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:

$${\displaystyle \frac{1}{t}}\times t{y}^{\prime }+{\displaystyle \frac{1}{t}}\times (t+1)y={\displaystyle \frac{1}{t}}\times t$$

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

$$y^{\prime }+{\displaystyle \frac{(t+1)}{t}}y=1$$

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

$${\displaystyle \frac{dy}{dx}}+{\displaystyle \frac{(t+1)}{t}}y=1$$

We also know that:

$$\$p(t)={\displaystyle \frac{(t+1)}{t}};q(t)=1\$$$

Putting these in our equation, we will have:

$${\displaystyle \frac{dy}{dx}}+p(t)y=q(t)$$

Now let us suppose:

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

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

$$u(t)=e^{\int {\displaystyle \frac{(t+1)}{t}}dt}$$

Integrating the power of $e$:

$$u(t)=e^{\int {\displaystyle \frac{t}{t}}dt+{\displaystyle \frac{1}{t}}dt}$$

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

Now we will simplify the exponential equation as follows:

$$u(t)=t{e}^t$$

From the second law of logarithm:

$$u(t)=e^{lnt{e}^t}$$

Take log on both sides of the equation:

$$lnu(t)=ln{e}^{lnt{e}^t}$$

$$lnu(t)=lnt{e}^t$$

$$u(t)=t{e}^t$$

We know that:

$$y(x)={\displaystyle \frac{\int u(t)q(t)dt}{u(t)}}$$

$$y(x)={\displaystyle \frac{\int (t{e}^t)(1)dt}{t{e}^t}}$$

$$y(x)={\displaystyle \frac{\int t{e}^{t}dt}{t{e}^t}}$$

Using integration by parts:

$$\int t{e}^{t}dt=t{e}^t–e^t+c$$

$$y(x)={\displaystyle \frac{t{e}^t-e^t+c}{t{e}^t}}$$

$$y(x)={\displaystyle \frac{t{e}^t}{t{e}^t}}–{\displaystyle \frac{e^t}{t{e}^t}}+{\displaystyle \frac{c}{t{e}^t}}$$

$$y(x)=1-{\displaystyle \frac{1}{t}}+{\displaystyle \frac{c}{t{e}^t}}$$

Putting the initial condition:

$$1=1-{\displaystyle \frac{1}{\ln 2}}+{\displaystyle \frac{c}{\ln 2e^t}}$$

$${\displaystyle \frac{1}{\ln 2}}={\displaystyle \frac{c}{\ln 2e^t}}$$

$${\displaystyle \frac{\ln 2e^t}{\ln 2}}={\displaystyle \frac{c}{1}}$$

$$e^{\ln 2}=c$$

$$c=2$$

Substituting the value of $c$ in the equation:

$$y(x)=1-{\displaystyle \frac{1}{t}}+{\displaystyle \frac{c}{t{e}^t}}$$

$$y(x)=1-{\displaystyle \frac{1}{t}}+{\displaystyle \frac{2}{t{e}^t}}$$

Numerical Result

$$y(x)=1-{\displaystyle \frac{1}{t}}+{\displaystyle \frac{2}{t{e}^t}}$$

Example

Integrate the following function:

$$\int {\displaystyle \frac{1}{x}}dx$$

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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