CALCULUS: If dy/dx = (y)(sec x)^2 and y=5 when x=0 then y = .....

IF dy/dx = (y)(sec x)^2 and y=5 when x=0 then y = ..... ???

Thank you!

dy/dx = y sec^2 x
dy/y = sec^2 x dx
ln y = tan x + C
ln 5 = tan 0 + C = C
ln y = tan x + ln 5
ln y - ln 5 = tan x
ln (y/5) = tan x
y/5 = e^(tan x)
y = 5 e^(tan x)

dy/dx = (y)(secx)^2 and y = 5 when x = 0, then y = ?

Let's separate the variables and integrate.
Doing so, gives:

dy/y = (sec)^2 dx
∫ dy/y = ∫ (sec)^2 dx
ln(y) + C = tanx + C
ln(y) = tanx + C
ln(5) = tan(0) + C
ln(5) = 0 + C
ln(5) = C.

Therefore,

ln(y) = tanx + ln(5)
y = e^(tanx + ln(5)).

Hope this helped much.