The solution of the differential equation (1+ | vedclass

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(A) Given the differential equation: $(1+x) \frac{dy}{dx} - xy = 1-x$. Divide by $(1+x)$ to write it in the standard form $\frac{dy}{dx} + P(x)y = Q(x

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$y(1+x) = x + ce^x$,where $c$ is the constant of integration

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(A) Given the differential equation: $(1+x) \frac{dy}{dx} - xy = 1-x$. Divide by $(1+x)$ to write it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$: $\frac{dy}{dx} - \frac{x}{1+x}y = \frac{1-x}{1+x}$. Here,$P(x) = -\frac{x}{1+x} = -\frac{x+1-1}{1+x} = -1 + \frac{1}{1+x}$ and $Q(x) = \frac{1-x}{1+x} = \frac{2-(1+x)}{1+x} = \frac{2}{1+x} - 1$. The integrating factor $IF = e^{\int P(x) dx} = e^{\int (-1 + \frac{1}{1+x}) dx} = e^{-x + \ln(1+x)} = (1+x)e^{-x}$. The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$. $y(1+x)e^{-x} = \int (\frac{1-x}{1+x}) (1+x)e^{-x} dx + c = \int (1-x)e^{-x} dx + c$. Using integration by parts: $\int (1-x)e^{-x} dx = (1-x)(-e^{-x}) - \int (-1)(-e^{-x}) dx = -(1-x)e^{-x} - \int e^{-x} dx = (x-1)e^{-x} + e^{-x} + c = xe^{-x} + c$. So,$y(1+x)e^{-x} = xe^{-x} + c$. Multiplying by $e^x$,we get $y(1+x) = x + ce^x$.

The solution of the differential equation $(1+x) \frac{dy}{dx} - xy = 1-x$ is

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