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  • Confused! kindly explain, - Differential equations - JEE Main-3

If y=ae^{x}+be^{2x}where a and b are arbitrary constants then corresponding differential equation will be

  • Option 1)

    \frac{d^{2}x}{dx^{2}}+3\frac{dy}{dx}-2y=0

  • Option 2)

    \frac{d^{2}x}{dx^{2}}-3\frac{dy}{dx}-2y=0

  • Option 3)

    \frac{d^{2}x}{dx^{2}}-3\frac{dy}{dx}+2y=0

  • Option 4)

    \frac{d^{2}x}{dx^{2}}+3\frac{dy}{dx}+2y=0

 

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

 

 Differentiating both the sides we have

\frac{dy}{dx}= ae^{x}+2be^{2x}= \left ( ae^{x}+be^{2x} \right )+be^{2x}= y+be^{2x}......\left ( 2 \right )

Again differentiating we get,

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

\Rightarrow \frac{d^{2}y}{dx^{2}}=3\frac{dy}{dx}-2y

 


Option 1)

\frac{d^{2}x}{dx^{2}}+3\frac{dy}{dx}-2y=0

Option 2)

\frac{d^{2}x}{dx^{2}}-3\frac{dy}{dx}-2y=0

Option 3)

\frac{d^{2}x}{dx^{2}}-3\frac{dy}{dx}+2y=0

Option 4)

\frac{d^{2}x}{dx^{2}}+3\frac{dy}{dx}+2y=0

Posted by

Himanshu

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