The solution curve of the differential equati | vedclass

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(A) Given differential equation: $(2 y - 5) \frac{dy}{dx} = -3$.Separating the variables,we get: $(2 y - 5) dy = -3 dx$.Integrating both sides: $\int

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$2 x + 3 y = 9$

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(A) Given differential equation: $(2 y - 5) \frac{dy}{dx} = -3$.Separating the variables,we get: $(2 y - 5) dy = -3 dx$.Integrating both sides: $\int (2 y - 5) dy = \int -3 dx$.$y^2 - 5 y = -3 x + C$.Since the curve passes through $(0, 1)$,substitute $x = 0$ and $y = 1$: $(1)^2 - 5(1) = -3(0) + C \Rightarrow C = -4$.So,the equation of the curve is $y^2 - 5 y + 3 x + 4 = 0$.Rearranging to standard form: $y^2 - 5 y = -3 x - 4$.Completing the square: $(y - \frac{5}{2})^2 = -3 x - 4 + \frac{25}{4} = -3 x + \frac{9}{4} = -3(x - \frac{3}{4})$.The vertex of the parabola is $(\frac{3}{4}, \frac{5}{2})$.Checking the options for the line $2 x + 3 y = k$: $2(\frac{3}{4}) + 3(\frac{5}{2}) = \frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9$.Thus,the vertex lies on the line $2 x + 3 y = 9$.

The solution curve of the differential equation $2 y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx}$,passing through the point $(0, 1)$,is a conic whose vertex lies on the line:

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