As you said, you want that your wave function is normalized.

So, it is not correct that the solution outside the wall is $$\psi=Cexp(\mu x)+Dexp(-\mu x).$$ In fact this function diverges for $x\rightarrow\pm\infty.$

In order to keep your wave function normalized, you must use one solution for $x<-a$ and another one for $x>a.$

For $x<-a$, you have

$$\psi_-=Cexp(\mu x).$$

While for $x>a$ you get

$$\psi_+=Dexp(-\mu x).$$

Your wavefunction is thus:

$$\begin{cases} x<-a & \psi(x)=Cexp(\mu x) \\ -a<x<a & \psi(x)=Acos(\lambda x)+Bsin(\lambda x) \\ x>a& \psi(x)=Dexp(-\mu x) \end{cases}$$ In this way you imposed the normalizability request and you can just ask continuity and derivability in $x=\pm a.$

As you said, you want that your wave function is normalized.

So, it is not correct that the solution outside the wall is $$\psi=Cexp(\mu x)+Dexp(-\mu x).$$ In fact this function diverges for $x\rightarrow\pm\infty.$

In order to keep your wave function normalized, you must use one solution for $x<-a$ and another one for $x>a.$

For $x<-a$, you have

$$\psi_-=Cexp(\mu x).$$

While for $x>a$ you get

$$\psi_+=Dexp(-\mu x).$$

Your wavefunction is thus:

$$\begin{cases} x\in[-\infty,-a] & \psi(x)=Cexp(\mu x) \\ x\in[-a,a] & \psi(x)=Acos(\lambda x)+Bsin(\lambda x) \\ x\in[a,\infty] & \psi(x)=Dexp(-\mu x) \end{cases}$$ In this way you imposed the normalizability request and you can just ask continuity and derivability in $x=\pm a.$

As you said, you want that your wave function is normalized.

So, it is not correct that the solution outside the wall is $$\psi=Cexp(\mu x)+Dexp(-\mu x)$$

In fact this function diverges for $x\rightarrow\pm\infty.$

You can use two different wave functions outside the box. For $x<-a$, you have

$$\psi_-=Cexp(\mu x),$$

while for $x>a$ you have

$$\psi_+=Dexp(-\mu x),$$

In this way you imposed the normalizability request and you can just ask continuity and derivability in $x=\pm a.$

As you said, you want that your wave function is normalized.

So, it is not correct that the solution outside the wall is $$\psi=Cexp(\mu x)+Dexp(-\mu x).$$ In fact this function diverges for $x\rightarrow\pm\infty.$

In order to keep your wave function normalized, you must use one solution for $x<-a$ and another one for $x>a.$

For $x<-a$, you have

$$\psi_-=Cexp(\mu x).$$

While for $x>a$ you get

$$\psi_+=Dexp(-\mu x).$$

Your wavefunction is thus:

$$\begin{cases} x<-a & \psi(x)=Cexp(\mu x) \\ -a<x<a & \psi(x)=Acos(\lambda x)+Bsin(\lambda x) \\ x>a& \psi(x)=Dexp(-\mu x) \end{cases}$$ In this way you imposed the normalizability request and you can just ask continuity and derivability in $x=\pm a.$