Question

In Exercises \(1-14\) find a particular solution. $$ y^{\prime \prime}-4 y^{\prime}+4 y=e^{2 x}\left(1-3 x+6 x^{2}\right) $$

Short answer

Question: Find the general solution of the given second-order linear differential equation: $$ y'' - 4y' + 4y = e^{2x}(1 - 3x + 6x^2). $$ Answer: The general solution for the given differential equation is: $$ y(x) = C_1e^{2x} + C_2xe^{2x} + \frac{1}{6}e^{2x} - \frac{1}{6}e^{2x}x + \frac{3}{4}e^{2x}x^2, $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Finding the Complementary Solution

First, we need to find the complementary solution for the homogeneous differential equation: $$ y'' - 4y' + 4y = 0. $$ To do this, we will make a characteristic equation. Let \(m\) be the characteristic root, then the characteristic equation is given by: $$ m^2 - 4m + 4 = 0. $$ This equation can be factored as: $$ (m - 2)^2 = 0. $$ The double root is \(m = 2\). So, the complementary solution is given by: $$ y_c(x) = C_1e^{2x} + C_2xe^{2x}. $$

Finding the Particular Solution

We will now use the method of undetermined coefficients to find the particular solution of the given differential equation: $$ y'' - 4y' + 4y = e^{2x}(1 - 3x + 6x^2). $$ Since the right-hand side of the equation is of the format \(e^{ax}(Ax^2 + Bx + C)\), we assume the particular solution is of the form: $$ y_p(x) = A(x)e^{2x} + B(x)e^{2x}x + C(x)e^{2x}x^2. $$ Let’s find the first and second derivatives of \(y_p(x)\): $$ y_p'(x) = Ae^{2x}(1 + 2x) + Be^{2x}x(1 + 2x) + 2Cxe^{2x}(1 + 2x) + 4Cx^2e^{2x}. $$ $$ y_p''(x) = 2Ae^{2x}(1 + 2x) + 2Ae^{2x}(2) + 2Be^{2x}(1 + 2x) + 4Be^{2x}x + 8Ce^{2x}(1 + 2x) + 12Ce^{2x}x + 8Cxe^{2x}(2). $$ Now, we will plug \(y_p(x)\), \(y_p'(x)\), and \(y_p''(x)\) back into the given differential equation, and match corresponding terms to solve for \(A\), \(B\), and \(C\). $$ (2A+2B+8C)e^{2x}(1+2x) + (4A+4B+24C)e^{2x} = e^{2x}(1-3x+6x^2). $$ After simplifying the equation we get: $$ (6A-2B+8C)e^{2x} + (16B+24C)e^{2x}x = (1 - 3x + 6x^2)e^{2x}. $$ Now we can match coefficients: $$ 6A - 2B + 8C = 1 $$ $$ 16B + 24C = -3 $$ $$ 8C = 6 $$ Solving this system of equations, we find: $$ A = \frac{1}{6} $$ $$ B = -\frac{1}{6} $$ $$ C = \frac{3}{4} $$ Therefore, the particular solution is: $$ y_p(x) = \frac{1}{6}e^{2x} - \frac{1}{6}e^{2x}x + \frac{3}{4}e^{2x}x^2. $$

Combining the Complementary and Particular Solutions

Now, we will add the complementary and particular solutions to get the general solution for the given differential equation: $$ y(x) = y_c(x) + y_p(x) = C_1e^{2x} + C_2xe^{2x} + \frac{1}{6}e^{2x} - \frac{1}{6}e^{2x}x + \frac{3}{4}e^{2x}x^2. $$

Key concepts

Particular Solution

The particular solution of a differential equation is a function that satisfies the non-homogeneous equation. This means it includes not just the influence of the homogeneous part, but also captures the behavior dictated by any external forces or inputs. In the given exercise, the differential equation is:\[ y'' - 4y' + 4y = e^{2x}(1 - 3x + 6x^2). \]The solution comprises two parts: the complementary solution and the particular solution. The particular solution specifically addresses the external input on the right side of the equation, which in this case, is a polynomial multiplied by an exponential function.The particular solution is typically found using methods like undetermined coefficients or variation of parameters. Here, we use the method of undetermined coefficients to propose a form for the solution and determine its coefficients by plugging it into the differential equation.This results in:\[ y_p(x) = \frac{1}{6}e^{2x} - \frac{1}{6}e^{2x}x + \frac{3}{4}e^{2x}x^2.\]This function, when substituted back in the equation, balances the left and right sides, showing the impact of the non-homogeneous part.

Complementary Solution

The complementary solution refers to the part of the solution to a differential equation that addresses the homogeneous aspect of the equation. In mathematical terms, the homogeneous equation derived from the exercise is:\[ y'' - 4y' + 4y = 0. \]To find this, you solve the characteristic equation derived from the homogeneous equation. For this exercise, the characteristic equation is:\[ m^2 - 4m + 4 = 0. \]Factoring the equation gives:\[ (m - 2)^2 = 0. \]Here, we have a repeated root \( m = 2 \). With repeated roots, the complementary solution is formed using combinations of exponential functions:

• \( C_1e^{2x} \) addresses the first root.
• \( C_2xe^{2x} \) accommodates the second, repeated root.

Thus, the complementary solution, representing the system's natural response without external inputs, is:\[ y_c(x) = C_1e^{2x} + C_2xe^{2x}. \]

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find the particular solution of linear differential equations with constant coefficients. It works well when the non-homogeneous part of the differential equation is of a suitable form, such as polynomial, exponential, sine, cosine, or a combination of these.In this exercise, the non-homogeneous term \( e^{2x}(1 - 3x + 6x^2) \) suggests a trial solution of the form:\[ y_p(x) = A e^{2x} + Bxe^{2x} + Cx^2e^{2x}. \]After proposing this form, differentiate it to get \( y_p'(x) \) and \( y_p''(x) \). Substitute these back into the original differential equation. By equating the coefficients of like terms from both sides, solve for \( A, B, \) and \( C \). This analysis is crucial because it ensures the particular solution fits precisely into the given equation.

Characteristic Equation

The characteristic equation is a polynomial equation that provides the roots needed to determine the complementary solution of a linear differential equation. For a second-order linear differential equation of the form:\[ ay'' + by' + cy = 0, \]we assume a solution of the form \( y = e^{mx} \), and upon substituting into the equation, we derive:\[ am^2 + bm + c = 0. \]This is the characteristic equation where \( m \) represents characteristic roots.In the example exercise, the characteristic equation is:\[ m^2 - 4m + 4 = 0. \]Solving this yields a double root \( m = 2 \). Repeated roots affect the complementary solution structure, as it implies additional terms, like \( xe^{2x} \), to ensure all possible solutions are covered.Thus, roots from the characteristic equation directly inform the structure of the complementary solution, crucial for solving differential equations effectively.