Answer

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Hint: In this question we have to solve the given differential equation, converting this into standard differential form of $\dfrac{{dy}}{{dx}} + py = q$ and then using the concept of $y\left( {I.F} \right) = \int {\left( {I.F} \right)qdx} $ Where I.F is $ = {e^{\int {pdx} }}$ will help in getting the solution of this differential equation. Then simple substitution of 1 in place of x will help in getting the value of y(1).

Complete step-by-step answer:

The given differential equation is

$\left( {1 + t} \right)\dfrac{{dy}}{{dt}} - ty = 1$

Divide by (1 + t) in above equation we have,

$\dfrac{{dy}}{{dt}} - \dfrac{t}{{\left( {1 + t} \right)}}y = \dfrac{1}{{\left( {1 + t} \right)}}$

Now the general differential equation is $\dfrac{{dy}}{{dx}} + py = q$, the solution of this differential equation is,

$y\left( {I.F} \right) = \int {\left( {I.F} \right)qdx} $ Where I.F is$ = {e^{\int {pdx} }}$.

Now in above equation $p = \dfrac{{ - t}}{{1 + t}}$ and $q = \dfrac{1}{{1 + t}}$.

So, first calculate integrating factor (I.F)

$ \Rightarrow I.F = {e^{\int {\dfrac{{ - t}}{{1 + t}}dx} }}$

$ \Rightarrow I.F = {e^{\int {\left( {\dfrac{1}{{1 + t}} - 1} \right)dx} }}$

Now integrate it as we know $\dfrac{1}{{1 + t}}$ integration is log (1 + t) so, apply this,

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

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

Now as we know ${e^{\log a}} = a$ so, use this property we have,

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

So, the solution of the differential equation is

\[y\left( {I.F} \right) = \int {\left( {I.F} \right)qdt} \]

$

\Rightarrow y\left( {1 + t} \right){e^{ - t}} = \int {\left( {1 + t} \right){e^{ - t}}\dfrac{1}{{1 + t}}dt} \\

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

$

Now as we know integration of $\int {{e^{ - t}}} = \dfrac{{{e^{ - t}}}}{{ - 1}} + c$ so use this in above equation we have

\[ \Rightarrow y\left( {1 + t} \right){e^{ - t}} = \dfrac{{{e^{ - t}}}}{{ - 1}} + c\] (Where c is some arbitrary integration constant)

\[ \Rightarrow y\left( {1 + t} \right){e^{ - t}} = - {e^{ - t}} + c\]……………… (1)

Now it is given that $y\left( 0 \right) = - 1$

So substitute t = 0 and y = -1 in above equation we have,

$

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

\Rightarrow - 1 = - 1 + c \\

\Rightarrow c = 0 \\

$

Therefore from equation (1) we have

\[ \Rightarrow y\left( {1 + t} \right){e^{ - t}} = - {e^{ - t}} + 0\]

\[

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

\Rightarrow y\left( {1 + t} \right) = - 1 \\

\Rightarrow y\left( t \right) = \dfrac{{ - 1}}{{1 + t}} \\

\]

Now we have to find out the value of y(1)

So put t = 1 in above equation we have,

\[ \Rightarrow y\left( 1 \right) = \dfrac{{ - 1}}{{1 + 1}} = \dfrac{{ - 1}}{2}\]

Hence option (a) is correct.

Note: Whenever we face such types of problems the key concept is to convert the differential equation into standard form in order to evaluate the integrating factor. The integration factor will help us obtain the solution required for the given differential equation.

Complete step-by-step answer:

The given differential equation is

$\left( {1 + t} \right)\dfrac{{dy}}{{dt}} - ty = 1$

Divide by (1 + t) in above equation we have,

$\dfrac{{dy}}{{dt}} - \dfrac{t}{{\left( {1 + t} \right)}}y = \dfrac{1}{{\left( {1 + t} \right)}}$

Now the general differential equation is $\dfrac{{dy}}{{dx}} + py = q$, the solution of this differential equation is,

$y\left( {I.F} \right) = \int {\left( {I.F} \right)qdx} $ Where I.F is$ = {e^{\int {pdx} }}$.

Now in above equation $p = \dfrac{{ - t}}{{1 + t}}$ and $q = \dfrac{1}{{1 + t}}$.

So, first calculate integrating factor (I.F)

$ \Rightarrow I.F = {e^{\int {\dfrac{{ - t}}{{1 + t}}dx} }}$

$ \Rightarrow I.F = {e^{\int {\left( {\dfrac{1}{{1 + t}} - 1} \right)dx} }}$

Now integrate it as we know $\dfrac{1}{{1 + t}}$ integration is log (1 + t) so, apply this,

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

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

Now as we know ${e^{\log a}} = a$ so, use this property we have,

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

So, the solution of the differential equation is

\[y\left( {I.F} \right) = \int {\left( {I.F} \right)qdt} \]

$

\Rightarrow y\left( {1 + t} \right){e^{ - t}} = \int {\left( {1 + t} \right){e^{ - t}}\dfrac{1}{{1 + t}}dt} \\

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

$

Now as we know integration of $\int {{e^{ - t}}} = \dfrac{{{e^{ - t}}}}{{ - 1}} + c$ so use this in above equation we have

\[ \Rightarrow y\left( {1 + t} \right){e^{ - t}} = \dfrac{{{e^{ - t}}}}{{ - 1}} + c\] (Where c is some arbitrary integration constant)

\[ \Rightarrow y\left( {1 + t} \right){e^{ - t}} = - {e^{ - t}} + c\]……………… (1)

Now it is given that $y\left( 0 \right) = - 1$

So substitute t = 0 and y = -1 in above equation we have,

$

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

\Rightarrow - 1 = - 1 + c \\

\Rightarrow c = 0 \\

$

Therefore from equation (1) we have

\[ \Rightarrow y\left( {1 + t} \right){e^{ - t}} = - {e^{ - t}} + 0\]

\[

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

\Rightarrow y\left( {1 + t} \right) = - 1 \\

\Rightarrow y\left( t \right) = \dfrac{{ - 1}}{{1 + t}} \\

\]

Now we have to find out the value of y(1)

So put t = 1 in above equation we have,

\[ \Rightarrow y\left( 1 \right) = \dfrac{{ - 1}}{{1 + 1}} = \dfrac{{ - 1}}{2}\]

Hence option (a) is correct.

Note: Whenever we face such types of problems the key concept is to convert the differential equation into standard form in order to evaluate the integrating factor. The integration factor will help us obtain the solution required for the given differential equation.

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