AP Calculus Help Five Questions

1. Find the particular solution to y " = 2sin(x) given the general solution y = -2sin(x) + Ax + B and the initial conditions y(pi/2) = 0 and y'(pi/2) = -2.

2. What function is a solution to the differential equation y ' - y = 0?

3. If dy/dx = cos^2(pi*y/4) and y = 1 when x = 0, find the value of x when y = 3.

4. The differential equation dy/dx = (x+1)/(x-2):
I.produces a slope field with horizontal tangents at x = -1
II.produces a slope field with vertical tangents at x = 2
III.produces a slope field with rows of parallel segments
Select those that are true.

5. If dy/dx = x*y^2 and if y = 1 when x = 0, then what is x when y = 3?

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  1. #1 is just like your other post, only you have to use two steps.
    y" = 2sin(x)
    y' = -2cosx+A
    -2 = -2cos(π/2) = 0+A so A=-2
    y' = -2cosx-2
    y = -2sinx-2x+B
    0 = -2sin(π/2)-2(π/2)+B = -2-π+B so B=2+π
    y = -2sinx-2x+π+2

    #2 This you can do by inspection: you know that d/dx(e^x) = e^x
    Or, you can go through the trouble of finding the integrating factor.

    dy/dx = cos^2(pi*y/4)
    dy = cos^2(π/4 y) dx
    dy/cos^2(π/4 y) = dx
    sec^2(π/4 y) dy = dx
    (4/π)tan(π/4 y) = x+c
    since y(0)=1,
    (4/π)tan(π/4) = 0+c
    4/π = c
    (4/π)tan(π/4 y) = x + 4/π
    when y=3, we have
    (4/π)tan(π/4*3) = x+4/π
    (4/π)(-1) = x+4/π
    x = -8/π

    is (x+1)/(x-2)=0 at x = -1?
    is it undefined when x=2?
    does the slope change with changes in y?

    #5 is just like #3
    dy/dx = x*y^2
    dy/y^2 = x dx
    -1/y = 1/2 x^2+c
    y = -1/(1/2 x^2+c) = -2/(x^2+c) (different c but who cares?)
    now just plug in (0,1) to find c

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