# Solve differential equation ty’+(t+1)y=t , y(ln2)=1 , t>0

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: $=x$
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