Question

In problems 1-24 Find the general solution of the given differential equation. Give the largest interval I over which the general solution is defined. Determine whether there are any transient terms in the general solution.

$\frac{dp}{dt}+2tP=P+4t-2$

Short answer

The general solution and interval is$P=2+C{e}^{-\left(t^2-t\right)};\text{I}:(-\infty ,\infty )$

Step-by-step solution

Concept of Linear Differential equation

$\frac{dy}{dx}+py=q$, where p and q can be constants or variables.

Converting a given differential equation into standard form

$\frac{dP}{dt}+(2t-1)P=4t-2$

Finding an integrating factor

$e^{\int (2t-1)dt}=e^{t^2-t}$

Finding a General Solution

$$\begin{gathered} {}^{t^2-t}\left[\frac{dP}{dt}+(2t-1)P\right]=e^{t^2-t}[4t-2] \\ \frac{d}{dt}\left[e^{t^2-t}P\right]=e^{t^2-t}(4t-2) \end{gathered}$$

Integrate the above equation on both sides.

$$\begin{gathered} \int d\left[e^{t^2-t}P\right]=\int (4t-2)e^{t^2-t}dt \\ {e}^{t^2-t}P=2e^{t^2-t}+C \\ P=2+C{e}^{-\left(t^2-t\right)} \end{gathered}$$

Finding intervals

The solution is defined at $(-\infty ,\infty )$

Hence, the general solution and interval is$P=2+C{e}^{-\left(t^2-t\right)};\text{I}:(-\infty ,\infty )$