Thursday 17 April 2014

`(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 =...

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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