If y(t) is asolution of $(1+t)\frac{dy}{dt}-ty=1$ and $y(0)=1, \ then \: y(1) \: is \: equal \: to$ Option 1) $-\frac{1}{2}$ Option 2) $e+\frac{1}{2}$ Option 3) $e-\frac{1}{2}$ Option 4) $\frac{1}{2}$

Using

Linear Differential Equation -

$\frac{dy}{dx}+Py= Q$

- wherein

P, Q are functions of x alone.

$\frac{dy}{dt} - \frac{t}{t+1}y= \frac{1}{t+1}$

$\Rightarrow P= \frac{-t}{t+1} = -\left ( \frac{t+1-1}{t+1} \right ) =\frac{1}{t+1}-1$

$\Rightarrow \int P dt=\log \left ( t+1 \right )-t$

$\Rightarrow I.f = e^{\int Pdt} = e^{\log \left ( t+1 \right )-t}=e^{\log \left ( t+1 \right )}e^{-t}$

$=\left ( t+1 \right ).e^{-t}$

Solution is:

$y.\left ( t+1 \right ).e^{-t}=\int \frac{\left ( t+1 \right )e^{-t}}{t+1}dt = \int e^{-t}dt=-e^{-t}+c$

Put t = 0, y = 1

$1\left ( 0+1 \right )e^{-0} = -e^{-0}+c = -1+c$

$1=-1+c \ \ \ \therefore c = 2$

$\Rightarrow y\left ( t+1 \right )e^{-t} =-e^{-t}+2$

Put t = 1

$2ye^{-1} = -e^{-1}+2 \ \ \ \therefore y = e-\frac{1}{2}$

Option 1)

$-\frac{1}{2}$

This option is incorrect

Option 2)

$e+\frac{1}{2}$

This option is incorrect

Option 3)

$e-\frac{1}{2}$

This option is correct

Option 4)

$\frac{1}{2}$

This option is incorrect

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