1) The student asks for help solving two differential equations for their differential equations class. 2) The first equation is y' = 1 + x + y^2 + xy^2. The solution is y = tan(x + x^2/2 + C) where the uniqueness and existence theorems are satisfied when x + x^2/2 + C does not equal integer multiples of π/2. 3) The second equation is y' = (cos2(x)) (cos2(2y)). The solution is 2tan2y - 2x - sin2x = C if cos2y is not 0, or y = ±(2n+1)/4

1. Please help me with my Assignment!I have no idea how to solve this problem.
This is the very first lesson in my Differential Equation class. I am taking a TBA class so I have
never had any instruction. If someone can help me, I would REALLY appreciate it.
Solve the following differential equation. State the regions of the xy-plane in which the
condition of the fundamental existence and uniqueness theorem are satisfied.
#1. y' = 1 + x + y2 + xy2
#2. y' = (cos2(x)) (cos2(2y))
#2 answer is 2tan2y - 2x - sin2x= C if cos2y is not = 0; also y=±(2n+1)/4 for any integer n;
everywhere
Solution
Here's #1
y' = 1+x +y^2 (1+x) Separate y's and x's
y'= (1+x) (1+y^2)
y'/(1+y^2)= 1+x
arc tan y= x +x^2/2 +c (take antiderivative of both sides)
y= tan (x+x^2/2 +C)
Uniqueness and existence theorems are satisfied wherever the tangent function is defined and
continuous which means x +x^2/2 +C cannot equal /2, 3/2, and so on.
#2
y' sec^2 2y = 1/2 +cos 2x/2 (separate x and y, use power reducing formula for x)
1/2 tan 2y = 1/2 x +sin 2x /4 +C (take antiderivative of both sides)
C= 2tan 2y - 2x - sin 2x (multiply both sides by 4; C= 4C)