Question

In Exercises \(15-19\) find the general solution. $$ y^{\prime \prime}+6 y^{\prime}+9 y=e^{2 x}(3-5 x) $$

Short answer

Question: Find the general solution of the given second-order linear non-homogeneous differential equation: $$y'' + 6y' + 9y = e^{2x}(3 - 5x).$$ Answer: The general solution to the given differential equation is $$y(x) = C_1 e^{-3x} + C_2 xe^{-3x} + \left(-\frac{5}{9}x^2 + \frac{7}{6}x -\frac{7}{9}\right)e^{2x}.$$

Step-by-step solution

Find the complementary function

First, we need to solve the homogeneous equation: $$ y'' + 6y' + 9y = 0 $$ The characteristic equation for this DE is: $$ r^2 + 6r + 9 = 0 $$ This is a quadratic equation, and we can solve it using the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Plugging in a=1, b=6 and c=9, we get: $$ r = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = -3 \pm \sqrt{0} $$ So, the roots are equal and real, r = -3, with multiplicity 2. In this case, the complementary function is: $$ y_c(x) = C_1 e^{-3x} + C_2 xe^{-3x} $$

Find a particular solution using undetermined coefficients

The forcing term is a non-homogeneous term of the form: $$ g(x) = e^{2x}(3 - 5x) $$ We can guess a particular solution of the form: $$ y_p(x) = (Ax^2 + Bx + C)e^{2x} $$ Taking the first derivative w.r.t x: $$ y'_p(x) = (2Ax+B)e^{2x} + (Ax^2+Bx+C)(2e^{2x}) $$ Taking the second derivative w.r.t x: $$ y''_p(x) = 4Ae^{2x} + 4(Ax^2+Bx+C)e^{2x} + 4(2Ax+B)e^{2x} $$ Substituting \(y_p\), \(y'_p\), and \(y''_p\) into the non-homogeneous equation: $$ (Ax^2+Bx+C)e^{2x}(9) + (2Ax+B)e^{2x}(6) + 4Ae^{2x}= e^{2x}(3-5x) $$ Now we can equate the coefficients of the powers of x: $$ 9Ae^{2x} = -5e^{2x} \Rightarrow A = -\frac{5}{9} $$ $$ 6B+18A = 3 \Rightarrow B = \frac{7}{6} $$ $$ 9C + 6B = 0 \Rightarrow C = -\frac{7}{9} $$ So, the particular solution is: $$ y_p(x) = \left(-\frac{5}{9}x^2 + \frac{7}{6}x -\frac{7}{9}\right)e^{2x} $$

Find the general solution

Finally, we can add the complementary function \(y_c\) and the particular solution \(y_p\) to obtain the general solution: $$ y(x) = y_c(x) + y_p(x) = C_1 e^{-3x} + C_2 xe^{-3x} + \left(-\frac{5}{9}x^2 + \frac{7}{6}x -\frac{7}{9}\right)e^{2x} $$ This is the general solution to the given differential equation.

Key concepts

Characteristic Equation

In the study of differential equations, the characteristic equation plays a crucial role in finding solutions to homogeneous linear differential equations with constant coefficients. Let's take a simple equation such as y'' + 6y' + 9y = 0. To solve it, we look for solutions of the form e^{rx}, where r is a number we need to determine. Substituting e^{rx} for y into the homogeneous equation results in the characteristic equation r^2 + 6r + 9 = 0.

By solving this quadratic equation, typically using the quadratic formula r = \frac{-b \(\pm\) \sqrt{b^2 - 4ac}}{2a}, we can find the roots which are the values of r. These roots tell us the type of functions that will make up the complementary function, which represents the solution to the homogeneous part of the differential equation.

Complementary Function

Once we have the characteristic equation's roots, we can write the complementary function, denoted by y_c(x). It is a linear combination of functions that correspond to the roots obtained from the characteristic equation. In the provided example, the roots are real and equal, both being -3. Hence, the complementary function takes on a special form due to the multiplicity of the roots: y_c(x) = C_1 e^{-3x} + C_2 xe^{-3x}.

C_1 and C_2 are arbitrary constants which will be determined by the initial conditions once given. The complementary function is essentially the general solution to the homogeneous differential equation and will be part of the overall solution.

Particular Solution

The particular solution, denoted by y_p(x), complements the complementary function to form the general solution for a non-homogeneous differential equation, like y'' + 6y' + 9y = e^{2x}(3-5x). To find y_p(x), one typically makes an educated guess based upon the form of the non-homogeneity, which in this case is e^{2x}(3-5x). A good guess for y_p(x) would have the same form of x and e^{2x}, hence (Ax^2 + Bx + C)e^{2x} is chosen.

Once we have this guess, we can differentiate it to match the order of the differential equation and substitute into the non-homogeneous equation to equate coefficients. This process yields the specific values of A, B, and C, which determine our particular solution. This solution addresses the unique effects of the non-homogeneous part of the equation.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find the particular solution by making an educated guess about its form, just as we did with y_p(x) = (Ax^2 + Bx + C)e^{2x}. This method works well when the non-homogeneous part, g(x), is relatively simple, such as polynomials, exponentials, sines, cosines, or any combination of these.

The next step is to substitute this guessed form into the differential equation and to determine the values of the unknown coefficients that make the equation true. We achieve this by comparing the coefficients of corresponding terms on both sides of the equation. The 'undetermined' part of the name comes from the unknown constants that are solved for, like in our example where A, B, and C were calculated. This method is a systematic way to tackle non-homogeneous equations and find the particular solution.