Question

In Exercises \(15-19\) find the general solution. $$ y^{\prime \prime}-2 y^{\prime}+y=e^{x}(2-12 x) $$

Short answer

Question: Determine the general solution of the following second-order linear nonhomogeneous differential equation: $$y^{\prime\prime} - 2y^{\prime} + y = e^{x}(2-12x)$$ Answer: The general solution is $$y(x) = C_1e^x + C_2xe^x + 2xe^{x}(x+\frac{1}{2})$$, where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Find the complementary solution

To find the complementary solution, we need to first determine the general solution of the associated homogeneous equation: $$ y^{\prime\prime} - 2y^{\prime} + y = 0 $$ This is a second-order linear homogeneous differential equation with constant coefficients. To solve this type of equation, we need to find the characteristic equation. The characteristic equation can be written as: $$ r^2 - 2r + 1 = 0 $$

Solve the characteristic equation

Now, we need to solve this quadratic equation for \(r\). By factoring or using the quadratic formula, we obtain: $$ (r - 1)^2 = 0 $$ The roots are \(r_1 = 1\) and \(r_2 = 1\). Both roots are equal, so our complementary solution is: $$ y_c(x) = C_1e^{x} + C_2xe^{x} $$ Where \(C_1\) and \(C_2\) are constants.

Choose a form for the particular solution

Next, we need to determine a form for the particular solution. The nonhomogeneous term of the given equation is: $$ e^{x}(2-12x) $$ Considering this form, we can assume that our particular solution will have the following form: $$ y_p(x) = Axe^{x}(x+B) $$ Where \(A\) and \(B\) are coefficients to be determined.

Determine the coefficients in the particular solution

To determine the coefficients in the particular solution, we need to find \(y_p'(x)\) and \(y_p''(x)\), then substitute them back into the given nonhomogeneous equation. $$ y_p'(x) = A(e^x + xe^x) + A(B-1)e^xx + ABe^x $$ $$ y_p''(x) = A(e^x + 2xe^x) + A(B-1)e^x(1+(B-1)x) $$ Now, plug \(y_p(x)\), \(y_p'(x)\) and \(y_p''(x)\) into the given equation: $$ A(e^x + 2xe^x) + A(B-1)e^x(1+(B-1)x) - 2(A(e^x + xe^x) + A(B-1)e^xx + ABe^x) + Axe^{x}(x+B) = e^x(2-12x) $$ Simplify to find \(A\) and \(B\): $$ Ae^x = 2e^x \;\Rightarrow\; A = 2 $$ $$ B = \frac{1}{2} $$ Thus, the particular solution is: $$ y_p(x) = 2xe^{x}(x+\frac{1}{2}) $$

Obtain the general solution

The general solution is the sum of the complementary solution and the particular solution: $$ y(x) = y_c(x) + y_p(x) $$ Therefore, the general solution is: $$ y(x) = C_1e^x + C_2xe^x + 2xe^{x}(x+\frac{1}{2}) $$

Key concepts

Characteristic Equation

In the context of second-order linear homogeneous differential equations, the characteristic equation is a pivotal concept. This algebraic equation is derived from the ordinary differential equation (ODE) by assuming solutions of the form e^{rx}, where r is the unknown to be solved for. For the equation y'' - 2y' + y = 0, the characteristic equation is r^2 - 2r + 1 = 0. Solving the characteristic equation helps us find the roots which are essential in constructing the complementary solution. For example, if the characteristic equation yields real and distinct roots, the complementary solution will consist of exponential functions with those roots as exponents. If the roots are real and repeated, as in the given exercise, the solution involves both an exponential function and its product with x. Consequently, in our exercise, we obtain r_1 = r_2 = 1, leading to a complementary solution of the form C_1e^{x} + C_2xe^{x}.

Complementary Solution

The complementary solution, usually denoted by y_c(x), is derived from the general solution of the homogeneous version of our given ODE. In simple terms, it's the part of the general solution that doesn't include any influence from the non-homogeneous part, e^{x}(2-12x), of the original equation. The complementary solution purely depends on the roots of the characteristic equation. When those roots are equal (r_1 = r_2), as in our exercise, the complementary solution has a specific expression involving a linear combination of an exponential function and its multiplication by x. This is due to the principle of superposition, which states that if we have two solutions to a linear homogeneous differential equation, any linear combination of these solutions is also a solution. Hence, the complementary solution we derived, C_1e^{x} + C_2xe^{x}, will satisfy the homogeneous equation for any constants C_1 and C_2.

Particular Solution

The particular solution y_p(x) addresses the specific influence of the non-homogeneous part of the differential equation on the general solution. In essence, we're looking for a function that, when substituted into the non-homogeneous differential equation, yields the non-homogeneous part — the right side of our ODE, e^{x}(2-12x). The process begins with an educated guess based on the form of the non-homogeneous term. In this case, since the non-homogeneous term contains an exponential function multiplied by a polynomial, we assume the particular solution will also mimic this form but with undetermined coefficients: Axe^{x}(x+B). By differentiating this assumed solution and substituting y_p(x), y'_p(x), and y''_p(x) back into the original non-homogeneous ODE, we isolate these coefficients to find their exact values. Through this method, we obtained A = 2 and B = 1/2 for the particular solution, resulting in 2xe^{x}(x+1/2). This particular solution is unique because it accounts for the specific external force or function applied to the system being modeled by the differential equation.

Second-order Linear Homogeneous Differential Equations

Understanding second-order linear homogeneous differential equations is pivotal for students delving into differential equations. These equations have the general form ay'' + by' + cy = 0, where a, b, and c are constants, and the function y depends on x. The 'second-order' denotes that the highest derivative is the second derivative, 'linear' means that both y and its derivatives appear to the first power and are not multiplied together, and 'homogeneous' indicates that the equation is set equal to zero (which also means the absence of an external forcing term).

To solve these equations, one must first find the complementary solution, which involves determining the characteristic equation and solving for its roots. The unique solutions that come from different root cases - real and distinct, real and repeated, or complex - form the basis of the complementary solution. Finally, in a non-homogeneous equation like the one from the exercise, we combine the complementary and particular solutions to construct the general solution that addresses the complete behavior of the system described by the equation.