Question

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

Solution:

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

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

we get,

`square  x + 2Bydy/dx = 0`

`Ax + By dy/dx = 0`      ...[II]

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

we get, 

A + B `square = 0`     ...[III]

Since equations (I), (II), and (III) are consistent in A and B.

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

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

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

Answer 1

The given equation is Ax² + By² = 1     ...(I)

Differentiate w.r.t.x:

we get,

\(\boxed{2A}\space x+2By\frac{dy}{dx}=0\)

`Ax + By dy/dx = 0`      ...[II)

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

we get, 

\(\mathrm{A}+\mathrm{B}\space\boxed{\left(y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right)}=0\)    ...(III)

Since equations (I), (II), and (III) are consistent in A and B.

∴ \[\begin{vmatrix} x^2 & y^2 & 1 \\ x & y\frac{dy}{dx} & 0 \\ 1 & \boxed{\left(y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right)} & 0 \end{vmatrix}=0\]

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

\[\boxed{xy\frac{d^2y}{dx^2}}+x\left(\frac{dy}{dx}\right)^2-y\frac{dy}{dx}=0\]