/*! 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 74 Solve the differential equation ... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the differential equation using either the method of undetermined coefficients or the variation of parameters. $$y^{\prime \prime}+6 y^{\prime}+9 y=e^{-x}$$

Short Answer

Expert verified
The solution is \( y = C_1 e^{-3x} + C_2 x e^{-3x} + \frac{1}{4}xe^{-x} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y'' + 6y' + 9y = e^{-x} \). It is a second-order linear non-homogeneous differential equation with constant coefficients.
02

Find the Complementary Solution

To find the complementary solution, solve the associated homogeneous equation \( y'' + 6y' + 9y = 0 \). The characteristic equation is \( r^2 + 6r + 9 = 0 \), which can be written as \((r+3)^2 = 0 \). Thus, \( r = -3 \) is a repeated root.
03

Write the Complementary Solution

For a repeated root \( r = -3 \), the complementary solution is \( y_c = C_1 e^{-3x} + C_2 x e^{-3x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.
04

Identify the Method of Undetermined Coefficients

Since the right-hand side of the equation is \( e^{-x} \), a form that is suitable for the method of undetermined coefficients, choose this method to find the particular solution.
05

Choose the Form of the Particular Solution

Propose a particular solution of the form \( y_p = Axe^{-x} \) because \( e^{-x} \) is part of the complementary solution's form. Adjust the form by multiplying by \( x \) due to the overlap.
06

Differentiate the Particular Solution

Differentiate \( y_p = Axe^{-x} \) to find \( y_p' = A(e^{-x} - xe^{-x}) \) and \( y_p'' = A(-2e^{-x} + xe^{-x}) \).
07

Substitute into the Original Equation

Substitute \( y_p, y_p', y_p'' \) into the differential equation: \( (A(-2e^{-x} + xe^{-x})) + 6(A(e^{-x} - xe^{-x})) + 9(Axe^{-x}) = e^{-x} \).
08

Simplify and Solve for A

Simplify the expression: \( -2A e^{-x} + Axe^{-x} + 6A e^{-x} - 6Axe^{-x} + 9Axe^{-x} = e^{-x} \). Collect like terms: \( (4A)e^{-x} + (4Ax)e^{-x} = e^{-x} \). Equate coefficients: \( 4A = 1 \) and \( 4Ax = 0 \). Thus, \( A = \frac{1}{4} \).
09

Write the Particular Solution

Thus, the particular solution is \( y_p = \frac{1}{4}xe^{-x} \).
10

Write the General Solution

The general solution of the differential equation is the sum of the complementary and particular solutions: \( y = C_1 e^{-3x} + C_2 x e^{-3x} + \frac{1}{4}xe^{-x} \).

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Key Concepts

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

Method of Undetermined Coefficients
The method of undetermined coefficients is a technique used to find a particular solution to a non-homogeneous linear differential equation. This method works well when the non-homogeneous term (right-hand side of the equation) is a simple function, such as a polynomial, an exponential, or a trigonometric function. The key idea is to assume a form for the particular solution that resembles the non-homogeneous term. For instance, if the non-homogeneous term is of the form \(e^{-x}\), then the guessed particular solution \(y_p\) could initially be \(Ae^{-x}\), where \(A\) is a constant to be determined.
However, if a term like \(e^{-x}\) already appears in the complementary solution, as is the case here, we adjust by incorporating a factor of \(x\) to prevent overlap, leading to the form \(y_p = Axe^{-x}\). The coefficients such as \(A\) are then determined by substituting the proposed solution into the original differential equation and solving for the unknown coefficients.
Complementary Solution
The complementary solution of a differential equation refers to the solution of the associated homogeneous equation. This type of solution specifically addresses the equation without the non-homogeneous term, allowing us to understand the natural behavior of the system without external forces.
For the given homogeneous equation \(y'' + 6y' + 9y = 0\), we solve the characteristic equation derived from this differential equation. The characteristic equation is formed as \(r^2 + 6r + 9 = 0\). On solving, it simplifies to \((r+3)^2 = 0\), indicating that \(r = -3\) is a repeated root.
In cases of repeated roots, the complementary solution is expressed as \(y_c = C_1 e^{-3x} + C_2 x e^{-3x}\), where \(C_1\) and \(C_2\) are constants determined by initial conditions. The inclusion of the term \(C_2 x e^{-3x}\) compensates for the repeated nature of the root, ensuring linear independence between terms.
Particular Solution
Finding the particular solution to a differential equation involves addressing the non-homogeneous part of the equation. The particular solution accounts for the external inputs or forces reflected in the non-homogeneous term. In our example, the non-homogeneous component is \(e^{-x}\). By employing the method of undetermined coefficients, we proposed a particular solution of the form \(y_p = Axe^{-x}\), suitably adjusted to accommodate repetition in the complementary function.
To establish the particular solution, substitute \(y_p\) along with its derivatives back into the original differential equation to determine the unknown coefficients. Simplifying and equating coefficients yields the constant \(A = \frac{1}{4}\). This gives us the particular solution \(y_p = \frac{1}{4}xe^{-x}\).
Finally, the general solution of the differential equation integrates both the complementary and particular solutions, written as \(y = C_1 e^{-3x} + C_2 x e^{-3x} + \frac{1}{4}xe^{-x}\). This comprehensive solution encompasses both natural and forced responses of the system.

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