The differential equation whose solution is Ax² + By² = 1 where A and B are arbitrary constants is of

Correct Answer is : second order and first degree

Correct Answer is : second order and first degree

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BITSAT 2012 Paper

Section: Mathematics

Question : 112 of 150

Marks: +1, -0

Ax^2 + By^2

= 1 where A and B are arbitrary constants is of

Solution:

Ax^2+By^2

= 1 ... (1)

Ax + By

\frac{dy}{dx}

= 0 ... (2)

A + By

\frac{d^2y}{dx^2}

+ B

\left( \frac{dy}{dx} \right)^2

= 0 ... (3)

From (2) and (3)

\{ -By\frac{d^2y}{dx^2} - B\left(\frac{dy}{dx}\right)^2 \}

+ By

\frac{dy}{dx}

= 0

Dividing both sides by –B, we get

\frac{d^2y}{dx^2}

+ x

\left( \frac{dy}{dx} \right)^2

- y

\frac{dy}{dx}

= 0

Which is a DE of order 2 and degree 1

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