Question

Use the Method of Variation of Parameters to determine the general solution for the following problems. a. \(y^{\prime \prime}+y=\tan x\). b. \(y^{\prime \prime}-4 y^{\prime}+4 y=6 x e^{2 x}\).

Short answer

Part a of the problem cannot be solved by the method of variation of parameters due to zero Wronskian. The solution for part b is \(y = e^{2x}(c_1 + c_2 x) - 3e^{2x}(x^2 e^{2x} - 2x e^{2x} + 2e^{2x}) + 3xe^{2x}(x e^{2x} - e^{2x})\).

Step-by-step solution

Solving Part a

The complementary function for the equation \(y^{\prime \prime}+y=\tan x\) is given by the solution of the homogeneous differential equation \(y^{\prime \prime}+y=0\), which is \(y = c_1 \cos x + c_2 \sin x\). Then, calculate the Wronskian, \(W = y_1y_2^{\prime} - y_2y_1^{\prime} = \sin x \cos x - \cos x \sin x = 0 \), which is not suitable for Variation of Parameters formula, thus, this equation can't be solved through Variation of Parameters.

Solving Part b

For the equation \(y^{\prime \prime}-4 y^{\prime}+4 y=6 x e^{2 x}\), the complementary function is given by the solution of the homogeneous differential equation \(y^{\prime \prime}-4 y^{\prime}+4 y=0\), which is \(y = e^{2x}(c_1 + c_2 x)\). Calculate the Wronskian, \(W=y_1y_2'-y_2y_1'\), which results in \(W = e^{4x}\). The method of variation of parameters requires us to compute the functions \(v_1\) and \(v_2\), which are given as \(v_1 = -\int\frac{y_2f}{W}dx\) and \(v_2 = \int\frac{y_1f}{W}dx\). For this problem, we then have \(v_1 = -\int\frac{x e^{4x} 6x e^{2x}}{e^{4x}} dx = -3 \int x^2 e^{2x} dx\) and \(v_2 = \int\frac{e^{4x} 6x e^{2x}}{e^{4x}} dx= 3\int x e^{2x} dx\). The result after integration by parts on \(v_1\) and \(v_2\) is \(v_1 = -3(x^2 e^{2x} - 2x e^{2x} + 2e^{2x})\), and \(v_2 = 3(x e^{2x} - e^{2x})\). The general solution would be \(y = y_1v_1 + y_2v_2 = e^{2x}(c_1 + c_2 x) - 3e^{2x}(x^2 e^{2x} - 2x e^{2x} + 2e^{2x}) + 3xe^{2x}(x e^{2x} - e^{2x})\).

Key concepts

Variation of Parameters

The method of variation of parameters provides a technique for finding particular solutions to non-homogeneous differential equations. It's particularly useful when other methods, like undetermined coefficients, are not applicable. Here’s how it works in broad terms:

• Firstly, identify the complementary function, which solves the associated homogeneous differential equation.
• Second, calculate the Wronskian of the solutions forming the complementary function.
• Finally, find the particular solution by using integrals that involve the Wronskian and the differential equation's non-homogeneous part.

The power of variation of parameters lies in its general applicability, but it does require solving potentially complex integrals. This method is an essential tool for tackling equations where non-homogeneous terms don't fit simpler systematic approaches. It's important to note: if the Wronskian is zero, like in part a of the exercise, the variation of parameters method cannot be used.

Complementary Function

The complementary function (CF) is a crucial component in solving linear differential equations. It addresses the homogeneous part of a differential equation. To find the CF, solve the homogeneous differential equation, which has the same left-hand side as the original equation but equated to zero.

For instance, in problem a, the homogeneous equation is given by:
\[ y'' + y = 0 \]
This leads to two linearly independent solutions, typically involving exponential functions or trigonometric functions depending on the equation’s form, forming the CF. In this case, the CF is:
\[ y = c_1 \cos x + c_2 \sin x \]

The constants \( c_1 \) and \( c_2 \) are determined by initial conditions or boundary values. Understanding the CF is essential because it defines the behavior of the solution in absence of external forces represented by the non-homogeneous part.

Homogeneous Differential Equation

A homogeneous differential equation is one where every term depends on the unknown function or its derivatives, but not on independent variables alone. In mathematical terms, an equation is homogeneous if written as:
\[ L(y) = 0 \]
\(L(y)\) being a linear differential operator.

The homogeneous differential equation represents a system without external or forcing inputs. Solutions to these equations are crucial as they form the basis for deriving the complementary function, central to solving non-homogeneous equations.

Homogeneous solutions often reflect the system's natural response based on its inherent properties. For example, for the equation \(y'' - 4y' + 4y = 0\), exponential functions appear naturally in its solutions, indicating growth or decay dynamics.

Wronskian

The Wronskian is a determinant used to assess the linear independence of solution functions for differential equations. If functions are linearly independent, their Wronskian, when evaluated, does not equal zero. This property is vital for verifying the independence of solutions in forming a complementary function and for using the variation of parameters method.

To compute the Wronskian, consider two solutions \(y_1\) and \(y_2\) of a second-order differential equation. The Wronskian \(W\) is given by the determinant:
\[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y'_1 & y'_2 \end{vmatrix} = y_1y'_2 - y_2y'_1 \]

In the problem, the complementary solutions are derived, and the Wronskian calculated, helps ascertain their suitability for parameter variation. When the Wronskian equals zero, as in problem a, it indicates a degenerative case where variation of parameters might not proceed as expected due to lack of a basis for determining unique solutions.