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(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 $

Answer

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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 $

For example, the order of the differential $ \dfrac{{{d^3}y}}{{d{x^3}}} $ is 3.

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