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  • Please,please help me - Differential equations - JEE Main-2

If y=ae^{x}+be^{-x} where a and b are arbitrary constants then corresponding differential equation will be

  • Option 1)

    \left ( \frac{d^{2}y}{dx^{2}} \right )=0

  • Option 2)

    \left ( \frac{d^{2}y}{dx^{2}} \right )=y

  • Option 3)

    \left ( \frac{d^{2}y}{dx^{2}} \right )+y=0

  • Option 4)

    \left ( \frac{d^{2}y}{dx^{2}} \right )=x

 

Answers (1)

best_answer

As we learnt

 

Formation of Differential Equations -

 

Let y and x be the dependent and the independent variables respectively. The equation of (x, y, c) = 0 , is family of curves

- wherein

C is the arbitrary Constant

 

 \frac{dy}{dx}= ae^{x}-be^{-x}

Again differentiating we get \frac{d^{2}y}{dx^{2}}= ae^{x}+be^{-x}= y

 


Option 1)

\left ( \frac{d^{2}y}{dx^{2}} \right )=0

Option 2)

\left ( \frac{d^{2}y}{dx^{2}} \right )=y

Option 3)

\left ( \frac{d^{2}y}{dx^{2}} \right )+y=0

Option 4)

\left ( \frac{d^{2}y}{dx^{2}} \right )=x

Posted by

Himanshu

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