The differential equation whose solution is A | vedclass

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(D) Given equation is $Ax^2 + By^2 = 1 \quad \dots(1)$Differentiating with respect to $x$:$2Ax + 2By \frac{dy}{dx} = 0 \implies Ax + By \frac{dy}{dx}

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second order and first degree

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(D) Given equation is $Ax^2 + By^2 = 1 \quad \dots(1)$Differentiating with respect to $x$:$2Ax + 2By \frac{dy}{dx} = 0 \implies Ax + By \frac{dy}{dx} = 0 \quad \dots(2)$Differentiating again with respect to $x$:$A + B \left( y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} ight)^2 ight) = 0 \quad \dots(3)$From $(2)$,$A = -\frac{By}{x} \frac{dy}{dx}$. Substituting this into $(3)$:$-\frac{By}{x} \frac{dy}{dx} + By \frac{d^2y}{dx^2} + B \left( \frac{dy}{dx} ight)^2 = 0$Dividing by $B$ (assuming $B 
eq 0$):$-\frac{y}{x} \frac{dy}{dx} + y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} ight)^2 = 0$Multiplying by $x$:$xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} ight)^2 - y \frac{dy}{dx} = 0$The highest order derivative is $\frac{d^2y}{dx^2}$,so the order is $2$. The power of the highest order derivative is $1$,so the degree is $1$.

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

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