If (x^2-3x+2) e^{frac{y}{x-1}}=x+2, then find | vedclass

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(B) Given equation: $(x^2-3x+2) e^{\frac{y}{x-1}} = x+2$. Factorizing the quadratic term: $(x-1)(x-2) e^{\frac{y}{x-1}} = x+2$. Taking natural logarit

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$-2$

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(B) Given equation: $(x^2-3x+2) e^{\frac{y}{x-1}} = x+2$. Factorizing the quadratic term: $(x-1)(x-2) e^{\frac{y}{x-1}} = x+2$. Taking natural logarithm on both sides: $\ln((x-1)(x-2)) + \frac{y}{x-1} = \ln(x+2)$. Rearranging for $y$: $y = (x-1) [\ln(x+2) - \ln((x-1)(x-2))]$. At $x=0$: $y = (0-1) [\ln(2) - \ln((-1)(-2))] = -1 [\ln(2) - \ln(2)] = 0$. Now, differentiate the original equation with respect to $x$: $(2x-3) e^{\frac{y}{x-1}} + (x^2-3x+2) e^{\frac{y}{x-1}} \cdot \frac{d}{dx}(\frac{y}{x-1}) = 1$. Using quotient rule for $\frac{d}{dx}(\frac{y}{x-1}) = \frac{(x-1)y' - y}{(x-1)^2}$. Substitute $x=0$ and $y=0$: $(0-3) e^{\frac{0}{-1}} + (0-0+2) e^{\frac{0}{-1}} \cdot \frac{(-1)y' - 0}{(-1)^2} = 1$. $-3(1) + 2(1) (-y') = 1$. $-3 - 2y' = 1$. $-2y' = 4$. $y' = -2$. Thus, $(\frac{dy}{dx})_{x=0} = -2$.

If $(x^2-3x+2) e^{\frac{y}{x-1}}=x+2$, then find the value of $(\frac{dy}{dx})_{x=0}$.

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