91Ó°ÊÓ

An initial value problem is a type of differential equation that comes with a specified condition, known as the initial condition. This initial condition helps to find a unique solution to the differential equation. In the given exercise, we have the differential equation:
\( t y^{\prime}+2 y=t^{2}-t+1 \)
and the initial condition:
\( y(1)=\frac{1}{2} \).

• The equation is solved for a specific value of the function \( y(t) \) at \( t = 1 \).
• The purpose of the initial condition is to find the particular solution that fits the specific requirement of \( y \) when \( t \) equals the given initial value.

The role of the initial condition is crucial as it ensures that among the infinite possible solutions of the differential equation, we identify the one that satisfies \( y(1)=\frac{1}{2} \). Once this specific condition is applied, you can determine the constant of integration, providing a precise and unique answer to the problem.

The integrating factor simplifies solving first-order linear differential equations. For the equation to be in this form:
\( y^{\prime} + P(t)y = Q(t) \),
An integrating factor \( \mu(t) \) is calculated, transforming the differential equation into an exact equation.

• Begin by identifying \( P(t) \) from the equation.
• Calculate the integrating factor, \( \mu(t) = e^{\int P(t) \, dt} \).

In the example, the equation becomes
\( y^{\prime} + \frac{2}{t}y = \frac{t^{2}-t+1}{t} \),
where \( P(t) = \frac{2}{t} \).
The integrating factor hence is:
\( \mu(t) = e^{\int \frac{2}{t} \, dt} = t^2 \).
Multiplying the equation with this factor facilitates the integration of both sides and transforms the equation into an exact differential, allowing easier integration.

An exact differential equation is an equation derived after applying the integrating factor, where the left-hand side is expressed as a derivative of a single function. Once this form is achieved, the equation can simply be integrated.
For the equation
\( t^2 y^{\prime}+2t y = t^{3}-t^{2}+t \),
an integrating factor has transformed it to:
\( \frac{d}{dt}(t^2 y) = t^3 - t^2 + t \).

• The left-hand side \( \frac{d}{dt}(t^2 y) \) represents the derivative of the product \( t^2 y \).
• This makes the equation straightforward to solve by integrating both sides with respect to \( t \).

Upon integration, you get:
\( t^2 y = \frac{1}{4}t^4 - \frac{1}{3}t^3 + \frac{1}{2}t^2 + C \).
This integration step works because the differential form reflects an exact derivative, thus easing up the task of finding the particular solution by direct integration.

A particular solution to a differential equation is a specific solution that meets the given initial conditions. Once the differential equation is solved and integrated, you arrive at a general solution that typically contains an arbitrary constant.

• In our example, the general solution was found to be:
• \( y(t) = \frac{1}{4}t^2 - \frac{1}{3}t + \frac{1}{2} + \frac{C}{t^2} \).

To determine the particular solution, use the initial condition \( y(1) = \frac{1}{2} \):
\[ \frac{1}{2} = \frac{1}{4} - \frac{1}{3} + \frac{1}{2} + C \].
By solving for \( C \), you find \( C = 0 \), giving the particular solution:
\( y(t) = \frac{1}{4}t^2 - \frac{1}{3}t + \frac{1}{2} \).
The unique aspect of the particular solution is that it satisfies both the equation and the initial condition, providing a specific application of the general solution for a scenario.