Question

Solve the initial value problem. $$ y^{\prime \prime}-2 y^{\prime}+2 y=-e^{x}(6 \cos x+4 \sin x), \quad y(0)=1, y^{\prime}(0)=4 $$

Short answer

Question: Solve the initial value problem (IVP) $$y^{\prime \prime}-2 y^{\prime}+2 y=-e^{x}(6\cos x+4\sin x)$$ with initial conditions $$y(0) = 1$$ and $$y^{\prime}(0) = 4$$. Answer: The specific solution for the given IVP is $$y(x) = e^x(\cos x + B\sin x) + xe^x(4\cos x + 3\sin x)$$ where B has no effect on the initial conditions and can be treated as a free parameter.

Step-by-step solution

Find the complementary function

First, we solve the homogeneous part of the given differential equation, which is: $$ y^{\prime \prime}-2 y^{\prime}+2 y=0 $$ The corresponding auxiliary equation is: $$ m^2 - 2m + 2 = 0 $$ Solve for m: $$ m = \frac{-(-2) \pm \sqrt{(-2)^2 - 4\cdot1\cdot2}}{2\cdot1} = 1 \pm i $$ Since we have complex roots, the complementary function is: $$ y_c = e^x(A\cos x + B\sin x) $$

Find the particular solution using undetermined coefficients

We have the inhomogeneous part of the given differential equation: $$ y^{\prime \prime}-2 y^{\prime}+2 y=-e^{x}(6\cos x+4\sin x) $$ Guess a particular solution in the form: $$ y_p = xe^x(C\cos x + D\sin x) $$ Now, find the first and second derivatives of y_p: $$ y_p^{\prime} = e^x[(C\cos x + D\sin x)+x((-C\sin x+D\cos x)] $$ $$ y_p^{\prime \prime} = e^x[(-C\sin x+D\cos x)+(-C\sin x+D\cos x)+x(-C\cos x-D\sin x)] $$ Substituting y_p, y_p', and y_p'' into the given differential equation, we get: $$ -e^x[(-C\cos x-D\sin x)]-2 e^x[(C\cos x + D\sin x)+x((-C\sin x+D\cos x)] + 2e^x(x(C\cos x + D\sin x)) = -e^x(6\cos x+4\sin x) $$ Solve for C and D: $$ C = 4, \qquad D = 3 $$ So, the particular solution is: $$ y_p = xe^x(4\cos x + 3\sin x) $$

Find the general solution

The general solution is the sum of the complementary function and the particular solution: $$ y(x) = y_c + y_p = e^x(A\cos x + B\sin x) + xe^x(4\cos x + 3\sin x) $$

Use the initial conditions to find the specific solution

Now, we will use the initial conditions y(0) = 1 and y'(0) = 4 to find the constant coefficients A and B: Substitute x = 0 into y(x) equation: $$ y(0) = 1 = e^0(A\cos 0 + B\sin 0) + 0 = A $$ So, A = 1. For the first derivative, we already found: $$ y^{\prime} = e^x[(C\cos x + D\sin x)+x((-C\sin x+D\cos x)] $$ Substitute x = 0 into y'(x) equation: $$ y^{\prime}(0) = 4 = e^0\left[4\cdot \cos(0) + 3\cdot \sin(0)\right] = 4 $$ Since it is equal to 4, there is no need to find B because the equation already holds true. The specific solution is: $$ y(x) = e^x(\cos x + B\sin x) + xe^x(4\cos x + 3\sin x) $$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In essence, they represent the relationship between a function and the rate of change of that function. The equation given in the exercise,
\( y'' - 2y' + 2y = -e^{x}(6\cos x + 4\sin x) \),
is a second-order linear differential equation with non-constant coefficients on the right-hand side, which signifies some external force influencing the system. To solve such a differential equation, you need to find a function \( y(x) \) whose derivatives satisfy the equation for all \( x \).

The solution process typically involves finding a general solution that handles the homogeneous equation (without the external force part), and then a particular solution that is designed to fit the non-homogeneous part of the equation. The 'initial value problem' aspect of this means you're given conditions (\(y(0) = 1, y'(0) = 4\)) that the solution must satisfy at a particular point (here \(x = 0\)).

Complementary Function

A complementary function, denoted as \( y_c \), is the solution to the homogeneous part of a differential equation. In our exercise, this refers to the equation \( y'' - 2y' + 2y = 0 \).
The purpose of finding the complementary function is to capture the behavior of the differential equation's variable \( y \) without the influence of the external inhomogeneous part.

The approach for finding \( y_c \) involves solving the characteristic equation and using the roots to construct general solutions. If the roots are complex, as they are in this case, the complementary function will include trigonometric functions and the base function \( e^{x} \) due to Euler's formula \( e^{ix} = \cos(x) + i\sin(x) \). The complementary function here is \( y_c = e^{x}(A\cos x + B\sin x) \) where \( A \) and \( B \) are constants determined by the initial conditions.

Particular Solution

The particular solution, denoted as \( y_p \), is the solution to the non-homogeneous part of the differential equation which in this case includes the term \( -e^{x}(6\cos x + 4\sin x) \).
Whereas the complementary function captures the inherent behavior of the system described by the differential equation, the particular solution accounts for the external influence, represented by the non-homogeneous part. To find the particular solution, one typically makes an educated guess about the form it will take, based on the form of the inhomogeneous part.

In our problem, due to the presence of \( e^{x}\cos x \) and \( e^{x}\sin x \) in the non-homogeneous term, an appropriate guess is \( y_p = xe^{x}(C\cos x + D\sin x) \), with new constants \( C \) and \( D \). This extra \( x \) is necessary to accommodate the similarity between the non-homogeneous part and the complementary function. After calculating derivatives and substituting them back into the original equation, the actual values of \( C \) and \( D \) are obtained.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions to non-homogeneous linear differential equations. This method works well when the non-homogeneous part is a linear combination of functions whose derivatives are also linear combinations of the same set of functions.

To apply this method, you propose a particular solution with undetermined coefficients, that is similar to the function in the non-homogeneous part of the equation. Then, you substitute this guess, and its derivatives, into the differential equation and solve for the unknown coefficients.
In our case, to solve for the constants \( C \) and \( D \) in the expression \( y_p = xe^x(C\cos x + D\sin x) \), we substitute \( y_p \) and its derivatives into the original differential equation and equate coefficients of like terms. This enables us to solve for \( C \) and \( D \) as real numbers that make the equation valid, which leads us to the correct form of the particular solution.