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  • Help me solve this - Differential equations - JEE Main-2

The differential equation whose solution is Ax^{2}+By^{2}=1,\; where\; A\; and\; B are arbitrary constants is of

  • Option 1)

    second order and second degree

  • Option 2)

    first order and second degree

  • Option 3)

    first order and first degree

  • Option 4)

    second order and first degree

 

Answers (1)

best_answer

As we learnt in 

Formation of Differential Equations -

A differential equation can be derived from its equation by the process of differentiation and other algebraical process of elimination

-

 

 Ax^{2}+ By^{2}= 1

\Rightarrow 2Ax+2By\frac{dy}{dx}= 0

\Rightarrow Ax+By\frac{dy}{dx}= 0

\Rightarrow A.1+B\left[\frac{dy}{dx}.\frac{dy}{dx}+y.\frac{d^2 y}{dx^2} \right ] = 0

\Rightarrow A+B\left[ \left(\frac{dy}{dx} \right )^{2}+y\frac{d^{2}y}{dx^{2}}\right ]=0

By eliminating A and B, 

xyy''+x(y')^{2}- yy'=0

 

 


Option 1)

second order and second degree

Incorrect option

Option 2)

first order and second degree

Incorrect option

Option 3)

first order and first degree

Incorrect option

Option 4)

second order and first degree

Correct option

Posted by

Aadil

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