If x^m + y^m = k where (m neq 1) and frac{d^2 | vedclass

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(C) Given $x^m + y^m = k$. Differentiating with respect to $x$:$mx^{m-1} + my^{m-1} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x^{m-1}}{y^{m-1}

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$3$

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(C) Given $x^m + y^m = k$. Differentiating with respect to $x$:$mx^{m-1} + my^{m-1} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x^{m-1}}{y^{m-1}}$.Differentiating again with respect to $x$ using the quotient rule:$\frac{d^2y}{dx^2} = -\left[ \frac{(m-1)x^{m-2}y^{m-1} - x^{m-1}(m-1)y^{m-2} \frac{dy}{dx}}{(y^{m-1})^2} ight]$.Substitute $\frac{dy}{dx} = -\frac{x^{m-1}}{y^{m-1}}$:$\frac{d^2y}{dx^2} = -\left[ \frac{(m-1)x^{m-2}y^{m-1} - x^{m-1}(m-1)y^{m-2} (-\frac{x^{m-1}}{y^{m-1}})}{y^{2m-2}} ight] = -\frac{(m-1)x^{m-2}y^{2m-2} + (m-1)x^{2m-2}y^{m-2}}{y^{3m-3}}$.This simplifies to $\frac{d^2y}{dx^2} = -\frac{(m-1)x^{m-2}y^{m-2}(y^m + x^m)}{y^{3m-3}} = -\frac{(m-1)x^{m-2}k}{y^{2m-1}}$.Comparing with $\frac{ax^b}{y^c}$, we have $a = -k(m-1)$, $b = m-2$, $c = 2m-1$.Given $a + b + c = 0$, then $-k(m-1) + (m-2) + (2m-1) = 0$.$-k(m-1) + 3m - 3 = 0 \implies -k(m-1) + 3(m-1) = 0$.Since $m 
eq 1$, we divide by $(m-1)$ to get $-k + 3 = 0$, so $k = 3$.

If $x^m + y^m = k$ where $(m \neq 1)$ and $\frac{d^2y}{dx^2} = \frac{ax^b}{y^c}$ such that $a + b + c = 0$, then the value of $k$ is...

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