Question

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?

Answer 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.

#3
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

#4
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

Answer 2

-1-2cosx