Question

Consider the initial value problem $$ y^{\prime}-\frac{3}{2} y=3 t+2 e^{t}, \quad y(0)=y_{0} $$ Find the value of \(y_{0}\) that separates solutions that grow positively as \(t \rightarrow \infty\) from those that grow negatively. How does the solution that corresponds to this critical value of \(y_{0}\) behave as \(t \rightarrow \infty\) ?

Short answer

Answer: There is not a single critical value of \(y_0\) that solely separates solutions growing positively or negatively as \(t \rightarrow \infty\). The long-term behavior depends on a combination of factors and the given initial conditions.

Step-by-step solution

Find the integrating factor

To solve the first-order linear ODE, we will first find an integrating factor, which is given by the exponential of the integral of the coefficient of the dependent variable, in this case, \(-\frac{3}{2}\). Let's compute the integrating factor: $$ I(t) = e^{\int{-\frac{3}{2} dt}} = e^{-\frac{3}{2}t} $$

Multiply the ODE by the integrating factor

We now multiply the entire ODE by the integrating factor \(I(t) = e^{-\frac{3}{2}t}\): $$ e^{-\frac{3}{2}t} (y' - \frac{3}{2} y) = e^{-\frac{3}{2}t} (3t + 2e^{t}) $$

Find the general solution by integrating both sides

Now we can integrate both sides of the equation to find the general solution: $$ \int e^{-\frac{3}{2}t}(y'-\frac{3}{2}y)dt = \int e^{-\frac{3}{2}t}(3t+2e^t)dt $$ The integral on the left-hand side becomes: $$ y(t)e^{-\frac{3}{2}t} $$ and for the right-hand side, we'll perform an integration by parts and simplification: $$ \int e^{-\frac{3}{2}t}(3t+2e^t)dt = \frac{3}{4}e^{-\frac{3}{2}t}t^2 - \frac{5}{2}e^{-\frac{3}{2}t}t - 7e^{-\frac{3}{2}t} +2e^{-\frac{1}{2}t} + C $$ Equating both sides, we get the general solution of the ODE: $$ y(t)= e^{\frac{3}{2}t}\left(\frac{3}{4}t^2 - \frac{5}{2}t -7 +2e^{t} + C \right) $$

Determine the behavior of the solution as \(t \rightarrow \infty\)

To find the behavior of the solution as \(t \rightarrow \infty\), we need to examine the individual terms. We note that the main contributing terms as \(t \rightarrow \infty\) are: $$ t^2 e^{\frac{3}{2}t}, \quad -\frac{5}{2}t e^{\frac{3}{2}t}, \quad 2e^{\frac{1}{2}t} e^{\frac{3}{2}t} $$ The first term is positive and becomes dominant as \(t \rightarrow \infty\), the second term is negative, and the third term is positive. By controlling the coefficients, we can determine the direction of the growth.

Find the value of \(y_{0}\) for critical behavior

We are given the initial condition \(y(0) = y_0\). To find the separating value for \(y_{0}\), we need to determine under which conditions the behavior will change. From the general solution, $$ y(0) = y_0 = \frac{3}{4}(0)^2 - \frac{5}{2}(0) - 7 + 2e^{(0)} + C $$ Which can be simplified to: $$ y_0 = -7 + 2 + C $$ Now, let's consider the dominant term \(t^{2}e^{\frac{3}{2}t}\) in the solution. If the coefficient \(\frac{3}{4}\) is positive, the solution will grow positively in the long run. If the coefficient is negative, the solution will grow negatively. If the coefficient is zero, however, the solution becomes critical. The critical value of \(y_{0}\) occurs when \(\frac{3}{4} = 0\), thus no value of \(y_0\) will make this term vanish. It is important to note that, although no value of \(y_0\) will make the dominant term vanish, there is a balance when considering the other terms too. Thus, we cannot find a single critical value of \(y_0\) that separates solutions that grow positively as \(t \rightarrow \infty\) from those that grow negatively. The behavior will depend on a combination of factors and the given initial conditions.

Key concepts

Integrating factor

When dealing with a first-order linear differential equation, finding the integrating factor is a crucial step. This factor, typically denoted as \(I(t)\), transforms the equation into a more easily solvable form. The integrating factor is calculated using the formula \(e^{\int P(t) \, dt}\), where \(P(t)\) is the coefficient of the dependent variable, \(y\), in its linear differential equation form \(y' + P(t)y = Q(t)\). In the given problem, \(P(t) = -\frac{3}{2}\), so the integrating factor becomes \(I(t) = e^{-\frac{3}{2}t}\). This integrating factor plays a role similar to a multiplier, preparing the equation for integration and ultimately leading to the solution of the differential equation. By multiplying both sides of the differential equation by \(I(t)\), the left-hand side reduces to the derivative of \(y(t)I(t)\). This simplification is a key aspect in solving the equation.

Initial value problem

In differential equations, an initial value problem specifies the value of the unknown function at a given point, adding a condition to an otherwise general solution. For our exercise, we're given \(y(0) = y_0\). This means that when \(t = 0\), the value of \(y\) should satisfy this initial condition. Solving the initial value problem involves finding a particular solution that passes through this point, tailoring the general solution to fit the specific requirements. The general solution we derived must be combined with this initial condition to determine the integration constant \(C\). This incorporation is significant in many applications, as it aligns mathematical models with real-world conditions or experimental data. Without it, the solutions would be hypothetical and lack practical relevance.

Long-term behavior

The long-term behavior of solutions to differential equations reveals how solutions behave as the independent variable approaches infinity. In this exercise, we analyze how \(t \rightarrow \infty\) affects our solutions. The dominant terms in the general solution indicate which parts of the equation will predominate over others as \(t\) becomes very large. In our solution, \(t^2 e^{\frac{3}{2}t}\) emerges as a dominant term, growing rapidly and overshadowing other components. If the coefficient for this term is positive, the solution grows positively; if negative, it grows negatively. Identifying growth behavior helps in understanding system stability and potential outcomes. Although finding a critical \(y_0\) in this situation is tricky due to the consistent dominance of \(t^2 e^{\frac{3}{2}t}\), understanding these dynamics is key in qualitative studies of differential equations.

General solution of ODE

The general solution of an ordinary differential equation (ODE) provides a complete representation of the infinite set of possible solutions. For this first-order linear ODE, it appears as \(y(t) = e^{\frac{3}{2}t}(\frac{3}{4}t^2 - \frac{5}{2}t - 7 + 2e^t + C)\). This solution includes an arbitrary constant \(C\), reflecting the infinite number of curves that the equation can describe. Solving an ODE involves determining this general solution, often by integrating after multiplying by the integrating factor. General solutions form the basis of many mathematical analyses, offering frameworks for understanding dynamics that can be adjusted for particular conditions through specific values of \(C\). This flexibility is crucial, allowing solutions to be tailored to various initial conditions, much like the one provided in our initial value problem.