In this question, we have to find the** Integration** of the given function $ t y^\prime + ( t + 1) y = t $ by using different **integration rules**.

The basic concept behind this question is the knowledge of **derivatives, integration,** and the **rules** such as the **product **and** quotient integration rules**.

## Expert Answer

Given function we have:

\[ t y^\prime + ( t + 1) y = t \]

First, divide $t$ into both sides of the equation and then we will get:

\[ \dfrac { 1}{ t} \times t y^\prime + \dfrac { 1}{ t} \times ( t + 1) y = \dfrac { 1}{ t} \times t \]

Canceling $t $ in the **numerator** with the **denominator** we get:

\[ y^\prime +\dfrac { ( t + 1) }{ t} y = 1 \]

We know that here $y^\prime = \dfrac { dy }{ dx }$, putting in the equation:

\[ \dfrac { dy }{ dx } +\dfrac { ( t + 1) }{ t} y = 1 \]

We also know that:

\[$p(t) = \dfrac { ( t + 1) }{ t} \space ; \space q (t) = 1$\]

Putting these in our equation, we will have:

\[ \dfrac { dy }{ dx } + p(t) y = q (t) \]

Now let us suppose:

\[ u (t) = e^{\int p(t) dt}\]

After putting the value of $p(t) $ here then we will have:

\[ u (t) = e^{\int \dfrac { ( t + 1) }{ t} dt}\]

**Integrating** the **power** of $e$:

\[ u (t) = e^{\int \dfrac { t }{ t } dt + \dfrac { 1}{ t} dt }\]

\[ u (t) = e^{ t + \ln(t) }\]

Now we will simplify the **exponential equation** as follows:

\[ u (t) =te^t\]

From the **second law of logarithm**:

\[ u (t) = e^{ ln t e^t}\]

Take **log** on both sides of the equation:

\[ln u(t)= ln e^{ ln t e^t}\]

\[ln u(t)= ln t e^{t}\]

\[u(t)= t e^{t}\]

We know that:

\[ y(x) = \dfrac{\int u(t) q(t ) dt}{ u(t) } \]

\[ y(x) = \dfrac{\int (t e^{t }) (1) dt}{t e^{t }} \]

\[ y(x) = \dfrac{\int t e^{t } dt}{t e^{t}} \]

Using **integration by parts**:

\[ \int t e^{t} dt = te^t – e^t + c\]

\[ y(x) = \dfrac{ te^t -e^t+c}{t e^{t}} \]

\[ y(x) = \dfrac{ te^t }{t e^{t}} – \dfrac{e^t}{t e^{t}} +\dfrac{c}{t e^{t}} \]

\[ y(x) = 1- \dfrac{1}{t}+ \dfrac{c}{t e^{t}} \]

Putting the **initial condition**:

\[1=1-\dfrac{1}{\ln2}+ \dfrac{c}{\ln2 e^{t}} \]

\[ \dfrac{1}{\ln2}= \dfrac{c}{\ln2 e^{t}} \]

\[ \dfrac{\ln2 e^{t}}{\ln2}= \dfrac{c}{1} \]

\[ e^{\ln 2} =c\]

\[ c = 2\]

Substituting the value of $c$ in the equation:

\[ y(x) = 1- \dfrac{1}{t}+ \dfrac{c}{t e^{t}} \]

\[ y(x) = 1- \dfrac{1}{t}+ \dfrac{2}{t e^{t}} \]

## Numerical Result

\[ y(x) = 1- \dfrac{1}{t}+ \dfrac{2}{t e^{t}}\]

## Example

**Integrate** the following function:

\[\int \dfrac{1}{x} dx\]

**Solution:**

\[= \ln{\left|x \right|}\]

\[=e^{\ln{x}}\]

We know that $ e^{\ln{x}} = x $ so we have the above** equation** as:

\[=x\]