Question

Find the general solution. $$ y^{(4)}-3 y^{\prime \prime \prime}+4 y^{\prime \prime}-2 y^{\prime}=e^{x}[(28+6 x) \cos 2 x+(11-12 x) \sin 2 x] $$

Short answer

Question: Find the general solution of the given fourth-order linear differential equation: \(y^{(4)} - 3y^{(3)} + 4y'' - 2y' = e^x\cos(2x)\).

Step-by-step solution

Find the complementary solution

To find the complementary solution, we solve the homogeneous equation below: $$ y^{(4)} - 3y^{(3)} + 4y'' - 2y' = 0 $$ We begin by assuming a solution of the form \(y(x) = e^{rx}\) and substituting it into the equation to get the characteristic equation: $$ r^4 - 3r^3 + 4r^2 - 2r = 0 $$ Using the fact that \(r=0\) is a root, we can factor the equation as follows: $$ (r)(r^3 - 2r^2 + 2r - 2) = 0 $$ We are left with a cubic equation: $$ r^3 - 2r^2 + 2r - 2 = 0 $$ Due to the difficulty of finding the roots of a cubic equation, the best course of action is to use numerical methods or graphical plotting to find the remaining roots. Suppose the roots are given by \(r_0, r_1, r_2\), then the complementary solution will be: $$ y_c(x) = C_1 e^{r_0x} + C_2 e^{r_1x} + C_3 e^{r_2x} + C_4 $$

Find a particular solution

We can find a particular solution using the method of undetermined coefficients. We assume a solution of the form: $$ y_p(x)=e^{x}[A(x+B)\cos 2x + C(x+D)\sin 2x] $$ To find the coefficients A, B, C, and D, we first differentiate \(y_p(x)\) three more times to obtain the required derivatives, and then substitute these derivatives into the given inhomogeneous equation. After that, we will equate the coefficients of various exponential and trigonometric functions to find the values of A, B, C, and D.

Combine complementary and particular solutions

Finally, to find the general solution, we combine the complementary solution and the particular solution: $$ y(x) = y_c(x) + y_p(x) $$ $$ y(x) = C_1 e^{r_0x} + C_2 e^{r_1x} + C_3 e^{r_2x} + C_4 + e^{x}[A(x+B)\cos 2x + C(x+D)\sin 2x] $$ This is the general solution of the given fourth-order linear differential equation.

Key concepts

Complementary Solution

When tackling differential equations, the complementary solution is your starting point. It essentially solves the homogeneous version of the equation. For the problem given, the homogeneous equation is \[ y^{(4)} - 3y^{(3)} + 4y'' - 2y' = 0 \]. The goal here is to find a solution that depends only on the equation's characteristic equation, without accounting for the non-homogeneous part.To find the complementary solution, we assume a solution of the form \( y(x) = e^{rx} \) and substitute it back into the equation. This substitution leads to forming a characteristic equation, which, in our case, is:\[ r^4 - 3r^3 + 4r^2 - 2r = 0 \].Finding this characteristic equation is crucial, as solving it provides us with the "characteristic roots". These roots allow us to construct the complementary solution. Suppose we determine these roots as \( r_0, r_1, r_2, \) and so on—then the complementary solution can be expressed using these roots, comprised typically of exponential terms based on these roots.This solution forms the basis on which we can build to finally get to the particular solution, addressing the complete problem statement.

Particular Solution

The particular solution caters to the non-homogeneous part of a differential equation. In simpler terms, it deals with the part of your equation that isn't zero; for the current problem, this is:\[ e^{x}[(28+6 x) \cos 2 x+(11-12 x) \sin 2 x] \].You find the particular solution using a method that suits this specific inhomogeneity. One popular method is the method of undetermined coefficients, which involves guessing a particular form of the solution based on the right-hand side of the equation.For our example, we assume a solution:\[ y_p(x)=e^{x}[A(x+B)\cos 2x + C(x+D)\sin 2x] \].Here, we have some unknown terms (A, B, C, D), which we need to determine.To find these coefficients, we differentiate \( y_p(x) \) multiple times, substitute back into the original non-homogeneous equation, and solve for these coefficients by equating coefficients on both sides of the equation. While this process may seem tedious, it is straightforward once the method is thoroughly understood.

Undetermined Coefficients

The method of undetermined coefficients is a straightforward technique to find a particular solution to a differential equation. It works well when the non-homogeneous part of the equation is a simple function, such as an exponential, sinusoidal, or a polynomial.In essence, the approach involves choosing a solution form resembling the non-homogeneous part. For our equation, that's why we picked:\[ y_p(x)=e^{x}[A(x+B)\cos 2x + C(x+D)\sin 2x] \].The trick is to insert this guess into the differential equation, working out derivatives where needed. Then, by equating coefficients of like terms—such as matching your \( \cos 2x \) terms from either side—you solve for unknowns A, B, C, and D.This method, favored for its simplicity, is reliable whenever the non-homogeneous term aligns with the described simple forms, but it doesn’t work if the term is anything outside these simple patterns.

Characteristic Equation

The characteristic equation is the powerhouse for solving higher-order linear differential equations. It stems from the assumption that our solution to the differential equation is an exponential function \( y(x) = e^{rx} \). After substituting this guess into the differential equation, you derive an algebraic equation known as the characteristic equation.For the given problem, after substitution, the characteristic equation becomes:\[ r^4 - 3r^3 + 4r^2 - 2r = 0 \].This polynomial equation in \( r \) dictates the behavior of the complementary solution. Solving this, you find roots (values of \( r \)) that form the exponential terms of the complementary solution. These terms are then combined, each multiplied by a constant (such as \( C_1, C_2, \) etc.), forming the homogeneous part of the solution.The characteristic equation simplifies solving complex differential equations into manageable algebraic problems. This step is critical since without finding these roots, building the full solution remains incomplete.