Differential Equations

Given slope at (x,y) is
d

y

d

x=

y

x

+ sec (

y

x

)
let

y

x

=t ⇒ y = xt ⇒ ‌‌

dy

dx

=t+x

dt

dx

t + x

dt

dx

= t + sec(t)
∫ cot t dt = ∫

1

x

dx
sin t = lnx + c
sin (

y

x

) = lnx + c
This curve passes through (1,

π

6

)
sin (

π

6

) = ln(1) + c ⇒ c =

1

2

sin (

y

x

) = lnx +

1

2