If y(t) is a solution of (1 + t)frac{dy}{dt} | vedclass

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(A) The given differential equation is $(1 + t)\frac{dy}{dt} - ty = 1$. Dividing by $(1 + t)$,we get $\frac{dy}{dt} - \frac{t}{1 + t}y = \frac{1}{1 +

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$-\frac{1}{2}$

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(A) The given differential equation is $(1 + t)\frac{dy}{dt} - ty = 1$. Dividing by $(1 + t)$,we get $\frac{dy}{dt} - \frac{t}{1 + t}y = \frac{1}{1 + t}$. This is a linear differential equation of the form $\frac{dy}{dt} + P(t)y = Q(t)$,where $P(t) = -\frac{t}{1 + t}$ and $Q(t) = \frac{1}{1 + t}$. The integrating factor ($I$.$F$.) is $e^{\int P(t) dt} = e^{\int -\frac{t}{1 + t} dt} = e^{\int (-1 + \frac{1}{1 + t}) dt} = e^{-t + \ln(1 + t)} = (1 + t)e^{-t}$. The general solution is $y \cdot (I.F.) = \int Q(t) \cdot (I.F.) dt + C$. $y(1 + t)e^{-t} = \int \frac{1}{1 + t} \cdot (1 + t)e^{-t} dt + C = \int e^{-t} dt + C = -e^{-t} + C$. Given $y(0) = -1$,we substitute $t = 0$: $-1(1 + 0)e^{0} = -e^{0} + C \Rightarrow -1 = -1 + C \Rightarrow C = 0$. Thus,$y(1 + t)e^{-t} = -e^{-t}$,which simplifies to $y = -\frac{1}{1 + t}$. For $t = 1$,$y(1) = -\frac{1}{1 + 1} = -\frac{1}{2}$.

If $y(t)$ is a solution of $(1 + t)\frac{dy}{dt} - ty = 1$ and $y(0) = -1$,then $y(1)$ is equal to

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