dy/dx=cos(x+y), y(0)=PI/4
I'm trying to do this with substitution. z=x+y
after that the teacher lost me taking the derivitive of both sides is
=>1+dy/dx=dz/dx How?
You have z = x + y,
now differentiate both sides, with respect to x. Then
d(z)/dx = d(x + y)/dx
dz/dx = d(x)/dx + d(y)/dx
dz/dx = dx/dx + dy/dx
dz/dx = 1 + dy/dx (and you had 1 + dy/dx = dz/dx)
Now we substitute for z = x+y and dy/dx = cos (x + y) into the underlined equation above.
dz/dx = 1 + cos(x + y)
dz/dx = 1 + cos(z)
Using cos(2A) = 2cos^2(A) - 1, (double angle formula)
dz/dx = 1 + (2cos^2(z/2) - 1) = 2cos^2(z/2)
dz/dx = 2cos^2(z/2)
inverting the above,
dx/dz = (1/2)sec^2(z/2)
Now integrating the above,
x = tan(z/2) + Const
z = 2arctan((x + const)
Substituting back in for z = x + y,
x + y = 2arctan(x + Const)
y(x) = -x + 2arctan(x + Const)