Answer 1

`x+ydy/dx=sec(x^2+y^2)...........(i)`

`put x^2+y^2=t`

Differentiating w.r.t. x, we get

`2x+2ydy/dx=dt/dx`

`x+ydy/dx=1/2dt/dx`

`1/2 dt/dx=sect`

`dt/sect=2dx`

Integrating on both sides, we get

`intcostdt=2intdx`

sin t = 2x + c
 sin (x² + y²) = 2x + c [1]
When x = y = 0
sin (0 + 0) = 2 (0) + c
c = 0
Particular solution is sin (x² + y²) = 2x