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

A. √(e + 14)

Explanation:

`(dy)/(dx) = (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)