Question

In each of Problems 9 through 16 find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior as \(t\) increases. $$ y^{\prime \prime}+y^{\prime}-2 y=0, \quad y(0)=1, \quad y^{\prime}(0)=1 $$

Short answer

In summary, the unique solution to the given second-order linear differential equation with constant coefficients and initial conditions is \(y(t) = \frac{1}{3}e^{-2t} + \frac{4}{3}e^{t}\). As the parameter t increases, the overall behavior of the solution is that it will grow exponentially due to the dominance of the \(e^{t}\) term.

Step-by-step solution

Find the characteristic equation

Given the differential equation $$ y^{\prime \prime}+y^{\prime}-2 y=0, $$ we can find its characteristic equation by substituting \(y=e^{rt}\). The characteristic equation is a quadratic equation in the form: $$ r^2 + r - 2 = 0. $$

Solve the characteristic equation

Now, we need to solve for the roots of the characteristic equation: \(r^2 + r - 2 = 0\) Factoring the quadratic equation: \((r + 2)(r - 1) = 0\) Now, we get the two roots, r1 and r2, for the characteristic equation: \(r_1 = -2\) and \(r_2 = 1\)

Write the general solution

Using the roots of the characteristic equation, we can write the general solution for the given differential equation: $$ y(t) = C_1 e^{-2t} + C_2 e^{t} $$ where \(C_1\) and \(C_2\) are constants to be determined using the given initial conditions.

Apply the initial conditions

Using the initial condition \(y(0) = 1\): $$ y(0) = C_1 e^{-2(0)} + C_2 e^{0} => 1 = C_1 + C_2. $$ Using the initial condition \(y^{\prime}(0) = 1\), we first need to find the derivative of \(y(t)\): $$ y^{\prime}(t) = -2C_1 e^{-2t} + C_2 e^{t}. $$ Now, apply the initial condition: $$ y^{\prime}(0) = -2C_1 e^{-2(0)} + C_2 e^{0} => 1 = -2C_1 + C_2. $$ Solving the system of equations for \(C_1\) and \(C_2\), we get \(C_1 = \frac{1}{3}\) and \(C_2 = \frac{4}{3}\). Our unique solution then becomes: $$ y(t) = \frac{1}{3}e^{-2t} + \frac{4}{3}e^{t}. $$

Sketch the graph and describe the behavior

To sketch the graph of the solution \(y(t) = \frac{1}{3}e^{-2t} + \frac{4}{3}e^{t}\), plot the function and observe its behavior as the parameter t increases. As the parameter t increases, the exponential term \(e^{t}\) dominates the behavior of the function, causing it to grow larger. The exponential term \(e^{-2t}\) will decay quickly, becoming negligible. As a result, the overall behavior of the solution as t increases is that it will grow exponentially. Please note that I cannot provide a hand-drawn graph here, but you can plot the function on a graphing calculator or online plotting tool for a visual representation.

Key concepts

Characteristic Equation

In differential equations, the characteristic equation is a powerful tool used to find solutions to linear differential equations with constant coefficients. By transforming the original differential equation into a simpler algebraic problem, we can identify key values that simplify its solution. For instance, let's consider the equation from the problem: \[ y'' + y' - 2y = 0 \]To find the characteristic equation, we substitute \( y = e^{rt} \). This substitution leads to the equation:\[ r^2 + r - 2 = 0 \]This is a quadratic equation in terms of \( r \) and is vital in pinpointing the nature of the solutions to the differential equation. Solving this equation provides us with values for \( r \), which help construct the general solution of the original differential equation.

Initial Value Problem

An initial value problem in differential equations is a task where the solution to the differential equation must satisfy specific conditions at a given point. This problem helps in determining particular solutions from the general solution, allowing us to pinpoint a unique path among many potential solutions.In the example we're examining, the initial conditions are given as:

• \( y(0) = 1 \)
• \( y'(0) = 1 \)

These conditions are crucial as they determine the constants in the general solution of the differential equation. Applying these conditions ensures that the solution fits the specific scenario described by the initial values.

Exponential Growth

Exponential growth is a concept often encountered in solutions to differential equations. It refers to the rapid increase of a function as a result of an exponent in its expression, often represented by terms like \( e^t \).In the problem example, the solution is:\[ y(t) = \frac{1}{3}e^{-2t} + \frac{4}{3}e^{t} \]Here, the term \( \frac{4}{3}e^{t} \) depicts exponential growth. This growth becomes more pronounced as \( t \) increases, while the other term \( \frac{1}{3}e^{-2t} \) will decay exponentially. As \( t \) continues to rise, the exponential growth term \( e^{t} \) will significantly influence the long-term behavior of the function, causing it to increase without bound.

General Solution

The general solution of a differential equation is the expression that includes all possible solutions, showing how they depend on arbitrary constants. These constants can later be adjusted based on given initial or boundary conditions, leading to a unique particular solution.For the differential equation:\[ y'' + y' - 2y = 0 \]with the characteristic roots \( r_1 = -2 \) and \( r_2 = 1 \), the general solution is formulated as:\[ y(t) = C_1 e^{-2t} + C_2 e^{t} \]Here, \( C_1 \) and \( C_2 \) are constants that represent the infinite family of solutions. The specific solution to the initial problem is found by using given conditions to set these constants, allowing for a complete description of the system described by the differential equation.