Question

Differential equations-need someone to double check my solution please
dy/dx=ysec^2x and y=5 when x=0, then y=?

Answer 1

I separated the terms and got
(1/y) dy=sec^2(x)dx
I integrate each side

Answer 2

ln(y)=tanx+C
I sub x=0 and y=5
got
ln5=tan0+c
ln5=c

Answer 3

applied e as a base to both side (not sure I am describing that correctly)
y=e^(tanx+ln5)
y=e^tanx (e^(ln5))
y=5e^(tanx)
that is my solution

Answer 4

@amistre64

Answer 5

you are correct...

Answer 6

thanks

Answer 7

pfft, i could have said precal was correct too .. ;)

Answer 8

thanks