Question

Analysis of a separable equation Consider the differential equation \(y y^{\prime}(t)=\frac{1}{2} e^{t}+t\) and carry out the following analysis. a. Find the general solution of the equation and express it explicitly as a function of \(t\) in two cases: \(y>0\) and \(y \leq 0\). b. Find the solutions that satisfy the initial conditions \(y(-1)=1\) and \(y(-1)=2\). c. Graph the solutions in part ( b) and describe their behavior as \(t\) increases. d. Find the solutions that satisfy the initial conditions \(y(-1)=-1\) and \(y(-1)=-2\). e. Graph the solutions in part (d) and describe their behavior as \(t\) increases.

Short answer

Question: Determine the general solution for the separable differential equation \(y y^{\prime}(t)=\frac{1}{2} e^{t}+t\) for both \(y>0\) and \(y \leq 0\). Then, find specific solutions for \(y(-1)=1\), \(y(-1)=2\), \(y(-1)=-1\), and \(y(-1)=-2\), and describe the behavior of the solutions as \(t\) increases. Answer: The general solution for \(y > 0\) is \(y(t) + C_1 = \frac{1}{2}e^{t} + t + \frac{C_2}{y(t)}\), and the general solution for \(y \leq 0\) is \(y(t)+ C_1 = -\frac{1}{2}e^{t} - t - \frac{C_2}{y(t)}\)). For the specific initial conditions, the solutions are: 1. For \(y(-1) = 1\): \(y(t) = \frac{1}{2}e^{t} + t + \frac{\frac{1}{2}e - 1}{y(t)} - \frac{3}{2} + \frac{1}{2}e\) 2. For \(y(-1) = 2\): \(y(t) = \frac{1}{2}e^{t} + t + \frac{2e - 4}{y(t)} - 2 + e\) 3. For \(y(-1) = -1\): \(y(t) = -\frac{1}{2}e^{t} - t - \frac{\frac{1}{2}e+1}{y(t)} + \frac{1}{2}e - 1\) 4. For \(y(-1) = -2\): \(y(t) = -\frac{1}{2}e^{t} - t - \frac{e+2}{y(t)} + 2 - \frac{1}{2}e\) For the solutions with positive initial conditions, they will grow without bound as \(t\) increases, with the first solution always below the second. For the solutions with negative initial conditions, they will decrease to negative infinity, with the third solution always above the fourth.

Step-by-step solution

Find the general solution for the differential equation

To solve the given differential equation, we need to rewrite it as a separable equation. The given equation is \(y y^{\prime}(t)=\frac{1}{2} e^{t}+t\). Divide both sides by \(y\) to get separable form: \(y^{\prime}(t) = \frac{1}{2} \frac{e^{t}}{y} + \frac{t}{y}\). Now, we can integrate with respect to \(t\) on both sides.

Integration

Integrate both sides with respect to \(t\): \(\int y^{\prime}(t) dt = \int \left(\frac{1}{2} \frac{e^{t}}{y} + \frac{t}{y}\right) dt\). The left side is just an integration of \(y'(t)\), which is \(y(t) + C_1\). The right side will require integrating both terms: \(\int \frac{1}{2} \frac{e^{t}}{y} dt + \int \frac{t}{y} dt\). We get \(y(t) + C_1 = \frac{1}{2}e^{t}y^{-1} + ty^{-1} + C_2\).

Rearrange to find y(t)

To isolate \(y(t)\), multiply both sides by \(y\). We have \(y^2(t) + C_1y(t) - \frac{1}{2} e^{t}y(t) - ty(t) = C_2\). Now, we will handle the general solution in two cases for \(y>0\) and \(y \leq 0\).

General solution for y>0 and y

For \(y>0\), we can divide both sides by \(y(t)\) again to simplify the equation: \(y(t) + C_1 = \frac{1}{2}e^{t} + t + \frac{C_2}{y(t)}\). Similarly, for \(y \leq 0\), we have \(y(t)+ C_1 = -\frac{1}{2}e^{t} - t - \frac{C_2}{y(t)}\).

Find solutions with initial conditions (part b)

To find the solutions that satisfy the initial conditions \(y(-1)=1\) and \(y(-1)=2\), we substitute these values into the appropriate cases of the general solution. For \(y(-1)=1\) and \(y>0\): \(1 + C_1 = \frac{1}{2}e^{-1} -1 + \frac{C_2}{1}\). For \(y(-1)=2\) and \(y>0\): \(2 + C_1 = \frac{1}{2}e^{-1} -1 + \frac{C_2}{2}\).

Solve for C_1 and C_2 in part b

Solving the equations in Step 5 for \(C_1\) and \(C_2\), we find the constants for the initial conditions in part b: For \(y(-1)=1\), \(C_1 = \frac{3}{2} - \frac{1}{2}e\) and \(C_2 = \frac{1}{2}e - 1\), and for \(y(-1)=2\), \(C_1 = 2 - e\) and \(C_2 = 2e - 4\).

Graph solutions and describe behavior for part b

Graph the solutions found in Step 6 and describe their behavior as \(t\) increases. As \(t\) increases, the solutions will grow for both initial conditions, each following its own path, with the \(y(-1) = 1\) solution always below the \(y(-1) = 2\) solution. The solutions will keep growing without bound as \(t\) increases.

Find solutions with initial conditions (part d)

Similar to part b, we will find the solutions that satisfy the initial conditions \(y(-1)=-1\) and \(y(-1)=-2\) using the \(y \leq 0\) case of the general solution. For \(y(-1)=-1\) and \(y \leq 0\): \(-1 + C_1 = -\frac{1}{2}e^{-1} +1 - \frac{C_2}{-1}\). For \(y(-1)=-2\) and \(y \leq 0\): \(-2 + C_1 = -\frac{1}{2}e^{-1} +1 - \frac{C_2}{-2}\).

Solve for C_1 and C_2 in part d

Solving the equations in Step 8 for \(C_1\) and \(C_2\), we find the constants for the initial conditions in part d: For \(y(-1)=-1\), \(C_1 = \frac{1}{2}e - 1\) and \(C_2 = \frac{1}{2}e + 1\), and for \(y(-1)=-2\), \(C_1 = 2 - \frac{1}{2}e\) and \(C_2 = e + 2\).

Graph solutions and describe behavior for part d

Graph the solutions found in Step 9, and describe their behavior as \(t\) increases. As \(t\) increases, the solutions will decrease for both initial conditions, with the \(y(-1)=-1\) solution always above the \(y(-1)=-2\) solution. The solutions will keep going to negative infinity as \(t\) increases.

Key concepts

General Solution of Differential Equations

The general solution to a differential equation provides a family of functions that all satisfy the equation. Specifically for separable differential equations, like the one in our exercise \(yy'(t) = \frac{1}{2} e^{t} + t\), the process involves separating variables and integrating both sides to find an implicit solution of \(y(t)\).

For \(y>0\) and \(y \leq 0\), the general solution takes different forms because of the absolute value associated with the integration of \(y^{-1}\). We can express the general solution in terms of \(t\) by integrating and applying the initial conditions to find specific constant values (\(C_1\) and \(C_2\)). This yields two different expressions for the function \(y(t)\) depending on the sign of \(y\).

Integration of Differential Equations

Integration is at the heart of solving separable differential equations. After rearranging the equation into a form where one side is in terms of \(t\) and the other in terms of \(y\), we perform integration on both sides separately. In this exercise, the integration process is crucial for moving from \(y'(t)\) to \(y(t)\), and involves integrating complex expressions like \(\frac{e^{t}}{y}\) and \(\frac{t}{y}\).

The challenge often comes with integrating the right-hand side, which may require u-substitution or partial fractions, and handling the absolute values correctly. After integration, constants of integration \(C_1\) and \(C_2\) are introduced, which will later be determined by initial conditions.

Initial Value Problems

Initial value problems involve solving differential equations with given starting conditions. These problems are pivotal because they produce a specific solution out of the general solution by utilizing initial conditions to solve for the constants of integration.

In this exercise, once the general solution is derived, we apply the initial conditions, such as \(y(-1)=1\) and \(y(-1)=2\), to determine the constants. It leads us to particular solutions that not only satisfy the differential equation but also pass through the given points on the \(ty\)-plane. Initial value problems exemplify how a differential equation can model a scenario where the starting state is known and the subsequent behavior is predictable.

Graphing Solutions of Differential Equations

Graphing the solutions of differential equations visually demonstrates the behavior of functions over time. In the problem we're discussing, students are tasked to graph solutions for different initial conditions. This graphical representation helps to compare the growth or decline of different solutions based on their initial values.

Graphing is a powerful tool for interpreting the long-term behavior of solutions. As \(t\) increases, the graph shows whether solutions approach infinity, a finite value, or negative infinity, and how they diverge or converge. In this case, solutions for positive initial values grow unboundedly, while for negative initial values, they plummet towards negative infinity, each reflecting the nature of the exponential and linear terms in the differential equation.