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Form the differential equation from $a{x^2} + b{y^2} = 1$ (A) $xy{y_2} + x{y_1}^2 - y{y_1} = 0$ (B) $xy{y_2} - x{y_1}^2 - y{y_1} = 0$ (C) $xy{y_2} + x{y_1}^2 + y{y_1} = 0$ (D) $xy{y_2} - x{y_1}^2 + y{y_1} = 0$

Last updated date: 13th Jun 2024
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Hint: In this question, we have to derive the differential equation from the given expression. There are two variables given in the expression $x$ and $y$ . According to the question, the differential equation has to be a second order differential equation in terms of $y$ which means that we have to differentiate the given expression two times w.r.t. $x$ .
We know that $\dfrac{{dy}}{{dx}} = {y_1}$
$\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = {y_2}$
And,
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Given: The expression given is-
$a{x^2} + b{y^2} = 1$
Where $a{\text{ and b}}$ are arbitrary constants.
In order to find the differential form of this expression we have to do the derivation. Since there are 2 arbitrary constants in the expression and to remove them, we have to do the differentiation twice.
So, differentiating both sides of the expression w.r.t. $x$ we get,
$\dfrac{d}{{dx}}\left( {a{x^2} + b{y^2}} \right) = \dfrac{d}{{dx}}\left( 1 \right)\\ a\dfrac{d}{{dx}}\left( {{x^2}} \right) + b\dfrac{d}{{dx}}\left( {{y^2}} \right) = 0$
Solving this we get,
$a \times 2x + b \times 2y \times \dfrac{{dy}}{{dx}} = 0$
We can write the first derivative $\dfrac{{dy}}{{dx}} = {y_1}$ in the equation. So, we get,
$2ax + 2by{y_1} = 0\\ 2\left( {ax + by{y_1}} \right) = 0$
We can write this as-
$ax + by{y_1} = 0$
This is our first equation.
Now differentiating both sides again w.r.t. $x$ we get,
$\dfrac{d}{{dx}}\left( {ax + by{y_1}} \right) = 0\\ a\dfrac{d}{{dx}}\left( x \right) + b\dfrac{d}{{dx}}\left( {y{y_1}} \right) = 0$
We can write
$\dfrac{d}{{dx}}\left( {y{y_1}} \right) = y\dfrac{d}{{dx}}\left( {{y_1}} \right) + {y_1}\dfrac{d}{{dx}}\left( y \right)\\ = y\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) + {y_1}\left( {\dfrac{{dy}}{{dx}}} \right)$
We know that $\dfrac{{dy}}{{dx}} = {y_1}$ and
$\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{{d^2}y}}{{d{x^2}}}\\ = {y_2}$
So, $\dfrac{d}{{dx}}\left( {y{y_1}} \right) = y \times {y_2} + {y_1} \times {y_1}\\ = y{y_2} + {y_1}^2$
Substituting these values in the equation we get,
$a \times 1 + b \times \left( {y{y_2} + {y_1}^2} \right) = 0\\ a + b\left( {y{y_2} + {y_1}^2} \right) = 0\\ a = - b\left( {y{y_2} + {y_1}^2} \right)$
Putting this value of $a$ in the first equation we get,
$- b\left( {y{y_2} + {y_1}^2} \right)x + by{y_1} = 0\\ y{y_1} = \left( {y{y_2} + {y_1}^2} \right)x\\ y{y_1} = xy{y_2} + x{y_1}^2$
We can write this equation as-
$xy{y_2} + x{y_1}^2 - y{y_1} = 0$
Therefore, the differential equation is $xy{y_2} + x{y_1}^2 - y{y_1} = 0$
So, the correct answer is “Option A”.

Note: The number of times we have to do the differentiation depends upon the numbers of arbitrary constants present in the expression. In the second-order differential equation, the highest order of the equation is 2. The order of the derivative is defined as the number of times of the derivation of the variable.
For example, the order of the differential $\dfrac{{{d^3}y}}{{d{x^3}}}$ is 3.