Problem 33
Question
Use the result of the preceding problem to determine a particular solution to the given differential equation. $$\left(D^{2}-4 D+13\right)(D-3) y=F(x)$$
Step-by-Step Solution
Verified Answer
The complementary solution of the given differential equation is
\[
y_c(x) = C_1e^{3x} + C_2e^{2x}\cos(3x) + C_3e^{2x}\sin(3x),
\]
where \(C_1, C_2\), and \(C_3\) are constants. For the particular solution, we need more information on the form of \(F(x)\). Once we have that, we can choose an appropriate trial function, substitute it into the differential equation, and solve for the coefficients. Then, the complete particular solution will be the sum of the complementary and particular solutions.
1Step 1: Identify the complementary equation and its solution
The given differential equation is
\[
\left(D^2 - 4D + 13\right)(D - 3)y = F(x),
\]
where \(D\) is the differential operator (\(Dy = y'\) and so on).
To find the complementary solution, we consider the homogeneous version of the given differential equation. It is given by
\[
\left(D^2 - 4D + 13\right)(D - 3)y = 0.
\]
Let's denote this differential equation as
\[
(D^3 - 7D^2 + 25D -39)y = 0.
\]
To find the complementary solution, we need to find the roots of the characteristic equation:
\[
r^3 - 7r^2 + 25r - 39 = 0.
\]
This cubic equation is challenging to solve. However, we know that the given equation can be factored into the form \(\left(D^2 - 4D + 13\right)(D - 3)\). Thus, one root is \(r_1 = 3\) (from the factor \(D-3\)). The other two roots, say \(r_2\) and \(r_3\), come from the quadratic factor, \(D^2 - 4D + 13\). We can use the quadratic formula to solve for these roots, which are complex conjugates: \(r_{2,3} = 2 \pm 3i\).
Now that we have found the roots \(r_1, r_2, r_3\), we can write down the complementary solution, which is given by
\[
y_c(x) = C_1e^{3x} + C_2e^{2x}\cos(3x) + C_3e^{2x}\sin(3x),
\]
where \(C_1, C_2\), and \(C_3\) are constants.
2Step 2: Identify the trial function for the particular solution
Since we don't have information about the form of the function \(F(x)\), we cannot explicitly choose a trial function for the particular solution. However, we can describe the general approach to this step.
In general, the trial function will be a function of the same form as \(F(x)\). If \(F(x)\) is a polynomial, the trial function will also be a polynomial. If \(F(x)\) is an exponential, trigonometric function, or a product of such functions, the trial function will be the same type of function, possibly with additional polynomial terms.
3Step 3: Obtain the complete particular solution
Since we cannot provide a specific trial function without knowing the form of \(F(x)\), we cannot complete this step in detail. However, once you have chosen the appropriate trial function based on \(F(x)\), you should substitute the trial function into the given differential equation and solve for the unknown coefficients. Finally, add the complementary solution and the particular solution to obtain the complete particular solution.
Key Concepts
Complementary SolutionCharacteristic EquationTrial Function
Complementary Solution
When solving a linear differential equation, the complementary solution (also known as the homogeneous solution) plays a crucial role. It represents the general solution to the associated homogeneous differential equation, meaning the equation with zero on the right-hand side.
A key step in finding the complementary solution is to identify the characteristic equation, which is derived from substituting an exponential function of the form erx into the homogeneous equation. The roots of this characteristic equation determine the form of the complementary solution. For a second-order differential equation, our characteristic equation will be quadratic, and for a third-order, it will be cubic, and so on.
In our problem, we were looking for the complementary solution of a cubic differential equation. We factored the equation, and through this, we identified one real root and two complex conjugate roots. We used these roots to construct the complementary solution, which is a linear combination of exponential functions and, if applicable, trigonometric functions that correspond to the real and complex roots of the characteristic equation, respectively.
Understanding this can significantly improve grasping the concept of complementary solutions, making it easier to deal with any linear differential equation.
A key step in finding the complementary solution is to identify the characteristic equation, which is derived from substituting an exponential function of the form erx into the homogeneous equation. The roots of this characteristic equation determine the form of the complementary solution. For a second-order differential equation, our characteristic equation will be quadratic, and for a third-order, it will be cubic, and so on.
In our problem, we were looking for the complementary solution of a cubic differential equation. We factored the equation, and through this, we identified one real root and two complex conjugate roots. We used these roots to construct the complementary solution, which is a linear combination of exponential functions and, if applicable, trigonometric functions that correspond to the real and complex roots of the characteristic equation, respectively.
Understanding this can significantly improve grasping the concept of complementary solutions, making it easier to deal with any linear differential equation.
Characteristic Equation
The characteristic equation is a pivotal concept in the study of linear differential equations. It is obtained by replacing the differential operator, usually denoted as D, with a variable (often r) and setting the equation equal to zero. This polynomial equation's roots are essential in determining the behavior of the differential equation's solutions.
For our exercise, we take the polynomial that results from the homogeneous equation: r3 - 7r2 + 25r - 39 = 0 and solve it for r. We identify one of the roots directly from the factoring and find the remaining roots using the quadratic formula. These roots represent the exponents in the exponential functions that make up the complementary solution.
To enhance the learning experience, practicing how to derive and solve the characteristic equation can provide deeper insights. This reinforces the understanding that the characteristic equation's roots directly influence the form and nature of the solution to the differential equation.
For our exercise, we take the polynomial that results from the homogeneous equation: r3 - 7r2 + 25r - 39 = 0 and solve it for r. We identify one of the roots directly from the factoring and find the remaining roots using the quadratic formula. These roots represent the exponents in the exponential functions that make up the complementary solution.
To enhance the learning experience, practicing how to derive and solve the characteristic equation can provide deeper insights. This reinforces the understanding that the characteristic equation's roots directly influence the form and nature of the solution to the differential equation.
Trial Function
In the context of solving non-homogeneous differential equations, the trial function (also known as the particular solution) is a function that is assumed to be a part of the complete solution to the equation. It specifically addresses the non-homogeneous part of the equation, i.e., the function on the right-hand side, denoted by F(x).
The form of the trial function is based on the nature of F(x). It often resembles F(x) but includes undetermined coefficients that we later solve for. If F(x) is a polynomial, the trial function will also be a polynomial of the same degree. For trigonometric or exponential functions F(x), the trial function will have a similar structure.
Without knowing F(x), we can't provide a specific form for the trial function in our exercise. However, the step-by-step explanation focuses on how to construct an appropriate trial function based on the given F(x), which aligns with best educational practices. This emphasizes the importance of understanding the trial function's role in solving differential equations and how it combines with the complementary solution to form the complete solution.
The form of the trial function is based on the nature of F(x). It often resembles F(x) but includes undetermined coefficients that we later solve for. If F(x) is a polynomial, the trial function will also be a polynomial of the same degree. For trigonometric or exponential functions F(x), the trial function will have a similar structure.
Without knowing F(x), we can't provide a specific form for the trial function in our exercise. However, the step-by-step explanation focuses on how to construct an appropriate trial function based on the given F(x), which aligns with best educational practices. This emphasizes the importance of understanding the trial function's role in solving differential equations and how it combines with the complementary solution to form the complete solution.
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