If (x - y)e^{x/(x - y)} = k,then: | vedclass

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Question

(B) Given the equation: $(x - y)e^{x/(x - y)} = k$. Taking the natural logarithm on both sides: $\ln(x - y) + \frac{x}{x - y} = \ln k$. Differentiatin

Vedclass · Accepted Answer

$y\frac{dy}{dx} + x - 2y = 0$

Vedclass · Answer

(B) Given the equation: $(x - y)e^{x/(x - y)} = k$. Taking the natural logarithm on both sides: $\ln(x - y) + \frac{x}{x - y} = \ln k$. Differentiating both sides with respect to $x$: $\frac{1}{x - y}(1 - \frac{dy}{dx}) + \frac{(x - y)(1) - x(1 - \frac{dy}{dx})}{(x - y)^2} = 0$. Multiply the entire equation by $(x - y)^2$: $(x - y)(1 - \frac{dy}{dx}) + (x - y) - x(1 - \frac{dy}{dx}) = 0$. $(x - y) - (x - y)\frac{dy}{dx} + x - y - x + x\frac{dy}{dx} = 0$. Combine the terms involving $\frac{dy}{dx}$: $(-x + y + x)\frac{dy}{dx} + (x - y + x - y - x) = 0$. $y\frac{dy}{dx} + x - 2y = 0$.

If $(x - y)e^{x/(x - y)} = k$,then:

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