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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Obtain the differential equation from the relation Ax2 + By2 = 1, where A and Bare constants. - Mathematics and Statistics

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प्रश्न

Obtain the differential equation from the relation Ax2 + By2 = 1, where A and Bare constants.

बेरीज
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उत्तर

The given equation is Ax2 + By2 = 1     ...[1]

Differentiate w.r.t.x:

we get,

`2 Ax + 2 By (dy)/dx = 0`

`Ax + By dy/dx = 0`      ...[2]

Differentiating equation (II) w.r.t.x,

we get, 

A + B `(y (d^2 y)/dx^2 + (dy/dx)^2) = 0`    ...[3]

Since equations (1), (2) and (3) are consistent in A and B.

∴ `|(x^2, y^2, 1),(x, y dy/dx, 0),(1, (y (d^2y)/dx^2 + (dy/dx)^2), 0)| = 0`

∴ `{x [y (d^2 y)/dx^2 + (dy/dx)^2] − y dy/dx} = 0`

∴ ` xy (d^2y)/dx^2 +  x(dy/dx)^2 − y dy/dx = 0`

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