Question

In Exercises \(15-19\) find the general solution. $$ y^{\prime \prime}-3 y^{\prime}+2 y=e^{3 x}(1+x) $$

Short answer

Question: Find the general solution for the given second-order linear differential equation: $$y^{\prime \prime}-3 y^{\prime}+2 y=e^{3 x}(1+x).$$ Answer: The general solution for the given differential equation is $$y(x)=C_1e^{x}+C_2e^{2x}+\frac{1}{5}xe^{3x}-\frac{1}{10}e^{3x},$$ where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Find the complementary solution

To find the complementary solution, we first set the non-homogeneous term (RHS) to zero and solve the corresponding homogeneous equation: $$ y^{\prime \prime}-3 y^{\prime}+2 y=0. $$The characteristic equation for this homogeneous equation is: $$r^2-3r+2=0.$$ By factoring, we obtain: $$(r-1)(r-2)=0,$$ which gives us the roots \(r_1=1\) and \(r_2=2.\) Thus, the complementary solution is: $$y_c(x)=C_1e^{x}+C_2e^{2x}.$$

Find the particular solution

For the particular solution, we notice that the non-homogeneous term is \(e^{3x}(1+x)\) which is a product of an exponential function and a polynomial. We assume a particular solution of the form: $$y_p(x)=Axe^{3x}+Be^{3x}.$$ Taking the first derivative, we get: $$y_p^{\prime}(x)=(3A+B)xe^{3x}+(A+3B)e^{3x}.$$ And the second derivative: $$y_p^{\prime \prime}(x)=(6A+6B)xe^{3x}+(6A+9B)e^{3x}.$$ Now, we substitute \(y_p(x)\), \(y_p^{\prime}(x)\), and \(y_p^{\prime \prime}(x)\) into the given non-homogeneous equation: $$[(6A+6B)xe^{3x}+(6A+9B)e^{3x}]-3[(3A+B)xe^{3x}+(A+3B)e^{3x}]+2[(Axe^{3x}+Be^{3x})]=e^{3x}(1+x).$$ Simplifying the equation, we get: $$e^{3x}[(2B-3A)x+(5A)]=e^{3x}(1+x).$$ Equating the coefficients, we get the following system of equations: $$\begin{cases} 2B-3A=1, \\ 5A=1. \end{cases}$$ Solving this system, we find that \(A=\frac{1}{5}\) and \(B=-\frac{1}{10}\). Thus, the particular solution is: $$y_p(x)=\frac{1}{5}xe^{3x}-\frac{1}{10}e^{3x}.$$

Find the general solution

Now, we can find the general solution by adding the complementary solution and the particular solution: $$y(x)=y_c(x)+y_p(x)=C_1e^{x}+C_2e^{2x}+\frac{1}{5}xe^{3x}-\frac{1}{10}e^{3x}.$$

Key concepts

Complementary Solution

The complementary solution (or homogeneous solution) is fundamental when solving linear differential equations with constant coefficients. It represents the solution to the simplified version of the equation where all non-homogeneous terms (i.e., sources or driving functions) are set to zero. Think of it as the solution that describes the system's behavior without any external influences.

The process to find it typically involves a few steps: First, we form the characteristic equation of the homogeneous differential equation by assuming a solution of the form \( e^{rx} \), where \( r \) is a constant to be determined. For the equation \( y^{''} - 3y' + 2y = 0 \), the characteristic equation is \( r^2 - 3r + 2 = 0 \). Solving this quadratic gives the roots, which then lead to the complementary solution. In our example, the roots are \( r_1 = 1 \) and \( r_2 = 2 \), so the complementary solution is \( y_c(x) = C_1e^{x} + C_2e^{2x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions.

Particular Solution

In contrast to the complementary solution, the particular solution addresses the external influence or the non-homogeneous part of the differential equation. It's the solution that specifically satisfies the non-homogeneous equation, and finding it usually requires some method of guessing based on the form of the non-homogeneous term.

A practical approach for guessing involves analyzing the non-homogeneous term and choosing a similar function with undetermined coefficients. This is what we did for the term \( e^{3x}(1+x) \) by assuming a particular solution of the form \( y_p(x) = Axe^{3x} + Be^{3x} \). We differentiate to find \( y_p^{'}(x) \) and \( y_p^{''}(x) \), substitute back into the equation, and equate coefficients to solve for A and B. This process helps us determine the particular solution, which in our case is \( y_p(x) = \frac{1}{5}xe^{3x} - \frac{1}{10}e^{3x} \).

Characteristic Equation

The characteristic equation is a tool used to find the complementary solution of a homogeneous differential equation with constant coefficients. It transforms the differential equation into an algebraic one, making it much easier to solve.

To formulate the characteristic equation, replace each derivative of \( y \) with a power of \( r \), corresponding to the degree of the derivative. For example, a second derivative \( y'' \) becomes \( r^2 \), and a first derivative \( y' \) becomes \( r \). For the given differential equation \( y^{''} - 3y' + 2y = 0 \), we get the characteristic equation \( r^2 - 3r + 2 = 0 \). It helps to remember that the roots of the characteristic equation indicate the exponential terms of the complementary solution. Real distinct roots, as in our exercise, lead to exponential functions, while complex roots would relate to trigonometric functions, which is an entirely different case for complementary solutions.