If X = x + h, Y = y + k transforms frac{dy}{d | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{2x + 3y - 7}{3x + 2y - 8}$ ... $(i)$ To transform this into a homogeneous differential equ

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$(2, 1)$

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{2x + 3y - 7}{3x + 2y - 8}$ ... $(i)$ To transform this into a homogeneous differential equation,we substitute $x = X - h$ and $y = Y - k$. Substituting these into $(i)$,we get: $\frac{dY}{dX} = \frac{2(X - h) + 3(Y - k) - 7}{3(X - h) + 2(Y - k) - 8} = \frac{2X + 3Y - (2h + 3k + 7)}{3X + 2Y - (3h + 2k + 8)}$. For the equation to be homogeneous,the constant terms in the numerator and denominator must be zero: $2h + 3k + 7 = 0$ ... (ii) $3h + 2k + 8 = 0$ ... (iii) Multiplying (ii) by $3$ and (iii) by $2$: $6h + 9k + 21 = 0$ $6h + 4k + 16 = 0$ Subtracting the equations: $(9k - 4k) + (21 - 16) = 0 \Rightarrow 5k + 5 = 0 \Rightarrow k = -1$. Wait,re-evaluating the signs: The original equation is $\frac{2x+3y-7}{3x+2y-8}$. Setting $2h+3k-7=0$ and $3h+2k-8=0$: $4h+6k-14=0$ $9h+6k-24=0$ Subtracting: $5h - 10 = 0 \Rightarrow h = 2$. Substituting $h=2$ into $2(2)+3k-7=0 \Rightarrow 4+3k-7=0 \Rightarrow 3k=3 \Rightarrow k=1$. Thus,$(h, k) = (2, 1)$.

If $X = x + h, Y = y + k$ transforms $\frac{dy}{dx} = \frac{2x + 3y - 7}{3x + 2y - 8}$ to a homogeneous differential equation,then $(h, k) =$

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