Question 1044364
By solving the differential equation, we easily get 

{{{y = 2x - x^2/2 +7/2}}},

as per the boundary condition.

Now, to get the slope of the tangent line at the point (0,7/2), simply substitute x = 0 into the original DE {{{dy/dx = 2-x}}}, 
from which we get {{{dy/dx = 2}}}.

===> The slope of the normal line to the curve at point P is -1/2 (i.e., the negative reciprocal of 2).

===> The equation of the normal line to the curve at point P is 

{{{y-7/2 = -x/2}}}, or {{{y = 7/2 - x/2}}}  <===THE NORMAL LINE

To find the other point Q on the curve where the normal line meets the curve again, solve

{{{7/2 - x/2 =  2x - x^2/2 +7/2}}},

===> {{{0 = 5x - x^2}}} ===> 0 = x(5-x)  ===> x = 0, 5.

When x = 0, y = 7/2, which is just the first point the normal meets the curve.

When x = 5, y = 1, which is the point (5,1), the second point where the normal meets the curve.

{{{graph( 300, 300, -3, 7, -3, 7, 2x-x^2/2+7/2, 7/2-x/2 , 7/2+2x)}}}