Answer

Verified

455.7k+ views

Hint: First of all, write the equation of the line passing through the origin by using formula \[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\]. Then differentiate both sides with respect to x and finally substitute the value of m in it.

Complete step-by-step answer:

Here, we have to find the differential equation of the family of all straight lines passing through the origin. We know that to find the differential equation of the family of curves, we have to find the equation of the curve first. So, now we will find the equation of the line passing through the origin.

We know that the equation of the line of slope m and passing through points \[\left( {{x}_{1}},{{y}_{1}} \right)\] is written as

\[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\]

So, by substituting \[{{y}_{1}}=0\] and \[{{x}_{1}}=0\], we get the equation of the line passing through the origin as,

\[\left( y-0 \right)=m\left( x-0 \right)\]

Or \[y=mx....\left( i \right)\]

By dividing m on both the sides, we get

\[\dfrac{y}{x}=m....\left( ii \right)\]

We know that according to the product rule of differentiation, \[\dfrac{d}{dx}\left( f.g \right)=g\left( \dfrac{df}{dx} \right)+f.\left( \dfrac{dy}{dx} \right)\]

So, by differentiating both sides of equation (i) with respect to x, we get,

\[\dfrac{dy}{dx}=m\left( \dfrac{dx}{dy} \right)+x\left( \dfrac{dm}{dx} \right)\]

We know that m is constant for a particular value of x and y, so \[\dfrac{dm}{dx}=0\]. So, we get,

\[\dfrac{dy}{dx}=m\left( 1 \right)+x\left( 0 \right)\]

\[\Rightarrow \dfrac{dy}{dx}=m\]

By substituting the value of m from equation (ii), we get,

\[\dfrac{dy}{dx}=\dfrac{y}{x}\]

By subtracting \[\dfrac{y}{x}\] from both sides of the above equation, we get,

\[\dfrac{dy}{dx}-\dfrac{y}{x}=0\]

By multiplying x dx on both the sides of the above equation, we get,

\[x\text{ }dy-y\text{ }dx=0\]

So, we get the differential equation of the family of all straight lines passing through the origin as

\[x\text{ }dy-y\text{ }dx=0\]

Note:

Students must note that to find the differential equation of any curve, they must eliminate all the constants from the equation like we eliminated ‘m’ in the above solution. Also, students can verify this differential equation by substituting the value of \[\dfrac{dy}{dx}\] in the differential equation and checking if the original equation of the curve is obtained or not.

Complete step-by-step answer:

Here, we have to find the differential equation of the family of all straight lines passing through the origin. We know that to find the differential equation of the family of curves, we have to find the equation of the curve first. So, now we will find the equation of the line passing through the origin.

We know that the equation of the line of slope m and passing through points \[\left( {{x}_{1}},{{y}_{1}} \right)\] is written as

\[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\]

So, by substituting \[{{y}_{1}}=0\] and \[{{x}_{1}}=0\], we get the equation of the line passing through the origin as,

\[\left( y-0 \right)=m\left( x-0 \right)\]

Or \[y=mx....\left( i \right)\]

By dividing m on both the sides, we get

\[\dfrac{y}{x}=m....\left( ii \right)\]

We know that according to the product rule of differentiation, \[\dfrac{d}{dx}\left( f.g \right)=g\left( \dfrac{df}{dx} \right)+f.\left( \dfrac{dy}{dx} \right)\]

So, by differentiating both sides of equation (i) with respect to x, we get,

\[\dfrac{dy}{dx}=m\left( \dfrac{dx}{dy} \right)+x\left( \dfrac{dm}{dx} \right)\]

We know that m is constant for a particular value of x and y, so \[\dfrac{dm}{dx}=0\]. So, we get,

\[\dfrac{dy}{dx}=m\left( 1 \right)+x\left( 0 \right)\]

\[\Rightarrow \dfrac{dy}{dx}=m\]

By substituting the value of m from equation (ii), we get,

\[\dfrac{dy}{dx}=\dfrac{y}{x}\]

By subtracting \[\dfrac{y}{x}\] from both sides of the above equation, we get,

\[\dfrac{dy}{dx}-\dfrac{y}{x}=0\]

By multiplying x dx on both the sides of the above equation, we get,

\[x\text{ }dy-y\text{ }dx=0\]

So, we get the differential equation of the family of all straight lines passing through the origin as

\[x\text{ }dy-y\text{ }dx=0\]

Note:

Students must note that to find the differential equation of any curve, they must eliminate all the constants from the equation like we eliminated ‘m’ in the above solution. Also, students can verify this differential equation by substituting the value of \[\dfrac{dy}{dx}\] in the differential equation and checking if the original equation of the curve is obtained or not.

Recently Updated Pages

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

10 examples of friction in our daily life

Trending doubts

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

10 examples of law on inertia in our daily life

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Change the following sentences into negative and interrogative class 10 english CBSE