91Ó°ÊÓ

Separable differential equations are a special type of differential equations where variables can be separated on different sides of the equation. For a differential equation to be separable, it must be possible to express it in the form:
\( \frac{dy}{dx} = g(x)h(y) \).
This means that we can rewrite the equation so that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other.

In the given exercise, the equation \( y' = -y^2 \, \text{sin}(t) \) is separable.
We can rewrite it as:
\( \frac{dy}{y^2} = -\text{sin}(t) \, dt \).

By separating the variables, we make it easier to integrate each side with respect to its own variable.

Integration is the process of finding the integral of a function, which is the reverse process of differentiation.

To solve separable differential equations, we use a variety of integration techniques.
In this exercise, we have to integrate \( \frac{dy}{y^2} \) and \( -\text{sin}(t) \) separately.

First, let’s focus on the integral of \( \frac{dy}{y^2} \):
\[ \frac{dy}{y^2} = y^{-2} \, dy \].
The integral of \( y^{-2} \) is:
\[ \boxed{-y^{-1}} + C_1 \].

Next, integrate \( -\text{sin}(t) \):
\[ \boxed{-\text{sin}(t)} \, dt \].
The integral of \( -\text{sin}(t) \) is:
\[ \boxed{\text{cos}(t) + C_2} \].

Combining these integration results, we have:
\[ -y^{-1} = \text{cos}(t) + C \].
Here, we assume \( C \) is the combined constant of integration.

When solving a differential equation, initial conditions help determine the specific solution that satisfies given values.

In our example, the initial condition is \( y(\frac{\text{Ï€}}{2}) = 1 \).
This means that when \( t = \frac{\text{Ï€}}{2} \), the value of \( y \) should be 1.

Using the result from our integration, \( -y^{-1} = \text{cos}(t) + C \), we apply the initial condition:
\[ -1 = \text{cos}\big(\frac{\text{Ï€}}{2}\big) + C \].

Since \( \text{cos}\big(\frac{\text{Ï€}}{2}\big) = 0 \), we end up with:
\[ -1 = 0 + C \]
Therefore, \( C = -1 \).

Substituting back into our general solution, we get:
\[ -y^{-1} = \text{cos}(t) - 1 \].
Solving for \( y \), we have:
\[ y = -\frac{1}{\text{cos}(t) - 1} \].

However, note that in this case \( \text{cos}\big(\frac{\text{Ï€}}{2}\big) \) simplifies to zero.
Thus, applying the initial condition leads to an inconsistency.
This indicates a problem in setting up or interpreting the differential equation, requiring some reconsideration of the initial parameters or model.