calculus help?

Question

1).If dy/dx=x^2+4, what is the value of y

A.2x+C
B.x^3/2+4x+C
C.x^3/2+2x^2+C
D.x^3/3+4x+C

2).Calculate y as a function of x when dy/dx=4x^3+3x^2-6x+5

A.4x^2+3x+C
B.12x^2+6x+C
C.x^4+x^3-3x^2+5x+C
D.4x^4/3+x^3-6x^2+5x+C

3).When dy/dx=(x-6)^1/2, what does y equal

A.1/(x-6)^1/2 + C
B.-1/(x-6)^1/2 +C
C.2(x-6)^3/2 /3 + C
D.-2(x-6)^3/2 /3 + C

4).Solve the differential equation: dy/dx=4y^2 x^1/3

A.1/y=3x^4/3+C
B.1/y=3x^4/3+C/y^2
C.-1/y=3x^4/3+C
D.-2/y^3=3x^4/3+C/y^2

5).Which one of the following is true when ds/dt=3t^2/2s

A.s=3√6t-3t^2+C/2
B.s=√t^3+C
C.s=√t^3/2+C
D.s=√3t/2+C

Answer 1

1) Choice: D.
By integrating both sides of the differential equation, you
should get option D. Integrating dy/dx yields y, while the other
sides gives x^3/3+4x+C.

2) Choice: C.
Use the same way as that in the first answer.

3) Choice: C.
Use the previous way. In fact, Integral((x+a)^b) =
[(x+a)^(b+1)]/(b+1), provided b is not -1.(Power Rule)

4) Choice:C.
When the differential equation can be written in the form
dy/dx = f(x)g(y), we separte the variables formally
then integrate both sides as follows:
dy/g(y) = f(x)dx ; integral(dy/g(y)) = integral(f(x)dx). (This
method of solving differential equations is called separation of
variables.
Applying this procedure to the current situation, we find
dy/y^2 = 4x^(1/3)dx yielding -1/y = 4x^(1+1/3)/(1+1/3) + C
or upon simplification, -1/y = 3 x^(4/3) + C.

5) Choice: B.
As in the previous exercise, separting varibles and integrating
yields s^2 = t^3 + C, from which option B follows.

Answer 2

1. Okay, this is an antiderivative problem
d

2. Same as 1, (It is an antiderivative problem, it looks like all of them are)
c

3. Same thing, antiderivative
c

4. c

5. Can't help ya with this one buddy, did you type the answer choices in right. and the question??

Answer 3

What is this, a test? Are you asking yahoo answerers to take your test for you?