/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Find the general solution. $$ ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the general solution. $$ \left(D^{2}+3 D-4\right) y=12 e^{2 x} $$

Short Answer

Expert verified
The general solution is \( y = C_1 e^{x} + C_2 e^{-4x} + 2e^{2x} \).

Step by step solution

01

Identify the Differential Equation Type

First, we recognize that the equation \( \left(D^{2}+3 D-4\right) y=12 e^{2 x} \) is a linear non-homogeneous differential equation with constant coefficients. The operator \( D \) represents differentiation with respect to \( x \).
02

Find the Complementary Solution \( y_c \)

To find the complementary solution, set the right-hand side of the equation to zero: \( (D^2 + 3D - 4)y = 0 \). Solve the characteristic equation \( r^2 + 3r - 4 = 0 \). The roots are \( r_1 = 1 \) and \( r_2 = -4 \), giving the complementary solution: \( y_c = C_1 e^{x} + C_2 e^{-4x} \).
03

Determine the Particular Solution \( y_p \)

Since the right-hand side is \( 12 e^{2x} \), we propose a particular solution of the form \( y_p = A e^{2x} \). Substitute \( y_p \) into the differential operator: \( (D^2 + 3D - 4)(Ae^{2x}) \). This results in \( (4A + 6A - 4A)e^{2x} = 12e^{2x} \), leading to \( 6A = 12 \). Solving this gives \( A = 2 \), so \( y_p = 2e^{2x} \).
04

Write the General Solution

Combine the complementary solution and particular solution to form the general solution of the differential equation: \( y = y_c + y_p = C_1 e^{x} + C_2 e^{-4x} + 2e^{2x} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations with Constant Coefficients
Linear differential equations with constant coefficients are a special kind of differential equations. In these equations, the coefficients of the terms are constants, meaning they do not change with respect to the independent variable, often denoted as \( x \). Such equations can be presented in the form \( (D^n + a_{n-1} D^{n-1} + ... + a_1 D + a_0)y = g(x) \), where \( D \) represents the differentiation with respect to \( x \).
Linear equations are defined as having terms including \( D \) raised to powers up to degree \( n \), but with no product of these terms. A non-homogeneous equation will have a function \( g(x) \) on the right-hand which isn't equal to zero; this contrasts with homogeneous equations where \( g(x) = 0 \).
Such equations often arise in real-world applications like physics and engineering, as they can model dynamic systems that respond predictably.
Complementary Solutions
The complementary solution, noted as \( y_c \), of a linear non-homogeneous differential equation, represents the solution to the associated homogeneous equation. For the equation \( (D^2 + 3D - 4)y = 0 \), we initially disregard the non-homogeneous part.
To find \( y_c \), one solves the corresponding homogeneous differential equation \( (D^2 + 3D - 4)y = 0 \), which simplifies into solving the characteristic equation. The characteristic equation derives from replacing \( D \) with \( r \), yielding \( r^2 + 3r - 4 = 0 \). By finding the roots of this equation, in this case, \( r_1 = 1 \) and \( r_2 = -4 \), one forms the complementary solution as a combination of exponential functions: \( y_c = C_1 e^{r_1 x} + C_2 e^{r_2 x} = C_1 e^x + C_2 e^{-4x} \).
This part of the solution accounts for the natural behavior of the system without external forces.
Particular Solutions
A particular solution, \( y_p \), of a linear non-homogeneous differential equation is one specific solution that satisfies the entire equation including its non-homogeneous aspect. It's found by making a suitable assumption for the form of \( y_p \).
In our example equation \( (D^2 + 3D - 4)y = 12 e^{2x} \), the non-homogeneous term is \( 12 e^{2x} \). Thus, we assume a particular solution of the form \( y_p = A e^{2x} \). This educated guess is known as the "method of undetermined coefficients."
Substituting \( y_p \) back into the differential equation and simplifying yields an equation where the coefficient \( A \) can be determined such that the terms on both sides match. For our case, substituting results in \( 6A e^{2x} = 12 e^{2x} \), which simplifies to \( A = 2 \). Hence, \( y_p = 2e^{2x} \), a solution particular to the given non-homogeneous differential equation.
Characteristic Equation
The characteristic equation is a foundational concept when solving linear differential equations. It is derived from the homogeneous part of the differential equation by taking the differential operator, here \( D \), and replacing it with \( r \).
This process converts the differential equation into an algebraic equation, often much simpler to solve. In our example, \( (D^2 + 3D - 4)y = 0 \) becomes \( r^2 + 3r - 4 = 0 \). Solving this quadratic equation gives us the roots which are critical in constructing the complementary solution.
The roots, \( r_1 \) and \( r_2 \), indicate the nature of the solutions. For real and distinct roots, the complementary solution is formed using exponential functions, \( e^{r x} \), for each root like \( C_1 e^{r_1 x} + C_2 e^{r_2 x} \). If there are repeated or complex roots, variations handle these specifics, but the core steps remain critically linked to the characteristic equation.

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