The equation of the family of the ellipses having foci on the y-axis and the centre at origin is as follows:
\(\frac{x2}{b2}+\frac{y^2}{a^2}=1...(1)\)

Differentiating equation(1)with respect to x,we get:
\(\frac{2x}{b^2}+\frac{2yy}{b^2}=0\)
\(⇒\frac{x}{b^2}+\frac{yy}{a^2}=0...(2)\)
Again, differentiating with respect to we get:
\(\frac{1}{b^2}+\frac{y.y+y.y}{a^2}=0\)
\(⇒\frac{1}{b^2}+\frac{1}{a^2}(y^2+yy)=0\)
\(⇒\frac{1}{b^2}=-\frac{1}{a^2}(y^2+yy)\)
Substituting this values in equation(2),we get:
\(x[-\frac{1}{a^2}((y)2+yy)]+\frac{yy}{a^2}=0\)
\(⇒-x(y)^2-xyy+yy=0\)
\(⇒xyy+x(y)2-yy=0\)
This is the required differential equation.
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).
A relation between involved variables, which satisfy the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called the general solution and the solution free from arbitrary constants is called particular solution.
Read More: Formation of a Differential Equation