Question

Form the differential equation by eliminating arbitrary constants from the relation \[\dfrac{{{x^2}}}{A} + \dfrac{{{y^2}}}{B} = 1\;\] or \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\].

Answer 1

Hint: To form a differential equation first of all we have to find the number of arbitrary constants present in the given equation. Now the differential equation can be formed by differentiating the given equation two times because two arbitrary constants are present in the equation.

Complete step-by-step answer:
Given: Equation of ellipse\[ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\]
Concept: Here we see that there are two arbitrary constants (a and b) in the equation of ellipse thus we have to get a differential equation of 2nd order for eliminating all constants.
 Differentiating given equation with respect to x, we get:
\[\dfrac{{2x}}{{{a^2}}} + \dfrac{{2y}}{{{b^2}}}\left( {\dfrac{{dy}}{{dx}}} \right) = 0\]
On simplifying the above equation we get
\[ \Rightarrow \dfrac{y}{x}\dfrac{{dy}}{{dx}} = \dfrac{{ - {b^2}}}{{{a^2}}}\]
Here we see that the arbitrary constant still present in the equation
therefore
Again differentiating with respect to x both side, we get:
\[ \Rightarrow \dfrac{y}{x}\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) + \dfrac{{x\dfrac{{dy}}{{dx}} - y}}{{{x^2}}}\dfrac{{dy}}{{dx}} = 0\]
Or,
\[ \Rightarrow xy\dfrac{{{d^2}y}}{{d{x^2}}} + x\left( {\dfrac{{dy}}{{dx}}} \right){^2} - y\dfrac{{dy}}{{dx}} = 0\]
which is the required differential equation of the given equation of ellipse.

Note: Formation of differential equation always means that eliminating the arbitrary constant present from the given equation. We generally differentiate the given equation that number of times equals the number of arbitrary constants. This is not the concept to find the differential equation, the concept is that we have to vanish the arbitrary constant from the equation. It may be the same number of times the differentiation is different.