91Ó°ÊÓ

Initial conditions are values given for the unknown function (and possibly its derivatives) at a specific point. These conditions are essential because they allow us to find the specific solution to a differential equation, instead of a general one. Without them, we would only be able to find a family of solutions. For instance, in our differential equation example, we are given the initial condition: \( y(0) = 1 \). This tells us that when \( t = 0 \), the value of \( y \) is 1.
Initial conditions help us determine the constant of integration (denoted as \( C \)) after we solve the differential equation. By plugging the initial condition into our solution, we resolve any unknown constants and obtain a unique solution tailored to the problem.

An integrating factor is a function used to transform a non-exact differential equation into an exact one, making it easier to solve. For first-order linear differential equations of the form \( y' + P(t)y = Q(t) \), the integrating factor \( \mu(t) \) is given by:
\[ \mu(t) = e^{\int P(t) \, dt} \].
For our problem, the integrating factor was found as:
\[ \mu(t) = e^{-5 \int t \, dt} = e^{-\frac{5}{2} t^2} \].
Using the integrating factor, we can multiply through the original differential equation to simplify it into a form that can be integrated more easily. This simplifies the problem significantly and allows us to find an explicit solution.

Solving a differential equation involves several steps that transform the equation into an integrated form. Let’s break down this process using our example:
- We started with\( y' = 5ty - 2t \) and rewrote it to match the standard form: \( y' - 5ty = -2t \).
- We determined the integrating factor: \( e^{-\frac{5}{2} t^2} \).
- Multiplying through by the integrating factor, we got:
\[ \left( e^{-\frac{5}{2} t^2} y \right)' = -2t e^{-\frac{5}{2} t^2} \].
- We then integrated both sides of the equation:
\[ \int \left( e^{-\frac{5}{2} t^2} y \right)' \, dt = \int -2t e^{-\frac{5}{2} t^2} \, dt \].
- This resulted in:
\[ e^{-\frac{5}{2} t^2} y = e^{-\frac{5}{2} t^2} + C \].
- By solving for \( y \), we obtained:
\[ y = 1 + C e^{\frac{5}{2} t^2} \].
- Applying the initial condition \( y(0) = 1 \), we found that \( C = 0 \).
The final solution is:
\[ y = 1 \].
This process demonstrates converting a differential equation into a manageable form by using integrating factors and initial conditions, ultimately leading to a clear and specific solution.