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  • I have a doubt, kindly clarify. - Ordinary Differential Equations - BITSAT

The degree of the differential equation corresponding to the family of curves y=a(x+a)^{2},where a is an arbitrary constant is

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    None of these

 

Answers (1)

best_answer

As we learnt

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 = a(x+a)^{2}

     = a(x^{2}+a^{2}+2ax)

 y = ax^{2}+a^{3}+2a^{2}x      

\therefore {y}' = 2ax+2a^{2}

   \therefore {y}'^{1} = 2a \therefore {y}'^{1} = 2a \therefore a= \frac{{y}'^{1}}{2}

Put a = {y}'^{1} in (i) = 2a \therefore a^{3}= \left | y'^{1} \right |^{3}

\therefore degree = 3

 

 


Option 1)

1

Incorrect

Option 2)

2

Incorrect

Option 3)

3

correct

Option 4)

None of these

Incorrect

Posted by

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