(1 point) The equation 3y′=3xy+2x2+2y2x2, (∗) can be written in the form y′=f(y/x), i. e., it is homogeneous, so we can use the substitution u=y/x to obtain a separable equation with dependent variable u=u(x).

Introducing this substitution and using the fact that y′=xu′+u we can write (∗) as y′=xu′+u=f(u) where f(u)= 1/3(3u+2u^2+2) .

Separating variables we can write the equation in the form g(u)du=dxx where g(u)= 3/(2(u^2+1)) .

An implicit general solution with dependent variable u can be written in the form ln(x)− 3/2arctan(u) =C.

Transforming u=y/x back into the variables x and y and using the initial condition y(1)=1 we find C= -(3pi)/8

Finally solve for y to obtain the explicit solution of the initial value problem y= xtan(2/3lnx+(3pi)/8) Note: You can earn partial credit on this problem.

the correct answers i think are b. and d.

give me brainliest if it is correct!

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

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6: 4= 3/2

18: 8= 9/4

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