Two Questions (antiderivative and diff eq.)

[tod88]

Homework Statement

(i've put my questions in bold)

The differential equation is

dy/dx = xy^2

I separated variables and came up with (solving for c)

c = -1/y - (x^2)/2

I then had to find the particular solution for y(0) = 1, so I came up with

c = -1 - 0 so c = -1

I then solved for y and got that y = -1/(1-(x^2)/2) ... is this the correct answer for the specific solution?

Finally, I had to determine what values of C gave the diff eq. vertical asymptotes. I know this happens when the denominator equals zero, so I said that it is the case when

c+(x^2)/2 = 0. This only happens when C = 0 and x = 0. Correct?

Homework Statement

The second question:

the function G(x) = G(-4) + antiderivative from -4 to x of f(t) dt over the interval
[-4,3]

I have the graph of f, however, so when I take G'(x) I get f(x), correct? What about the G(-2)? That's the part I am not sure of. I'm assuming that it doesn't matter since the interval runs from [-4,3] and the antiderivative from -4 to -4 is simply zero. Is this the right way of looking at it?

Thanks.

 

[HallsofIvy]

Homework Statement

(i've put my questions in bold)

The differential equation is

dy/dx = xy^2

I separated variables and came up with (solving for c)

c = -1/y - (x^2)/2

I then had to find the particular solution for y(0) = 1, so I came up with

c = -1 - 0 so c = -1

I then solved for y and got that y = -1/(1-(x^2)/2) ... is this the correct answer for the specific solution?

You can check it youself. Does y satisfy the original differential equation? Does it satisfy y(0)= -1?

Finally, I had to determine what values of C gave the diff eq. vertical asymptotes. I know this happens when the denominator equals zero, so I said that it is the case when

c+(x^2)/2 = 0. This only happens when C = 0 and x = 0. Correct?

Suppose c= -2? What would happen at x= 2 or -2? Are you told to assume that c cannot be negative?

Homework Statement

The second question:

the function G(x) = G(-4) + antiderivative from -4 to x of f(t) dt over the interval
[-4,3]

I have the graph of f, however, so when I take G'(x) I get f(x), correct? What about the G(-2)? That's the part I am not sure of. I'm assuming that it doesn't matter since the interval runs from [-4,3] and the antiderivative from -4 to -4 is simply zero. Is this the right way of looking at it?

Thanks.

What does "from -4 to -4" have to do with anything" $G(-2)= G(-4)+ \int_{-4}^{-2}f(x)dx$. But there was no mention of "G(-2)" in the statement of the question. Did you mean G(-4)? In fact, there is no question in the statement of the question! There is simply a definition of the function G(x). What is the question? To find the derivative of G? If so, then G(-4) is a constant and the derivative of a constant is 0.

 

[tod88]

Thanks very much for your help. I apologize if some of the questions were dumb / elementary...but I've figured out how I was thinking wrongly about the problem.