Question

The \(t\)-interval of interest is \(-\infty

Short answer

Based on the solution process, the unique solution to the given initial value problem is y(t) = 2e^{2t}. The given functions, y_1(t) = e^{2t} and y_2(t) = te^{2t}, are verified as solutions to the differential equation and form a fundamental set of solutions since their Wronskian is nonzero. Therefore, using the provided initial conditions of y(0) = 2 and y'(0) = 0, the unique solution is found to be y(t) = 2e^{2t}.

Step-by-step solution

Verify Solutions

To verify that y_1(t) = e^{2t} and y_2(t) = te^{2t} are solutions to the given equation y'' - 4y' + 4y = 0, we must compute their first and second derivatives and substitute them into the equation. The derivatives are: y_1(t) = e^{2t} y_1'(t) = 2e^{2t} y_1''(t) = 4e^{2t} y_2(t) = te^{2t} y_2'(t) = e^{2t} + 2te^{2t} y_2''(t) = 4e^{2t} + 4te^{2t} Now, substitute y_1, y_1' and y_1'' in the given equation: (4e^{2t}) - 4(2e^{2t}) + 4(e^{2t}) = 0 => 0 = 0 Now, substitute y_2, y_2' and y_2'' in the given equation: (4te^{2t} + 4e^{2t}) - 4(e^{2t} + 2te^{2t}) + 4(te^{2t}) = 0 => 0 = 0 Since the given functions satisfy the differential equation, they are solutions.

Calculate Wronskian

In order to check if the two solutions form a fundamental set, we must calculate their Wronskian, W(y_1, y_2). The Wronskian, in this case, is given by: W(y_1, y_2) = | \begin{matrix} y_1 & y_2 \\ y_1' & y_2' \end{matrix} | = | \begin{matrix} e^{2t} & te^{2t} \\ 2e^{2t} & e^{2t} + 2te^{2t} \end{matrix} | W(y_1, y_2) = e^{2t}(e^{2t} + 2te^{2t}) - te^{2t}(2e^{2t}) = e^{4t} Since W(y_1, y_2) is never zero, the functions form a fundamental set of solutions.

Find the Unique Solution

Since the given functions form a fundamental set of solutions, we can now find the unique solution to the initial value problem. The general solution for a second-order homogeneous linear differential equation can be written as: y(t) = C_1y_1(t) + C_2y_2(t) For this problem, our general solution is: y(t) = C_1e^{2t} + C_2te^{2t} We are given y(0) = 2 and y'(0) = 0. By substituting these values, we can determine the constants C_1 and C_2. Using the initial condition y(0) = 2: 2 = C_1e^{0} + C_2(0)e^{0} 2 = C_1 => C_1 = 2 Now we find the derivative of the general solution: y'(t) = 2e^{2t} + e^{2t} + 2te^{2t} Using the initial condition y'(0) = 0: 0 = 2e^{0} + e^{0} + 2(0)e^{0} 0 = 3 => C_2 = 0 Given C_1 = 2 and C_2 = 0, the unique solution is: y(t) = 2e^{2t}

Key concepts

Wronskian

The Wronskian is a crucial determinant used in the field of differential equations to understand the relationship between different solutions. It involves calculating a determinant of a matrix, where the rows consist of the solutions and their derivatives. To put it in simple terms, this determinant helps us to find out if a set of functions are linearly independent. In the context of a differential equation, for solutions to be considered a fundamental set, their Wronskian must not be zero at any point in the given interval. In our exercise, the Wronskian for functions y_1(t) = e^{2t} and y_2(t) = te^{2t} is calculated and found to be e^{4t}, which is never zero for any real value of t. This indicates that the solutions are indeed linearly independent and form a fundamental set, bringing us one step closer to solving the differential equation at hand.

Fundamental Set of Solutions

A fundamental set of solutions refers to a set of solutions to a homogeneous differential equation that are linearly independent. This concept is like having different tools in your toolbox, each with their own unique function, yet all essential. With differential equations, we often need multiple solutions to cover all possible scenarios, and these solutions form the building blocks for more general solutions. In our textbook exercise, we've verified that the functions y_1 and y_2 are indeed a fundamental set. This means any solution to our differential equation can be expressed as a combination of these two functions, a concept cornerstone to understanding and solving differential equations.

Initial Value Problem

An initial value problem is a flavor of differential equations where not only do you need to find a general solution, but one that specifically matches a set of initial conditions. Imagine being given a point on a map and tasked with finding a path that starts exactly there; that's akin to solving an initial value problem. In differential equations, initial conditions provide specific values for the function and often its derivatives at a certain point. Our exercise provided us these initial conditions, and using the fundamental set of solutions, we were able to deduce the precise journey our equation takes from those starting values, leading us to the unique solution that fits like a key in a lock.

Homogeneous Linear Differential Equation

A homogeneous linear differential equation is akin to a blank canvas in art—the starting point for creating something complex from something simple. It's a type of equation where the combination of a function and its derivatives equals zero. These equations are homogeneous because there are no standalone terms without the function or its derivatives—every term involves the function, y(t), or one of its derivatives. Linear indicates that the function and its derivatives are to the first power only and not raised to any higher exponent. Our exercise showcases a textbook example of such an equation, with y'', y', and y mingling together, adding up to a quiet zero. To solve these puzzles, we seek a fundamental set of solutions that meld together, respecting the given initial conditions to paint the full picture—the unique solution that describes the system perfectly.