91Ó°ÊÓ

When dealing with differential equations, it's important to recognize the type of equation, as it dictates the solution method. A first-order linear differential equation takes the form \(\frac{dy}{dt} + P(t) y = Q(t)\). Here, \(\frac{dy}{dt}\) is the first derivative of y with respect to t, P(t) is a function of t, and Q(t) is called the non-homogeneous term and is also a function of t.

In our exercise, the given equation is: \(t y^{\text{'}} + y = \sin t\). We can identify it as a first-order linear differential equation after rewriting it in the form \(\frac{dy}{dt}+\frac{1}{t}y = \frac{\sin t}{t}\). Recognizing this form helps us to proceed with appropriate solution techniques such as finding an integrating factor.

To solve a first-order linear differential equation, we typically use an integrating factor. This is a function we multiply through the equation to make the left-hand side an exact derivative, which simplifies integration.

For the general form \(\frac{dy}{dt} + P(t) y = Q(t)\), the integrating factor \(\mu(t)\) is defined as \(e^{\int P(t) dt}\).

In our problem, \(P(t) = \frac{1}{t}\), so the integrating factor \(\mu(t)\) is \(e^{\int \frac{1}{t} dt} = e^{\ln|t|}= t\).

We multiply the entire differential equation by this integrating factor to rewrite the left-hand side as an exact derivative and simplify the solving process.

After finding the integrating factor, we multiply it through the differential equation and use properties of derivatives to integrate both sides easily.

In our case, multiplying by \(t\) transforms the equation to \(\frac{d}{dt}(ty) = \sin t\), making the left side an exact derivative. We then integrate both sides: \(\int \frac{d}{dt}(ty) dt = \int \sin t dt\).

The solution to this integral is \(ty = -\cos t + C\), where C is a constant of integration.

Dividing by \(t\) gives the general solution: \(y = \frac{-\cos t}{t} + \frac{C}{t}\). At this point, the constant C remains undetermined until we apply the initial conditions.

An initial condition provides specific information to find the particular solution out of the general solution. It's usually given as \(y(t_0) = y_0\), which means when t equals \(t_0\), y equals \(y_0\).

In our problem, the initial condition is \(y\left(\frac{\pi}{2}\right) = 0\). We substitute these values into the general solution to find the constant C.

Using \(\frac{\pi}{2}\) for t and 0 for y, we get 0 = -\cos\left(\frac{\pi}{2}\right) + C. Since \cos\left(\frac{\pi}{2}\right)=0, this simplifies to C=0.

Therefore, substituting C back into our general solution gives us the particular solution: \(y = \frac{-\cos t}{t}\). This satisfies both the differential equation and the initial condition, making it the final solution to our initial value problem.