yy’-e^x=0, and y=4 when x=0, which means that:

y=x-lnx^2+4
y^2=4x^2+3
y=-2x+1/2x^3
y^2=2e^x+14
1/2ln|x^2+4|+6

Hmmm. Not so fast. I have no idea where that solution came from.

y y' = e^x
1/2 y^2 = e^x + c
y^2 = 2e^x + c
since y(0) = 4, 2+c = 16 ==> c = 14 and so
y^2 = 2e^x + 14
y = √2 (e^x + 7)

To find which of the given expressions represents the solution to the equation yy' - e^x = 0, we need to substitute the initial conditions y = 4 when x = 0 into each expression and see which one satisfies the given equation.

Let's go through each expression:

1) y = x - ln(x^2) + 4
Substitute x = 0: y = 0 - ln(0^2) + 4
Since ln(0^2) is undefined, we cannot determine the value of y for x = 0, so this expression does not represent the solution.

2) y^2 = 4x^2 + 3
Substitute x = 0: y^2 = 4(0^2) + 3
y^2 = 0 + 3
y^2 = 3
Since y^2 = 3 ≠ 4, this expression does not represent the solution.

3) y = -2x + 1/2x^3
Substitute x = 0: y = -2(0) + 1/2(0^3)
y = 0 + 0
y = 0
Since y = 0 ≠ 4, this expression does not represent the solution.

4) y^2 = 2e^x + 14
Substitute x = 0: y^2 = 2e^0 + 14
y^2 = 2(1) + 14
y^2 = 2 + 14
y^2 = 16
Taking the square root of both sides, we get y = ±4.

So, for x = 0, y can be either 4 or -4, and we have y = 4 when x = 0.

5) 1/2 ln|x^2 + 4| + 6
Substitute x = 0: 1/2 ln|(0^2) + 4| + 6
1/2 ln|4| + 6
1/2 ln(4) + 6
1/2(1.386) + 6
0.693 + 6
6.693
Since 6.693 ≠ 4, this expression does not represent the solution.

Therefore, out of the given expressions, only the fourth expression y^2 = 2e^x + 14 represents the solution to the equation yy' - e^x = 0, given the initial condition y = 4 when x = 0.