91Ó°ÊÓ

Differential equations are mathematical equations involving derivatives of an unknown function.
They describe how a specific quantity changes over time or in relation to another quantity.

There are several types of differential equations, but in this context, we're dealing with first-order linear differential equations.

First-order means the highest derivative in the equation is the first derivative (denoted as \(y'\)).
Linear indicates that the unknown function and its derivatives appear to the first power and are not multiplied together.
Understanding the type of differential equation helps in choosing the right method to solve it.

In the given problem, we have:

\[x^{2} y' + 3 x y = 1\]
Identifying this as a first-order linear differential equation is the first crucial step towards solving it. We rewrite it in the standard form:
\[y' + P(x)y = Q(x)\]
Making this identification simplifies the process, as it points us toward using the integrating factor technique.

The integrating factor is a function used to simplify the process of solving first-order linear differential equations.
The general form of such an equation is:
\[y' + P(x)y = Q(x)\]
The integrating factor, denoted as \(\mu(x)\), is given by:
\[\mu(x) = e^{\int P(x)\, dx}\]
It transforms the equation into an easily integrable form.

In our exercise, we derived the integrating factor for:
\[y' + \frac{3}{x} y = \frac{1}{x^{2}}\]
Here, \( P(x) = \frac{3}{x} \), and the integrating factor is:
\[\mu(x) = e^{\int \frac{3}{x}\, dx} = |x|^{3} = x^3\]
Don't forget that x must be positive, hence we consider \( x > 0 \).

Multiplying the entire differential equation by this integrating factor simplifies it to a form where you can directly see the derivative of a product.
This technique makes it easier to integrate both sides and find the unknown function.

The general solution of a differential equation is its solution in the most general form, representing all possible solutions.
For first-order linear differential equations, after finding the integrating factor and multiplying through the equation, we obtain:

\[\frac{d}{dx}(x^3 y) = x\]
Integrating both sides with respect to \(x\), we get:
\[x^3 y = \int x \, dx\]
This results in:
\[x^3 y = \frac{x^2}{2} + C\]
Solving for \(y\), we divide by \(x^3\):
\[y = \frac{x^2 + 2C}{2x^3} = \frac{1}{2} \cdot \frac{x^2}{x^3} + \frac{C}{x^3} = \frac{1}{2x} + \frac{C}{x^3}\]
This is the general solution:
\[y = \frac{1}{2x} + \frac{C}{x^3}\]
Here, \(C\) is the constant of integration, which describes the family of all possible solutions to the differential equation.
The general solution is essential as it includes every possible solution for different initial conditions.