Problem 36 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem: $$\begin{aligned} &\left(D^{2}+D-2\right) y=4 \cos x-2 \sin x,\\\ &y(0)=-1, y^{\prime}(0)=4. \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the given initial-value problem is: \(y(x) = \frac{5}{3}e^{x} + 2e^{-2x} -\frac{4}{3}\cos x + \sin x\)

Step 1: Solve the Homogeneous Equation

Find the solution for the homogeneous equation: \(\left(D^{2}+D-2\right) y = 0\) This equation is a quadratic equation of the form \(ar^2 + br + c = 0\), with the following coefficients: \(a = 1\), \(b = 1\), and \(c = -2\). We can find the roots of this quadratic equation using the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Substitute the coefficients into the formula and find the roots: \(r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)}\) \(r = \frac{-1 \pm \sqrt{9}}{2}\) Thus, the roots are given by \(r_1 = \frac{-1 + 3}{2} = 1\) and \(r_2 = \frac{-1 - 3}{2} = -2\). The complementary function (the solution to the homogeneous equation) is given by: \(y_c(x) = C_1e^{r_1x} + C_2e^{r_2x} = C_1e^{x} + C_2e^{-2x}\)

Step 2: Find a Particular Solution

We will now find a particular solution for the given non-homogeneous equation: \(\left(D^{2}+D-2\right) y = 4 \cos x - 2 \sin x\) We can assume a particular solution of the form: \(y_p(x) = A\cos x + B\sin x\) We must then find the first and second derivatives of \(y_p(x)\): \(y_p'(x) = -A\sin x + B\cos x\) \(y_p''(x) = -A\cos x - B\sin x\) Substitute the particular solution and its derivatives into the given non-homogeneous equation: \((-A\cos x - B\sin x) + (-A\sin x + B\cos x) - 2(A\cos x + B\sin x) = 4\cos x - 2\sin x\) Rearrange the equation: \((-3A\cos x - 2B\sin x) = 4\cos x - 2\sin x\) Now, equate the coefficients to find the values of \(A\) and \(B\): \(-3A = 4\) => \(A = -\frac{4}{3}\) \(-2B = -2\) => \(B = 1\) So, the particular solution is: \(y_p(x) = -\frac{4}{3}\cos x + \sin x\)

Step 3: Write the General Solution

Combine the complementary function and particular solution to write the general solution: \(y(x) = y_c(x) + y_p(x) = C_1e^{x} + C_2e^{-2x} -\frac{4}{3}\cos x + \sin x\)

Step 4: Apply Initial Conditions

Apply the given initial conditions to determine the constants \(C_1\) and \(C_2\): 1. \(y(0) = -1\) \(-1 = C_1e^{0} + C_2e^{-0} -\frac{4}{3}\cos(0) + \sin(0) = C_1 + C_2 - \frac{4}{3}\) 2. \(y'(0) = 4\) \(4 = e^{0}(C_1 - 2C_2) -\frac{4}{3}\sin(0) + \cos(0) = C_1 - 2C_2 + 1\) From the first condition, we can express \(C_2\) in terms of \(C_1\): \(C_2 = C_1 + \frac{1}{3}\) Substitute the expression for \(C_2\) into the second condition: \(4 = C_1 - 2(C_1 + \frac{1}{3}) + 1\) Solving for \(C_1\): \(C_1 = \frac{5}{3}\) Next, find \(C_2\): \(C_2 = C_1 + \frac{1}{3} = \frac{5}{3} + \frac{1}{3} = 2\)

Step 5: Write the Final Solution

Substitute the values of \(C_1\) and \(C_2\) into the general solution: \(y(x) = \frac{5}{3}e^{x} + 2e^{-2x} -\frac{4}{3}\cos x + \sin x\) This is the solution to the given initial-value problem.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In essence, they represent physical phenomena where the rate of change of one variable is proportional to another. For example, they can describe how populations grow, how heat moves, or how springs oscillate.

In the exercise given, the differential equation is \(D^{2}+D-2)y = 4 \cos x - 2 \sin x\), which is an example of a second-order linear differential equation with non-constant coefficients. This equation indicates that the acceleration of the system (represented by the second derivative) plus its velocity (first derivative) minus twice the position (the function itself) is equal to a forcing function comprising a mix of cosine and sine functions. Students typically work through these equations by first finding the general solution to the associated homogeneous equation and then finding a specific solution that accounts for the external forces (the non-homogeneous part).

Homogeneous Equation

A homogeneous equation is a differential equation where the terms without derivatives can be factored out, leaving no free-standing constants or functions. In other words, all terms are multiples of the function or its derivatives, like \(D^{2} + D - 2)y = 0\).

Often, solving these basic forms is the starting point for tackling more complex equations. The exercise demonstrates how to solve such an equation. First, we consider the characteristic equation, which, in this case, is a quadratic equation representing the exponentials in the solution. After finding the roots of this quadratic, two distinct exponentials emerge, which comprise the complementary or homogeneous solution to the differential equation. This solution represents the 'natural' behavior of the system without any external forces.

Particular Solution

The particular solution to a differential equation accounts for the non-homogeneous part, the external forcing or non-zero right-hand side. It's found by assuming a trial solution that shares a form with the non-homogeneous part. In our case, since the right-hand side is a linear combination of sine and cosine, we look for a particular solution that is also a linear combination of these functions.

In the step-by-step breakdown, a chosen form for the particular solution is substituted into the differential equation. The outcome allows the identification of constants which make the assumed form a valid solution. This solution \(y_p(x) = -\frac{4}{3}\cos x + \sin x\)) is combined with the complementary function to give a general solution to the original non-homogeneous equation. These steps outline a systematic approach to tackle initial-value problems involving differential equations.