The differential equation formed by eliminati | vedclass

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Question

(A) Given equation: $A x^2 + B y^2 = 1$ $(1)$Differentiating with respect to $x$: $2 A x + 2 B y \frac{d y}{d x} = 0 \implies A x + B y y' = 0$ $(2)$D

Vedclass · Accepted Answer

$x y \cdot \frac{d^2 y}{d x^2} + x \left(\frac{d y}{d x}\right)^2 = y \frac{d y}{d x}$

Vedclass · Answer

(A) Given equation: $A x^2 + B y^2 = 1$ $(1)$Differentiating with respect to $x$: $2 A x + 2 B y \frac{d y}{d x} = 0 \implies A x + B y y' = 0$ $(2)$Differentiating again with respect to $x$: $A + B (y')^2 + B y y'' = 0$ $(3)$From $(2)$,$A = -B \frac{y y'}{x}$. Substituting into $(3)$:$-B \frac{y y'}{x} + B (y')^2 + B y y'' = 0$Dividing by $B$ (assuming $B 
eq 0$):$-\frac{y y'}{x} + (y')^2 + y y'' = 0$Multiplying by $x$: $-y y' + x (y')^2 + x y y'' = 0$Rearranging: $x y y'' + x (y')^2 = y y'$Thus,the differential equation is $x y \frac{d^2 y}{d x^2} + x \left(\frac{d y}{d x}ight)^2 = y \frac{d y}{d x}$.

The differential equation formed by eliminating $A$ and $B$ from $A x^2 + B y^2 = 1$ is

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