Answer:
A. √(e + 14)
Step-by-step explanation:
[tex]\frac{dy}{dx} = \frac{e^{x} - 1}{2y}[/tex]
separating the variables, we have
2ydy = ([tex]e^{x}[/tex] - 1)dx
integrating, we have
∫ydy = ∫([tex]e^{x}[/tex] - 1)dx
∫ydy = ∫[tex]e^{x}[/tex]dx - ∫1dx + C
2y²/2 = [tex]e^{x}[/tex] - x + C
y² = [tex]e^{x}[/tex] - x + C
when x = 0, y = 4
So,
y² = [tex]e^{x}[/tex] - x + C
4² = [tex]e^{0}[/tex] - 0 + C
16 = 1 + C
16 = 1 + C
C = 16 - 1
C = 15
So,
y² = [tex]e^{x}[/tex] - x + 15
when x = 1, we find y
y² = [tex]e^{1}[/tex] - 1 + 15
y² = e + 14
y = √(e + 14)
