Answer
405.6k+ views
Hint: We will start by using the fact that the general equation of an ellipse with centre at (h, k) is $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$. Then we will write the equation of the ellipse for the centre at origin and differentiate it twice because for forming a differential equation we have removed arbitrary constants which are two in this case.
Complete step by step answer:
Now, we have to find the differential equation of all the ellipses whose centre is at origin and axes are along the coordinate axes.
Now, we know that the equation of an ellipse whose centre is at origin is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.
Now, since to form a differential equation we have to eliminate the arbitrary constant and to remove. Then we have to differentiate the equation as many time as the number of constant since the equation has 2 constant. So, we have to differentiate is two times.
$\begin{align}
& \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 \\
& \dfrac{d}{dx}\left( \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}} \right)=\dfrac{d}{dx}\left( 1 \right) \\
\end{align}$
Now, we know that $\dfrac{d}{dx}\left( \text{constant} \right)=0$
$\begin{align}
& \Rightarrow \dfrac{2x}{{{a}^{2}}}+\dfrac{2y}{{{b}^{2}}}\dfrac{dy}{dx}=0 \\
& \Rightarrow \dfrac{2x}{{{a}^{2}}}+\dfrac{2y}{{{b}^{2}}}\dfrac{dy}{dx}=0 \\
& \Rightarrow \dfrac{2y}{{{b}^{2}}}\dfrac{dy}{dx}=-\dfrac{2x}{{{a}^{2}}} \\
& \Rightarrow \dfrac{2y}{2x}\dfrac{dy}{dx}=-\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
& \Rightarrow \dfrac{y}{x}\dfrac{dy}{dx}=-\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
\end{align}$
Now again we will differentiate both sides with respect to x. So, we have,
\[\dfrac{d}{dx}\left( \dfrac{y}{x}.y' \right)=\dfrac{d}{dx}\left( -\dfrac{{{b}^{2}}}{{{a}^{2}}} \right)\]
Now we know that,
\[\begin{align}
& \dfrac{d}{dx}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{g\left( x \right)f'\left( x \right)-f\left( x \right)g'\left( x \right)}{g{{\left( x \right)}^{2}}} \\
& \dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=f\left( x \right)g'\left( x \right)+g\left( x \right)f'\left( x \right) \\
\end{align}\]
So, we have,
$\begin{align}
& \Rightarrow \dfrac{y}{x}\dfrac{d}{dx}\left( y' \right)+y'\dfrac{d}{dx}\left( \dfrac{y}{x} \right)=0 \\
& \Rightarrow \dfrac{y}{x}{{y}^{“}}+y'\left( \dfrac{xy'-y\times 1}{{{x}^{2}}} \right)=0 \\
& \Rightarrow \dfrac{y}{x}{{y}^{“}}+y'\left( \dfrac{xy'-y}{{{x}^{2}}} \right)=0 \\
& \Rightarrow xy{{y}^{‘’}}+x{{\left( y' \right)}^{2}}-y{{y}^{‘}}=0 \\
\end{align}$
is the required differential equation.
Note: To solve such types of questions it is important to note that we have differentiated the equation twice to remove all the arbitrary constants from the differential equation. Generally the number of times an equation has to be differentiated depends on the number of arbitrary constants it has.
Complete step by step answer:
Now, we have to find the differential equation of all the ellipses whose centre is at origin and axes are along the coordinate axes.
Now, we know that the equation of an ellipse whose centre is at origin is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.
![seo images](https://www.vedantu.com/question-sets/bee89550-0fb6-457f-b30f-8898f96fd62c440597666116630533.png)
Now, since to form a differential equation we have to eliminate the arbitrary constant and to remove. Then we have to differentiate the equation as many time as the number of constant since the equation has 2 constant. So, we have to differentiate is two times.
$\begin{align}
& \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 \\
& \dfrac{d}{dx}\left( \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}} \right)=\dfrac{d}{dx}\left( 1 \right) \\
\end{align}$
Now, we know that $\dfrac{d}{dx}\left( \text{constant} \right)=0$
$\begin{align}
& \Rightarrow \dfrac{2x}{{{a}^{2}}}+\dfrac{2y}{{{b}^{2}}}\dfrac{dy}{dx}=0 \\
& \Rightarrow \dfrac{2x}{{{a}^{2}}}+\dfrac{2y}{{{b}^{2}}}\dfrac{dy}{dx}=0 \\
& \Rightarrow \dfrac{2y}{{{b}^{2}}}\dfrac{dy}{dx}=-\dfrac{2x}{{{a}^{2}}} \\
& \Rightarrow \dfrac{2y}{2x}\dfrac{dy}{dx}=-\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
& \Rightarrow \dfrac{y}{x}\dfrac{dy}{dx}=-\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
\end{align}$
Now again we will differentiate both sides with respect to x. So, we have,
\[\dfrac{d}{dx}\left( \dfrac{y}{x}.y' \right)=\dfrac{d}{dx}\left( -\dfrac{{{b}^{2}}}{{{a}^{2}}} \right)\]
Now we know that,
\[\begin{align}
& \dfrac{d}{dx}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{g\left( x \right)f'\left( x \right)-f\left( x \right)g'\left( x \right)}{g{{\left( x \right)}^{2}}} \\
& \dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=f\left( x \right)g'\left( x \right)+g\left( x \right)f'\left( x \right) \\
\end{align}\]
So, we have,
$\begin{align}
& \Rightarrow \dfrac{y}{x}\dfrac{d}{dx}\left( y' \right)+y'\dfrac{d}{dx}\left( \dfrac{y}{x} \right)=0 \\
& \Rightarrow \dfrac{y}{x}{{y}^{“}}+y'\left( \dfrac{xy'-y\times 1}{{{x}^{2}}} \right)=0 \\
& \Rightarrow \dfrac{y}{x}{{y}^{“}}+y'\left( \dfrac{xy'-y}{{{x}^{2}}} \right)=0 \\
& \Rightarrow xy{{y}^{‘’}}+x{{\left( y' \right)}^{2}}-y{{y}^{‘}}=0 \\
\end{align}$
is the required differential equation.
Note: To solve such types of questions it is important to note that we have differentiated the equation twice to remove all the arbitrary constants from the differential equation. Generally the number of times an equation has to be differentiated depends on the number of arbitrary constants it has.
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)
Which are the Top 10 Largest Countries of the World?
![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)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![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)
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)
Change the following sentences into negative and interrogative class 10 english 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)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)