Question

From the differential equation of the family of parabolas having a vertex at the origin and axis along the positive y-axis.

Answer 1

Hint: First of all, write the equation of parabola having a vertex at origin and axis along the positive y-axis that is \[{{x}^{2}}=4ay\]. Now differentiate this equation with respect to x and substitute the value of 4a from one equation into another to get the differential equation.

Complete step by step answer:
In this question, we have to form a differential equation of the family of parabolas having a vertex at origin and axis along the positive y-axis. We know that the equation of the parabola having axis along the positive y-axis is given by \[{{\left( x-{{x}_{1}} \right)}^{2}}=4a\left( y-{{y}_{1}} \right)\] where \[\left( {{x}_{1}},{{y}_{1}} \right)\] are the coordinates of the vertex.
Here, we are given that vertex is at the origin, so by substituting \[{{x}_{1}}=0\] and \[{{y}_{1}}=0\], we get,
\[{{\left( x-0 \right)}^{2}}=4a\left( y-0 \right)\]
So, we get the equation of the family parabolas having a vertex at origin and axis along the positive y-axis as,
\[{{x}^{2}}=4ay....\left( i \right)\]
Now, to find the differential equation of this family of parabolas, we will differentiate both sides of the above equation with respect to x. We know that \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\] and \[\dfrac{d}{dy}\left( ky \right)=k\dfrac{dy}{dx}\left( k\text{ is constant} \right)\].
So by differentiating equation (i) with respect to x, we get,
\[\dfrac{d}{dx}\left( {{x}^{2}} \right)=\dfrac{d}{dx}\left( 4ay \right)\]
Here 4a is constant. So, we get,
\[2{{x}^{2-1}}=4a\dfrac{dy}{dx}\]
\[2x=4a\dfrac{dy}{dx}\]
By multiplying \[\dfrac{dx}{dy}\] on both the sides of the above equation, we get,
\[\left( 2x \right)\dfrac{dx}{dy}=4a\]
Now by substituting this value of 4a inequation (i), we get,
\[{{x}^{2}}=4ay\]
\[{{x}^{2}}=\left( 2x\dfrac{dx}{dy} \right)y\]
By multiplying \[\dfrac{dy}{dx}\] on both the sides of the above equation, we get,
\[{{x}^{2}}.\dfrac{dy}{dx}=2xy\]
\[\Rightarrow {{x}^{2}}\dfrac{dy}{dx}-2xy=0\]
By taking out the common from the above equation, we get,
\[x\left( x\dfrac{dy}{dx}-2y \right)=0\]
So, x = 0 or \[x\dfrac{dy}{dx}-2y=0\]
Let us take, \[\dfrac{dy}{dx}=y'\]
So, we get, xy’ – 2y = 0

So, we get the differential equation of the family of a parabola having a vertex at origin and axis along the y-axis as xy’ – 2y = 0.

Note: In this question, students can cross-check their answer by integrating the final differential equation and checking if it is giving the initial equation or not as integration and differentiation are the reverse operations of each other. Also, students must note that basically to find a differential equation of an equation, we have to eliminate constant terms from it like we have eliminated constant term 4a from \[{{x}^{2}}=4ay\] by differentiation.