Answer
397.2k+ views
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}} $
Complete step-by-step answer:
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.
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}} $
Complete step-by-step answer:
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)