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# Find the degree of differential equation of all curves having normal of constant length c-A) $1$ B) $3$ C) $4$ D) $2$

Last updated date: 23rd Jun 2024
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Hint: We can find the degree of differential equation by using the formula-
Length of normal=$y\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}$ where y is the given function and $\dfrac{{dy}}{{dx}}$ is the derivative of the function.

The degree is the power of the highest derivative .Here, we have to find the degree of differential equation of all curves having a normal of constant length c. We know that the if y=f(x) is any given function of a curve then at point $\left( {{{\text{x}}_1},{{\text{y}}_1}} \right)$ the length of normal is given as-
$\Rightarrow$ Length of normal=$y\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}$ where y is the given function and $\dfrac{{dy}}{{dx}}$ is the derivative of the function
$\Rightarrow {\text{c = }}y\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}$ --- (i)
$\Rightarrow {{\text{c}}^2} = {{\text{y}}^2}{\left( {1 + \dfrac{{dy}}{{dx}}} \right)^2}$
On simplifying and multiplying the function ${{\text{y}}^2}$ inside the bracket, we get-
$\Rightarrow {{\text{c}}^2} = {{\text{y}}^2}\left( {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2} + 2\dfrac{{dy}}{{dx}}} \right) \\ \Rightarrow {{\text{c}}^2} = {{\text{y}}^2} + {{\text{y}}^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} + 2{{\text{y}}^2}\dfrac{{dy}}{{dx}} \\$
Here the highest derivative is ${\left( {\dfrac{{dy}}{{dx}}} \right)^2}$ and its power is $2$ so the degree of the differential equation is also $2$