Question

(a) Draw a direction field for the given differential equation. How do solutions appear to behave as \(t \rightarrow 0 ?\) Does the behavior depend on the choice of the initial value \(a\) ? Let \(a_{0}\) be the value of \(a\) for which the transition from one type of behavior to another occurs. Estimate the value of \(a_{0}\). (b) Solve the initial value problem and find the critical value \(a_{0}\) exactly. (c) Describe the behavior of the solution corresponding to the initial value \(a_{0}\) - $$ t y^{\prime}+(t+1) y=2 t e^{-t}, \quad y(1)=a $$

Short answer

#Short Answer# The critical value of the initial condition, \(a\), is \(a_{0} = \frac{1}{2}e\). When \(a > a_{0}\), the solutions will grow as \(t \rightarrow 0\). When \(a < a_{0}\), the solutions will decay as \(t \rightarrow 0\). When \(a = a_{0}\), the solution will remain constant as \(t \rightarrow 0\).

Step-by-step solution

Rewrite as a standard first-order equation

To solve it, we can first divide all terms by \(t\) to put it into the standard form: $$ y' + \frac{t+1}{t}y = 2e^{-t} $$

Find the integrating factor

Next, we find the integrating factor, which is given by the exponential of the integral of the coefficient of \(y\): $ \mu(t) = e^{\int \frac{t+1}{t} dt} = e^{t + \ln{t}} = te^t $

Multiply by the integrating factor and simplify

Now, multiply the equation by the integrating factor and simplify: $$ te^ty' + (t+1)e^ty = 2t^2e^te^{-t} $$ $$ (te^ty)' = 2t^2e^t $$

Integrate and solve for y(t)

Integrate both sides with respect to t and then solve for \(y(t)\). $$ \int(te^ty)' dt = \int 2t^2e^t \, dt $$ Use integration by parts on the right side: Let \(u = t^2\) and \(dv = 2te^t dt\). So, \(du = 2t dt\), and by integrating by parts, we get \(v = (2t - 1)e^t\). Therefore, $$ te^ty = \int 2t^2e^t \, dt = (t^2 - t + \frac{1}{2})e^t + C $$ $$ y(t) = \frac{(t^2 - t + \frac{1}{2})e^t}{t} + \frac{C}{te^t} $$

Apply the initial condition and find the critical value \(a_{0}\)

Apply the initial condition \(y(1) = a\) and find \(C\): $$ a = \frac{(1^2 - 1 + \frac{1}{2})e^1}{1} + \frac{C}{1e^1} $$ $$ C = a - \frac{1}{2}e $$ So the solution for given initial value is: $$ y(t) = \frac{(t^2 - t + \frac{1}{2})e^t}{t} + \frac{a - \frac{1}{2}e}{te^t} $$ Now, let's analyze the behavior of the solutions as \(t\rightarrow 0\): As \(t\rightarrow 0\), the second term will dominate the solution. Therefore, the transition value \(a_{0}\) will be determined by the balance between the numerator and denominator in the second term: $$ a - \frac{1}{2}e = 0 $$ $$ a_{0} = \frac{1}{2}e $$ The critical value of \(a\) is \(a_{0} = \frac{1}{2}e\).

Describe the behavior of the solutions for different initial values

For \(a > a_{0}\), the solutions will grow as \(t \rightarrow 0\), as the numerator will dominate. For \(a < a_{0}\), the solutions will decay as \(t \rightarrow 0\), as the denominator will dominate. For \(a = a_{0}\), the solution will remain constant as \(t \rightarrow 0\). As we found earlier, the critical value is \(a_{0} = \frac{1}{2}e\).

Key concepts

Direction Field

A direction field, also known as a slope field, is a useful tool for visualizing differential equations. It consists of small line segments, or arrows, representing the slope of the solution at any given point in the plane. For the differential equation \[ t y^{\prime}+(t+1) y=2 t e^{-t}, \quad y(1)=a \] drawing a direction field can help us to anticipate the behavior of solutions without explicitly solving the equation. Each segment's slope indicates the rate of change of the solution at that point. By following the pattern of arrows, one can get an overall sense of how solutions will move along the graph. This is particularly useful when solving complex equations, as it provides a visual clue of how solutions might behave as time, denoted by \(t\), approaches a certain value such as zero.

Initial Value Problem

An initial value problem (IVP) specifies a differential equation along with an initial condition. The initial condition is usually given as a specific value of the function and its derivative at some point. In this exercise, the initial value problem is given by: \[ t y^{\prime}+(t+1) y=2 t e^{-t}, \quad y(1)=a \]. The solution to an IVP provides not only a function that satisfies the differential equation but also meets the initial condition provided. Solving an IVP typically involves integrating the differential equation while ensuring that the solution passes through the given initial point. This initial point "anchors" the solution, helping to specify it uniquely from potentially infinitely many possible solutions.

Integrating Factor

The integrating factor is a method used to solve first-order linear differential equations of the form: \[ y' + P(t)y = Q(t) \]. It involves multiplying the entire equation by a specific function, known as the integrating factor, which simplifies the process of finding a solution. In this exercise, the differential equation was transformed by an integrating factor: \[ \mu(t) = e^{\int \frac{t+1}{t} dt} = t e^t \]. This factor is derived by exponentiating the integral of the function multiplying \(y\) in the standard form of the linear differential equation. By applying it, the left side of the equation becomes the derivative of a product, making it straightforward to integrate both sides and find the solution for \(y(t)\). This is particularly effective as it systematically reduces complex problems to solvable expressions.

Boundary Behavior

Boundary behavior in differential equations refers to how solutions behave as they approach certain values, often at extremes like \(t \rightarrow 0 \) or \(t \rightarrow \infty \). In the exercise, it's crucial to understand how the solution behaves as \(t\) approaches zero, which helps identify critical values like \( a_0 \). For this specific problem, the behavior of the solution changes significantly depending on the value of \(a\).

• For \(a > a_0\), solutions grow as \(t \rightarrow 0\).
• For \(a < a_0\), solutions decay as \(t \rightarrow 0\).
• For \(a = a_0\), solutions remain constant as \(t \rightarrow 0\).

Understanding these behaviors allows for the proper assessment of stability and long-term trends of the solutions. The critical value, found to be \(a_0 = \frac{1}{2}e\), plays a critical role in anticipating and describing these solutions' transitions.