please solve y' = sec(x+y)?

Question

the final answer is --> y - tan[(x+y)/2] = c

it's in the separation by variables exercise.

thanks, if you don't know please don't answer

Answer 1

y' = sec(x+y)

let us substitute z = x+y so y ' = sec z
dz/dx = 1+y ' = 1+sec z
dz/dx = 1+sec z
dz /(1+sec z) = dx
cos (z) dz / (1+cos z) = dx

{2 cos^2 (z/2) - 1} dz / {2 cos^2 (z/2)} = dx

{2 cos^2 (z/2) - 1} dz / {2cos^2 (z/2)} = dx

{dz} - (1/2) {sec^2 (z/2)} dz = dx integrating

{z} - {tan (z/2)} = x +C

(x+y) - tan (x+y) /2 = x+C

y - tan (x+y) /2 = C answer

Answer 2

I don't think that the answer is as simple as the first person suggests. Integrating expressions with two variables is notoriously difficult. Where did you get what you say is the final answer? because I don't think that it is correct. If you write it as y = tan((x + y)/2) + c and differentiate you get dy/dx = sec^2 of ((x + y)/2) * (1 + dy/dx)/2 and I don't think that this rearranges to your first line. It's true that I've not spent too long on it, but instinct tells me that there's something wrong somewhere, maybe even in the question.

Answer 3

To solve this, understand that y' is also notated more commonly as dy/dx. I have no idea why people write y'. Anyway...

dy/dx = sec(x+y)
dy/dx= 1/cos(x+y) <-- Trig ID
dy/dx = 1/cos(x)cos(y) - sin(x)sin(y) <-- Trig ID
int(dy/dx) = int[1/(cosxcosy - sinxsiny)] <--- To get "y"
y = tan[(x + y) /2] + C <-- after integration and simplifcation
y - tan[(x + y) /2] = C