91Ó°ÊÓ

An integrating factor is a function used to solve linear ordinary differential equations (ODEs). It's particularly useful when the equation is in the form \(\frac{dy}{dt} + P(t) y = Q(t)\).

The integrating factor, typically denoted as \( \rho(t) = e^{\int P(t) dt} \), makes the equation easier to integrate. By multiplying through by this factor, the left-hand side of the equation becomes the derivative of a product of functions, simplifying the problem.

In our example, we identified \(P(t) = \frac{t}{6}\).
Thus, the integrating factor is calculated as:
\[ \rho(t) = e^{\int \frac{t}{6} dt} = e^{\frac{t^2}{12}}. \]
This factor helps transform the original ODE into a simpler form.

Ordinary Differential Equations (ODEs) involve functions and their derivatives. They are 'ordinary' because they contain functions of only one variable and its derivatives, as opposed to partial differential equations which involve multiple variables.

The ODE given in our exercise is:
\[6 y' + t y = t. \]

By rewriting it in standard form, we divide by 6:

\[ y' + \frac{t}{6} y = \frac{t}{6}. \]

This makes it easier to identify functions \(P(t)\) and \(Q(t)\), necessary for applying the integrating factor method.
Mastering how to transform and handle ODEs is crucial for solving real-world problems involving rates of change in physics, biology, economics, and engineering.

When solving ODEs, different integration techniques come into play, especially when integrating terms involving exponential functions or products of functions.

In the solution process:

• We found and applied the integrating factor \( e^{\frac{t^2}{12}} \).
• We then turned our differential equation into a form that allowed us to integrate both sides easily.
• Using substitution, where we let \( u = \frac{t^2}{12} \), simplified the integral on the right-hand side.

The integral evaluated as follows:
\[ \frac{d}{dt} \big( y e^{\frac{t^2}{12}} \big) = e^{\frac{t^2}{12}} \frac{t}{6} \]
Integrate both sides:
\[ \int \frac{d}{dt} \big( y e^{\frac{t^2}{12}} \big) dt = \int e^{\frac{t^2}{12}} \frac{t}{6} dt \]

The left-hand side becomes \( y e^{\frac{t^2}{12}} \), while the right-hand side needs substitution. Letting \( u = \frac{t^2}{12} \), then the integral simplifies:
\[ \int e^u du = e^u + C \]
Finally, substituting back:
\[ e^{\frac{t^2}{12}} + C. \]

Understanding and practicing various integration techniques ensures you are well-prepared for more complex ODEs.