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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 â€˜mâ€™ 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 â€˜mâ€™ 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.

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