Question

Consider the differential equation dy/dx= e^x-1/2y. If y = 4 when x = 0 what is a value of y when x = 1?

Answer 1

Answer:

A. √(e + 14)

Step-by-step explanation:

$\frac{dy}{dx} = \frac{e^{x} - 1}{2y}$

separating the variables, we have

2ydy = ($e^{x}$ - 1)dx

integrating, we have

∫ydy = ∫($e^{x}$ - 1)dx

∫ydy = ∫$e^{x}$dx - ∫1dx + C

2y²/2 = $e^{x}$ - x + C

y² = $e^{x}$ - x + C

when x = 0, y = 4

So,

y² = $e^{x}$ - x + C

4² = $e^{0}$ - 0 + C

16 = 1 + C

16 = 1 + C

C = 16 - 1

C = 15

So,

y² = $e^{x}$ - x + 15

when x = 1, we find y

y² = $e^{1}$ - 1 + 15

y² = e + 14

y = √(e + 14)