weird_dave:
https://www.arduino.cc/en/Main/ArduinoBoardDue"5V.This pin outputs a regulated 5V from the regulator on the board. The board can be supplied with power either from the DC power jack (7 - 12V), the USB connector (5V), or the VIN pin of the board (7-12V). Supplying voltage via the 5V or 3.3V pins bypasses the regulator, and can damage your board. We don't advise it."
It might work fine, or it might cause problems with the on-board regulator, difficult to tell from the datasheet.
Do you need the isolation from the main power supply?
When using the power plug to power the Due board, Power is applied through the 5V regulator. When powering the board through the USB port, the power is going through A MOS FET used as a diode. This diode protects the USB port, The power bypasses the 5V regulator And supplies power to the 3.3 V regulator. When supplying power through the 5V pin, you are essentially supplying power in the same manner as you would through the USB port. The only difference is you are not going through the MOS FET. No damage to the board will occur as long as you only apply 5V to the 5V output pin.
Joe.
PS, As an example of applying 5V regulated voltage, to the 5V pin.
http://www.elechouse.com/elechouse/index.php?main_page=product_info&cPath=148_211&products_id=2251
This is an adapter for 7 inch TFT's screen. This size screen takes a lot of the power away from the Due, And can tax the limits of the regulator if other items are connected. The solution is to have a 5V regulated power directly to the board.