First solve the equation

y'' - 4 y=0. The solution is y = u exp 2 x + v exp -2x. Normally u and v are constants. But here assume them to be function of and take the first derivative y = u(x) exp 2 x + v (x) exp -2x. The derivative is

y'= u' exp 2 x + 2 u exp 2x +v' exp -2 x - 2v exp -2x

Assume u' exp 2x + v' exp -2x=0

You have now y' = 2 u exp 2x - 2 v exp -2x. Take the second derivative

y'' =2 u' exp 2x + 4u exp 2x - 2 v' exp -2x +4v exp -2x

Put this in the equation (4u exp 2x + 4v exp -2x) - 2 ( u' exp 2x - v' exp -2x)

- 4( u exp 2x - v exp -2x) = 4x/ exp 2x

This gives

4(u exp 2x - v exp -2x) = 4x/exp 2x, You also have

u' exp 2x + v' exp -2x=0. We re write the above equation

u exp 4 x -v =x

u' exp 4x +v'=0, take the derivative of the first equation

u' exp 4x + 4u exp 4x -v'=1, u' exp 4x + 4 u exp 4x + 4 u exp 4x=1

u' + 8u = exp -4x. Multiply with exp 8x, d/dt ( u exp 8x)_= exp 4x, This gives

u exp 8x = 1/4 exp 4x, u= 1/4 exp -4x, Now we can find

v= u exp 4x -x= 1/4 exp 8x -x. The general solution is

y= 1/4 exp -4x exp 2x + ( 1/4 exp 8x -x) exp - 2x

Please check the result by redoing it your self.