(1, 1) , y'=-9x/(16y) Find an equation of the graph that passes through the point and has the given slope

To solve this equation, multiply by y and integrate:
yy' = -9/16 x,  int yy' dx = int (-9/16 x) dx,
y^2/2 = -9/32 x^2 + C, or  y = +-sqrt(C - 9/16 x^2),
where C is an arbitrary constant.
We need to find a suitable constant C using the given point. The condition is y(1) = 1, or
1 = +-sqrt(C - 9/16)  (+ is before the radical obviously).
This gives us  1 = C - 9/16, so  C = 25/16 and the final answer is
y(x) = +-sqrt(25/16 - 9/16 x^2).
 

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