $x_0 = x(0)$ is an initial condition, can I just say that is why it is constant. What about $\mu$?

$$

x' = a(t)x,\quad\quad a(t + T) = a(t),

$$

show that the solution is of the form

$$

x(t) = x_0e^{\mu t}p(t),

$$

where $\mu$ and $x_0$ are constants, and $p(t)$ is a $T$-periodic function.

How can I show the solution is of the form mentioned? Can I just say by Floquet Theory, the solution is of the form $X(t) = \sum\limits_{n=1}^{n}c_nx_n(t)$ where $x_n(t) = e^{\mu_nt}p_n(t)$ but then how do I get the $x_0$ for $c_1$?

By Floquet Theory, we can define $X_k = X(t)v_k$.Suppose $Bv_k = \lambda_kv_k$.

$$

X_k(t + T) = X(t + T)v_k = X(t)Bv_k = X(t)\lambda_kv_k = \lambda_kX(t)v_k = \lambda_kX_k(t)

$$

Let $\underbrace{\lambda_k}_{\text{characteristic multipliers}} = \text{exp}\left[\overbrace{\rho_k}^{\text{characteristic exponents}}T\right]$.

Define $p_k(t) = \frac{X_k(t)}{e^{\rho_k t}}$.

Then

\begin{alignat*}{3}

p_k(t + T) & = & \frac{X_k(t + T)}{\text{exp}\left[\rho_k(t + T)\right]}\\

& = & \frac{\lambda_kX_k(t)}{\text{exp}\left[\rho_k(t + T)\right]}\\

& = & \frac{\lambda_kX_k(t)}{\lambda_k\text{exp}\left[\rho_kt\right]}\\

& = & \frac{X_k(t)}{\text{exp}\left[\rho_kt\right]}\\

& = & p_k(t)

\end{alignat*}

Thus $p(t)$ is T periodic.

Since $x_0 = x(0)$ is an initial condition, can I just say that is why it is constant. What about $\mu$?

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