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upendra hembram Grade: 12
        

dy/dx +[(x2-1)(y2-1)]1/2/xy=0

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear upendra


dy/dx +[(x2-1)(y2-1)]1/2/xy=0


or  ydy/(y2-1)1/2      =    - dx (x2-1)1/2/x


or ydy/(y2-1)1/2      =    - dx (x2-1)/x(x2-1)1/2


or ydy/(y2-1)1/2      =   x dx /x2(x2-1)1/2     -xdx /(x2-1)1/2    


 


now intigrate


(y2-1)1/2    = tan-1(x2-1)1/2   - (x2-1)1/2    





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7 years ago
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