I've bought a new laptop on which I have to run kernel 5.1+ because my WiFi-chip (Intel® Wi-Fi 6 AX200) wouldn't be supported otherwise. Now, I'm running into problems with VirtualBox.
Kernel driver not installed (rc=-1908) The VirtualBox Linux kernel driver is either not loaded or not set up correctly. Please try setting it up again by executing '/sbin/vboxconfig' as root. If your system has EFI Secure Boot enabled you may also need to sign the kernel modules (vboxdrv, vboxnetflt, vboxnetadp, vboxpci) before you can load them. Please see your Linux system's documentation for more information. where: suplibOsInit what: 3 VERR_VM_DRIVER_NOT_INSTALLED (-1908) The support driver is not installed. On linux, open returned ENOENT.
What I tried
Please install the Linux kernel "header" files matching the current kernel for adding new hardware support to the system. The distribution packages containing the headers are probably: linux-headers-generic linux-headers-5.2.5-050205-generic
linux-headers-generic installed, and i can not install
linux-headers-5.2.5-050205-generic because it doesn't exist.
linux-headers-5.2.5-050205 exists and I have installed it, but it doesn't make a difference.
If your system is using EFI Secure Boot...I have Secure Boot disabled.
sudo modprobe vboxdrv
modprobe: FATAL: Module vboxdrv not found in directory /lib/modules/5.2.5-050205-generic
The directory exists, but the module is indeed not there.
I tried running
sudo apt install --reinstall linux-headers-$(uname -r) virtualbox-dkms dkms and reboot but that did not fix the problem
sudo apt remove virtualbox*and
sudo apt-get install build-essential linux-headers-'uname -r' dkms virtualbox-dkms
This again gives me the error
Package linux-headers-5.2.5-050205-generic is not available, but is referred to by another package., as with point 1.
sudo apt-get install dkmsand
sudo /etc/init.d/vboxdrv setup
sudo: /etc/init.d/vboxdrv: command not found
- Both VirtualBox 6 and VirtualBox 5.2 have this problem.
Edit : this post was marked as possible duplicate, which it is not, as I have explained above.