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Question

(C) Given the differential equation $\frac{dy}{dx} = \frac{y - x}{y - x - 1}$. Let $y - x = t$,then $\frac{dy}{dx} - 1 = \frac{dt}{dx}$,so $\frac{dy}{

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a parabola

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(C) Given the differential equation $\frac{dy}{dx} = \frac{y - x}{y - x - 1}$. Let $y - x = t$,then $\frac{dy}{dx} - 1 = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} + 1$. Substituting this into the equation: $\frac{dt}{dx} + 1 = \frac{t}{t - 1}$. $\frac{dt}{dx} = \frac{t}{t - 1} - 1 = \frac{t - (t - 1)}{t - 1} = \frac{1}{t - 1}$. Separating variables: $(t - 1) dt = dx$. Integrating both sides: $\int (t - 1) dt = \int dx \Rightarrow \frac{t^2}{2} - t = x + C$. Substituting $t = y - x$: $\frac{(y - x)^2}{2} - (y - x) = x + C$. $(y - x)^2 - 2(y - x) = 2x + 2C \Rightarrow (y - x)^2 - 2y + 2x = 2x + 2C \Rightarrow (y - x)^2 - 2y = K$. Given $y(-5) = -5$,so $(-5 - (-5))^2 - 2(-5) = K \Rightarrow 0 + 10 = K \Rightarrow K = 10$. The equation is $(y - x)^2 - 2y = 10$,which is $y^2 - 2xy + x^2 - 2y - 10 = 0$. The discriminant $h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$. Since the discriminant is $0$,the equation represents a parabola.

The solution of the differential equation $\frac{dy}{dx} = \frac{y - x}{y - x - 1}$,given $y(-5) = -5$,represents:

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