91Ó°ÊÓ

The integrating factor is a critical component in solving linear first-order differential equations. It serves as a tool that simplifies the process of finding a solution. In our given example, the differential equation is first rearranged into its standard form:

• \(\frac{d s}{d t} + P(t) s = Q(t)\),
• where \(P(t) = \frac{2}{t+1}\) and \(Q(t) = 3 + \frac{1}{(t+1)^3}\).

To compute the integrating factor \(\mu(t)\), we use the formula:

• \(\mu(t) = e^{\int P(t) \, dt}\),
• leading to \(\mu(t) = e^{\int \frac{2}{t+1} \, dt} = |t+1|^2\).

Since \(t > -1\), the absolute value simplifies, giving \(\mu(t) = (t+1)^2\). By using this integrating factor, we transform the differential equation, allowing the left side to become the derivative of \((t+1)^2 s\). This enables us to integrate both sides easily.

Putting a differential equation into its standard form is the first and crucial step in many solution processes. The standard form for a linear first-order differential equation is:

• \(\frac{d s}{d t} + P(t) s = Q(t)\)

In our exercise, the original equation is given as:

• \((t+1) \frac{d s}{d t} + 2 s = 3(t+1) + \frac{1}{(t+1)^2}\)

Dividing through by \((t+1)\) simplifies the equation, transforming it into standard form:

• \(\frac{d s}{d t} + \frac{2}{t+1} s = 3 + \frac{1}{(t+1)^3}\)

Getting the equation into standard form helps us identify the functions \(P(t)\) and \(Q(t)\), which are necessary for calculating the integrating factor.

Solving the differential equation becomes feasible once it's in standard form and we have the integrating factor. Here, we used the integrating factor method:
The equation \((t+1)\frac{d s}{d t} + 2s = 3(t+1) + \frac{1}{(t+1)^2}\) becomes \((t+1)^2 \frac{d s}{d t} + 2(t+1) s = 3(t+1)^2 + \frac{1}{(t+1)}\) after multiplying by the integrating factor \((t+1)^2\).

• The left-hand side is arranged to represent the derivative: \(\frac{d}{dt}((t+1)^2 s)\).
• This allows us to write: \(\frac{d}{dt}( (t+1)^2 s ) = 3(t+1)^2 + \frac{1}{(t+1)}\).

By integrating both sides, we find \( (t+1)^2 s = \int \left(3(t+1)^2 + \frac{1}{(t+1)}\right) \, dt \) resulting in:

• \((t+1)^2 s = t^3 + 3t^2 + 3t + \ln(t+1) + C\)

Finally, isolating \(s\), the solution becomes:

• \(s(t) = \frac{t^3 + 3t^2 + 3t + \ln(t+1) + C}{(t+1)^2}\).

Initial conditions are often provided to find a unique solution to a differential equation. They define the value of the solution at a specific point, enabling us to find the constant \(C\).
In many problems, after obtaining the general solution, an initial condition like \(s(t_0) = s_0\) is given. This can be applied to find the specific value of \(C\), ensuring the solution fits the real-world context or specific scenario being modeled. Without an initial condition, the solution remains general, containing an arbitrary constant, as seen in our solved example:

• \(s(t) = \frac{t^3 + 3t^2 + 3t + \ln(t+1) + C}{(t+1)^2}\).

Remember that the presence and use of initial conditions add substantial value by narrowing down the solutions to one that fully describes the given situation.