Which of the following is true for y(x) that | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = xy - 1 + x - y$.Rearranging the terms: $\frac{dy}{dx} = x(y + 1) - 1(y + 1) = (x - 1)(y + 1)$.Th

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$y(1) = e^{-\frac{1}{2}} - 1$

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(A) Given the differential equation: $\frac{dy}{dx} = xy - 1 + x - y$.Rearranging the terms: $\frac{dy}{dx} = x(y + 1) - 1(y + 1) = (x - 1)(y + 1)$.This is a variable separable differential equation: $\frac{dy}{y + 1} = (x - 1) dx$.Integrating both sides: $\int \frac{dy}{y + 1} = \int (x - 1) dx$.$\ln|y + 1| = \frac{x^2}{2} - x + C$.Using the initial condition $y(0) = 0$: $\ln|0 + 1| = \frac{0^2}{2} - 0 + C \Rightarrow \ln(1) = C \Rightarrow C = 0$.Thus,$\ln|y + 1| = \frac{x^2}{2} - x$,which implies $y + 1 = e^{\frac{x^2}{2} - x}$.So,$y(x) = e^{\frac{x^2}{2} - x} - 1$.To find $y(1)$,substitute $x = 1$: $y(1) = e^{\frac{1^2}{2} - 1} - 1 = e^{\frac{1}{2} - 1} - 1 = e^{-\frac{1}{2}} - 1$.

Which of the following is true for $y(x)$ that satisfies the differential equation $\frac{dy}{dx} = xy - 1 + x - y$ with the initial condition $y(0) = 0$?

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