Get Answers to all your Questions

header-bg qa

The differential equation representing the family of curves y^{2}=2c(x+\sqrt{c}),\; where \; c> 0, is a parameter, is of order and degree as follows

  • Option 1)

    order 1, degree 1

  • Option 2)

    order 1, degree 2

  • Option 3)

    order 2, degree 2

  • Option 4)

    order 1, degree 3

 

Answers (1)

As we learnt in 

Degree of a Differential Equation -

Degree of Highest order differential coefficient appearing in it, provided it can be expressed as a polynomial equation in derivatives

- wherein

\left ( \frac{dy}{dx} \right )^{2}+3\left ( \frac{dy}{dx} \right )-5=0

Degree = 2

 

Degree of a Differential Equation -

Degree of Highest order differential coefficient appearing in it, provided it can be expressed as a polynomial equation in derivatives

- wherein

\left ( \frac{dy}{dx} \right )^{2}+3\left ( \frac{dy}{dx} \right )-5=0

Degree = 2

 

 y^{2}= 2c(x+\sqrt{c})

y^{2}= 2cx+2c\sqrt{c})

2y.\frac{dy}{dx}=2c

\Rightarrow y.\frac{dy}{dx}=c

\Rightarrow y^{2}= 2 y\frac{dy}{dx}.x+ 2y(\frac{dy}{dx})^{\frac{3}{2}}

\Rightarrow \left(y^{2}- 2x y\frac{dy}{dx} \right )= 2y\left(\frac{dy}{dx}) \right )^{\frac{3}{2}}

\Rightarrow \left(y^{2}- 2x y\frac{dy}{dx} \right )^{2}= (2y)^{2}\left(\frac{dy}{dx} \right )^{3}

order 1, degree 3.

 


Option 1)

order 1, degree 1

Incorrect option

Option 2)

order 1, degree 2

Incorrect option

Option 3)

order 2, degree 2

Incorrect option

Option 4)

order 1, degree 3

Correct option

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE