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`        The differential equation of all circles which passes through the origin and whose centre lies on y axis is`
7 years ago

147 Points
```										Dear pallavi
let the center is (o,a)  and  radius is r
so equation of circle is
x2 + (y-a)2 =r2
differentiate
2x +2(y-a) dy/dx =0
x+ (y-a)dy/dx =0  ..............1

again differentiate
1+ (y-a)d2y/dx2 + dy/dx =0
put value of (y-a) from equation 1
1 -x(d2y/dx2)/(dy/dx) + dy/dx =0
(dy/dx )2 + dy/dx  - xd2y/dx2=0
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```
7 years ago
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