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.