SOLUTION: The differential equation dy/dx = y-3/x-3. produces a slope field with horizontal tangents at y = 3 produces a slope field with vertical tangents at y = 3 produces a slope fie

Algebra ->  Finance -> SOLUTION: The differential equation dy/dx = y-3/x-3. produces a slope field with horizontal tangents at y = 3 produces a slope field with vertical tangents at y = 3 produces a slope fie      Log On


   

Question 1030555: The differential equation dy/dx = y-3/x-3.
produces a slope field with horizontal tangents at y = 3
produces a slope field with vertical tangents at y = 3
produces a slope field with rows of parallel segments
I only
II only
I and II
III only
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
I only
----------------------
II cannot be true. The DE is dy%2Fdx+=+%28y-3%29%2F%28x-3%29. Letting y = 3 will give dy%2Fdx+=+0 (with x%3C%3E3), and hence corresponds to horizontal, not vertical tangents.
III cannot be true. Solving for the general solution of dy%2Fdx+=+%28y-3%29%2F%28x-3%29,
dy%2F%28y-3%29+=+dx%2F%28x-3%29 ==> ln(y-3) = ln(x-3) + lnk for arbitrary k.
==> ln(y-3) = ln(k(x-3)) ==> y-3 = k(x-3).
The general solution yields a family of straight lines that pass through the point (3,3) with the values of the slope varying through all the real numbers. In other words, drawing an arbitrary horizontal line on the cartesian plane will intersect this family at a row of infinite number of points where the (tangential) segments are not parallel, but in which the slopes vary from negative to positive infinity.