Question

Form the differential equation representing the family of curves given by $(x - a)^2 + 2y^2 = a^2$, where a is an arbitrary constant.

Answer 1

Answer:

The required differential equation is

2y d²y/dx² + 2 (dy/dx)² + 1 = 0.

Solution:

The given family of curves is

(x - a)² + 2y² = a²

or, x² - 2ax + a² + 2y² = a²

or, x² - 2ax + 2y² = 0

Differentiating both sides with respect to x, we get

d/dx (x²) - d/dx (2ax) + d/dx (2y²) = 0

or, 2x - 2a + 4y dy/dx = 0

or, x - a + 2y dy/dx = 0

or, 2y dy/dx + x = a

Again differentiating both sides with respect to x, we get

d/dx (2y dy/dx) + d/dx (x) = d/dx (a)

or, 2 (dy/dx)² + 2y d²y/dx² + 1 = 0

or, 2y d²y/dx² + 2 (dy/dx)² + 1 = 0

This is the required differential equation of the curve.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The differential equation representing the family of curves given by (x - a)^2 + 2y^2 = a^2

where a is an arbitrary constant

CALCULATION

$\sf{ {(x - a)}^{2} + 2 {y}^{2} = {a}^{2} \: \: } \: ........(1)$

Differentiating both sides with respect to x we get

$\displaystyle \: \sf{ 2(x - a) + 4y \frac{dy}{dx} = 0\: \: }$

$\implies \: \displaystyle \: \sf{ (x - a) + 2y \frac{dy}{dx} = 0\: \: }$

$\implies \: \displaystyle \: \sf{ a = x + 2y \frac{dy}{dx} \: \: }$

From Equation (1) we get

$\implies \: \displaystyle \: \sf{ {(x -x - 2y \frac{dy}{dx} )}^{2} + 2 {y}^{2} = { \bigg(x + 2y \frac{dy}{dx} \bigg)}^{2} }$

$\implies \: \displaystyle \: \sf{ { \bigg(2y \frac{dy}{dx} \bigg)}^{2} + 2 {y}^{2} = { \bigg(x + 2y \frac{dy}{dx} \bigg)}^{2} }$

$\implies \: \displaystyle \: \sf{ 4 {y}^{2} { \bigg( \frac{dy}{dx} \bigg)}^{2} + 2 {y}^{2} = {x }^{2} + 4xy \frac{dy}{dx} + 4 {y}^{2} { \bigg( \frac{dy}{dx} \bigg)}^{2} }$

$\implies \: \displaystyle \: \sf{{x }^{2} + 4xy \frac{dy}{dx} - 2 {y}^{2} = 0 }$

RESULT

The required differential equation is

$\boxed{ \: \displaystyle \: \sf{{x }^{2} + 4xy \frac{dy}{dx} - 2 {y}^{2} = 0 \: \: \: }}$

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