Question

(a) Verify that \(y_{1}=e^{x}\) and \(y_{2}=x e^{x}\) are solutions of $$y^{\prime \prime}-2 y^{\prime}+y=0$$ on \((-\infty, \infty)\) (b) Verify that if \(c_{1}\) and \(c_{2}\) are arbitrary constants then \(y=e^{x}\left(c_{1}+c_{2} x\right)\) is a solution of (A) on \((-\infty, \infty)\) (c) Solve the initial value problem $$y^{\prime \prime}-2 y^{\prime}+y=0, \quad y(0)=7, \quad y^{\prime}(0)=4$$ (d) Solve the initial value problem $$ y^{\prime \prime}-2 y^{\prime}+y=0, \quad y(0)=k_{0}, \quad y^{\prime}(0)=k_{1} $$

Short answer

Question: Verify the given solutions $y_1 = e^x$ and $y_2 = xe^x$ for the differential equation $y'' - 2y' + y = 0$. Verify the general solution $y = e^x (c_1 + c_2x)$. Then, solve the initial value problems: 1. $y(0) = 7, y'(0) = 4$ 2. $y(0) = k_0, y'(0) = k_1$ Answer: -verified- Both $y_1 = e^x$ and $y_2 = xe^x$ are solutions to the given differential equation. -verified- The general solution $y = e^x (c_1 + c_2x)$ is valid. Solution for initial value problem 1: $$y = e^x (7 - 3x).$$ Solution for initial value problem 2: $$y = e^x (k_0 + (k_1 - k_0)x).$$

Step-by-step solution

Verify the given solutions \(y_1 = e^x\) and \(y_2 = xe^x\)

To verify if \(y_1 = e^x\) and \(y_2 = xe^x\) are solutions to the differential equation, plug each of them into the equation and see if the equation holds. For \(y_1 = e^x\), let's first find its first derivative, \(y^{'}_1\), and second derivative, \(y^{''}_1\): $$ y^{'}_1 = \frac{d}{dx}(e^x) = e^x \\ y^{''}_1 = \frac{d^2}{dx^2}(e^x) = e^x. $$ Now plug \(y_1\), \(y^{'}_1\), and \(y^{''}_1\) into the equation: $$ (e^x) - 2(e^x) + (e^x) = 0. $$ This equation holds true, so \(y_1 = e^x\) is a solution to the given differential equation. For \(y_2 = xe^x\), let's find its first derivative, \(y^{'}_2\), and second derivative, \(y^{''}_2\): $$ y^{'}_2 = \frac{d}{dx}(xe^x) = e^x + xe^x \\ y^{''}_2 = \frac{d^2}{dx^2}(xe^x) = 2e^x + xe^x. $$ Now plug \(y_2\), \(y^{'}_2\), and \(y^{''}_2\) into the equation: $$ (2e^x + xe^x) - 2(e^x + xe^x) + (xe^x) = 0. $$ This equation holds true, so \(y_2 = xe^x\) is a solution to the given differential equation.

Verify the general solution \(y = e^x (c_1 + c_2x)\)

Having verified that \(y_1 = e^x\) and \(y_2 = xe^x\) are both solutions, we will now verify that the general solution \(y = e^x (c_1 + c_2x)\) also holds true. To do this, we will find the first and second derivatives of \(y\) and substitute them into the equation. First, find the first derivative, \(y'\), and the second derivative, \(y''\), of the given function \(y\): $$ y' = \frac{d}{dx}(e^x (c_1 + c_2x)) = e^x(c_1 + c_2x) + c_2e^x \\ y'' = \frac{d^2}{dx^2}(e^x (c_1 + c_2x)) = e^x(c_1 + c_2x) + 2c_2e^x. $$ Now, substitute \(y\), \(y'\), and \(y''\) into the differential equation: $$ (e^x(c_1 + c_2x) + 2c_2e^x) - 2(e^x(c_1 + c_2x) + c_2e^x) + (e^x(c_1 + c_2x)) = 0. $$ This equation holds true as well, so the general solution \(y = e^x (c_1 + c_2x)\) is valid.

Solve initial value problem 1

The initial conditions we are given for the first problem are: $$ y(0) = 7 \\ y'(0) = 4. $$ Apply the initial conditions to the general solution and its derivative, and solve for \(c_1\) and \(c_2\): $$ y(0) = e^0 (c_1 + c_2\cdot0) = 7 \Rightarrow c_1 = 7 \\ y'(0) = e^0(7 + 0) + 0 = 4 \Rightarrow c_2 = -3. $$ Thus, the particular solution for this initial value problem is: $$ y = e^x (7 - 3x). $$

Solve initial value problem 2

The initial conditions we are given for the second problem are: $$ y(0) = k_0 \\ y'(0) = k_1. $$ Apply the initial conditions to the general solution and its derivative, and solve for \(c_1\) and \(c_2\): $$ y(0) = e^0 (c_1 + c_2\cdot0) = k_0 \Rightarrow c_1 = k_0 \\ y'(0) = e^0(k_0 + 0) + 0 = k_1 \Rightarrow c_2 = k_1 - k_0. $$ Thus, the particular solution for this initial value problem is: $$ y = e^x (k_0 + (k_1 - k_0)x). $$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. When we talk about differential equations, we refer to the relationship between functions that represent physical quantities and their rates of change. In essence, these equations help us understand how a particular quantity changes over time or space.

Importance in Various Fields

Differential equations are pivotal in physics, engineering, economics, biology, and many other fields. They model everything from how populations grow to how heat moves through materials.

For instance, in physics, differential equations describe how the position of an object changes over time, giving us insights into velocity and acceleration—the foundation of Newton's laws of motion.

Homogeneous Equations

A homogeneous differential equation is a specific type of differential equation that possesses a unique trait: if you take any solution to the equation and multiply it by a constant, the result is also a solution to the equation.

Characteristics

Homogeneous equations maintain balance, or 'homogeneity,' meaning their terms are of the same order. The equation in our exercise, for example, is a second-order linear homogeneous differential equation because each term involves the function or its derivatives, and they all appear to the second order.

The solutions to these equations, as in the exercise, usually involve exponential functions or complex trigonometric functions. This ties back to the property that the sum of any two solutions to a homogeneous equation is also a solution—suggesting a linearity that is quintessential to linear algebra concepts.

Second Order Linear Differential Equation

A second order linear differential equation is a form of differential equation that includes the second derivative of a function and no higher derivatives. The general form is given by \( ay'' + by' + cy = 0 \), where \( a \), \( b \), and \( c \) are constants.

Initial Value Problems

When we pair such an equation with initial conditions, we create an initial value problem. Solving it involves integrating twice and applying the initial conditions to determine specific constant values. The initial value problems from the exercise illustrate the process—we start with our general solution, incorporate the initial values, and find a solution that fits these constraints—tailoring the general answer to a specific scenario.

Understanding these second order linear differential equations is vital in physics and engineering because they often model systems in equilibrium, such as springs, circuits, or pendulums.