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

Question

Dr. Abubacker · Accepted Answer

(c)
The given equation is $Ax^{2} + By^{2} = 1$
$\Rightarrow$ $2Ax + 2By\ \frac{dy}{dx} = 0$ …(i)
$\Rightarrow$ $2A + 2B\left\{ \left( \frac{dy}{dx} \right)^{2} + y\ \frac{d^{2}y}{dx^{2}} \right\} = 0$ …(ii)
Eliminating A and B from Eqs. (i) and (ii), we get
$y\ \frac{d^{2}y}{dx^{2}} + \left( \frac{dy}{dx} \right)^{2} - \frac{y}{x}.\ \frac{dy}{dx} = 0$
Here, order =2, degree =1

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

Explanation

Reviewed by Dr. Abubacker

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