# Find the degree of differential equation of all curves having normal of constant length c-

A) $1$

B) $3$

C) $4$

D) $2$

Answer

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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.

**Complete step-by-step answer:**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

So on putting the value of normal length, we get-

$ \Rightarrow {\text{c = }}y\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} $ --- (i)

We have to find degree of this differential equation so first we will square both side to remove the square-root,

$ \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 $2$

**Hence the answer is ‘D’.**

**Note:**Here the student may go wrong if they try to find the degree of differential equation in eq. (i) as the derivative is also under the square-root. So first we have to solve the eq. (i) and remove the square-root, only then can we easily find the degree.

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