Help....i have a calculus problem that im having trouble with...please help asap.?

Question

Consider the differential equation:

dy/dx = (y+1) / x, where x does not equal 0

Part One: On the axes provided sketch a slope field for the given differential equation at the eight points indicated.

The eight points given on the axes are: (-2,0) ; (-1,1); (-1,0); (-1,-1); (1,1); (1,0); (1,-1); (2,0)

So I am assuming you solve the equation and plug in the point values...but can someone show me how to do this correctly?

Part 2:
Find the particular solution y = f(x) to the differential equation with the intial condition f(-1)=1 and state its domain.

I have no clue how to do this .....please help.

Answer 1

The slope fields are kinda dumb really. You plug in values of x and y from each point into the differential equation to get the slope of the curve at that point. Then you make a little segment with that slope on the point. So you have a bunch of segments on your graph with different slopes. That's it.

Part 2, first separate the variables:

1/(y+1) dy = 1/x dx

Intergate both sides

ln |y+1| = ln |x| + c

Solve for c with the initial condition (-1,1)

ln |2| = ln |-1| + c
ln 2 = ln 1 + c
c = ln 2
So you have
ln |y+1| = ln |x| + ln 2
ln |y+1| = ln |2x|
|y+1| = |2x|
y+1 = ±|2x|
y = ±|2x| - 1 for x≠0 (from your original scenario)

Answer 2

The first answer above is correct but I can possibly make things a little clearer for you.
1. Slope field. At each point put the specific x and y values into the differential equation and work out the value of dy/dx.
Example: at (-2,0) dy/dx = (0 + 1)/-2 = -1/2 so you draw a little arrow at the point (-2,0) with gradient about -1/2 by eye.
2. When you integrate each side and put in a constant it will help to put the constant in as ln(c). This is a good idea in general; if you ever have to introduce a constant then introduce it in the way that best suits you.

PS. The second person could have left you something to do on your own.

Answer 3

Part One: Simply plot the points as given on the coordinate axes. The slope field, unless I am wrong, is the dy/dx as shown.

Part 2: Change the equation:
dy/(y+1) = dx/x
Integrate both sides and add a C constant on the RHS. Your solution will look like this:
y = f(x) + C
You evaluate C by the condition f(-1) = 1

Answer 4

dy/((y+1) = dx/x
so
ln I y+1I = lnI x I +C
so y+1 = KI x I
You have already the slopes at every point as
dy/dx (the slope) = (y+1)/x
At point (1,1) slope = 2/1 = 2

y+1=KIxI y = KIxI-1 with x=-1 and y=1 so
1=K-1 so K = 2 and
y=2IxI-1 Its domain are all real numbers except x=0