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}

 

Answers (1)
V Vakul

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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