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Chapter 3: Problem 16

Find a particular solution \(y_{p}\) of the given equation. In all these problems, primes denote derivatives with respect to \(x\). $$ y^{\prime \prime}+9 y=2 x^{2} e^{3 x}+5 $$

Short Answer

Expert verified
The particular solution is \( y_p = \frac{5}{9} + \left(\frac{2}{27}x^2 + \frac{4}{27}x + \frac{2}{81}\right)e^{3x} \).

Step by step solution

01

Identify the Homogeneous Equation

The given differential equation is \( y'' + 9y = 2x^2 e^{3x} + 5 \). The homogeneous part (with right-hand side set to zero) is \( y'' + 9y = 0 \).
02

Find Homogeneous Solution

The characteristic equation for \( y'' + 9y = 0 \) is \( r^2 + 9 = 0 \), which gives complex roots \( r = \, \pm 3i \). Thus, the homogeneous solution is \( y_h = c_1 \cos(3x) + c_2 \sin(3x) \).
03

Identify Structure of Particular Solution

To find a particular solution \( y_p \) for the given differential equation, observe the non-homogeneous part: \( 2x^2 e^{3x} + 5 \). Use the method of undetermined coefficients for each component separately.
04

Particular Solution for Polynomial Component

For the constant term \( 5 \), a particular solution of the form \( A \) is chosen. Differentiate and substitute into \( y'' + 9y = 5 \) to find \( A = \frac{5}{9} \).
05

Particular Solution for Exponential Component

Consider \( 2x^2 e^{3x} \). Since no term of the form involving \( \cos(3x) \) or \( \sin(3x) \) is a solution to the homogeneous equation, use \( (Bx^2 + Cx + D) e^{3x} \) for a particular solution. Differentiate and substitute into the differential equation to solve for coefficients \( B, C, D \).
06

Differentiate and Substitute

For \( (Bx^2 + Cx + D) e^{3x} \), calculate \( y'_p \) and \( y''_p \). Substitute into the left-hand side of the original differential equation and equate coefficients.
07

Solve for Coefficients

Solve the system of resulting equations to find \( B = \frac{2}{27} \), \( C = \frac{4}{27} \), \( D = \frac{2}{81} \).
08

Write the Particular Solution

Combine both particular solutions: \( y_p = \frac{5}{9} + (\frac{2}{27}x^2 + \frac{4}{27}x + \frac{2}{81})e^{3x} \).

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Key Concepts

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

Homogeneous Equations
In the world of differential equations, a homogeneous equation is a crucial concept. It is an equation where the right-hand side is zero. For example, let's consider the equation: \( y'' + 9y = 0 \). Here, we're dealing with a homogeneous equation.
  • Homogeneous equations often serve as the first step in solving more complex problems. They help us identify the structure of solutions.
  • The solutions to homogeneous equations are generally a combination of functions involving trigonometric functions (if you find complex roots) and exponential functions (for real roots).
The characteristic equation is key in solving homogeneous equations. It comes from substituting \( y = e^{rx} \) into the differential equation, leading to an equation usually involving powers of \( r \). For the example \( y'' + 9y = 0 \), the characteristic equation is \( r^2 + 9 = 0 \), leading to complex roots.
Particular Solution
After identifying the homogeneous equation, it's time to find a particular solution. A particular solution is a specific solution that fits the non-homogeneous equation. For the given differential equation, \( y^{\prime \prime}+9 y=2 x^{2} e^{3 x}+5 \), a particular solution would account for this non-homogeneous term.
  • The non-homogeneous part of the equation is often composed of polynomials, exponentials, or trigonometric functions.
  • These components require different approaches when finding a particular solution, adapting our methods to each component's nature.
The goal is to find a function \( y_p \) such that when substituted back into the differential equation, it satisfies the entire equation. It's all about crafting a specific solution to match the complexity or type of the non-homogeneous term.
Method of Undetermined Coefficients
One common method to find a particular solution is the method of undetermined coefficients. This method is like an educated guesswork, helping us find solutions for differential equations with specific non-homogeneous terms, like polynomials and exponentials.
  • The essential idea is to assume a form for \( y_p \) that incorporates undetermined coefficients, which we then solve for.
  • For polynomial components (like constants), start with a simple guess. For terms like \( 2x^2 e^{3x} \), assume a form that matches the structure, such as \( (Ax^2 + Bx + C)e^{3x} \).
After assuming a form, you will differentiate it and substitute it back into the differential equation. This substitution helps us solve for the unknown coefficients. Practically, this means setting up equations by matching coefficients on both sides of the equation.
Complex Roots
In solving homogeneous equations, the characteristic equation can sometimes have complex roots. This happens when the discriminant in the characteristic equation is negative. For example, solving \( r^2 + 9 = 0 \) gives roots \( r = \pm 3i \). These are complex roots because they include imaginary numbers.
  • Complex roots have the form \( a \pm bi \), where \( i \) is the imaginary unit \( \sqrt{-1} \).
  • The general solution for complex roots is a combination of trigonometric functions: \( y_h = c_1 \cos(bx) + c_2 \sin(bx) \).
These trigonometric functions come from Euler's formula, which relates complex exponentials to sin and cos functions. This relation helps transform the solutions into real, applicable terms. This understanding is essential for mastering the behavior of solutions to certain differential equations.

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Most popular questions from this chapter

Set up the appropriate form of a particular solution \(y_{p}\), but do not determine the values of the coefficients. $$ v^{(4)}+9 y^{\prime \prime}=\left(x^{2}+1\right) \sin 3 x $$

In each of Problems 11 through 14, find and plot both the steady periodic solution \(x_{\mathrm{sp}}(t)=C \cos (\omega t-\alpha)\) of the given differential equation and the transient solution \(x_{\mathrm{tr}}(t)\) that satisfies the given initial conditions. $$ x^{\prime \prime}+4 x^{\prime}+5 x=10 \cos 3 t ; x(0)=x^{\prime}(0)=0 $$

Find a particular solution \(y_{p}\) of the given equation. In all these problems, primes denote derivatives with respect to \(x\). $$ y^{\prime \prime}+2 y^{\prime}-3 y=1+x e^{x} $$

Solve the initial value problem. $$ y^{\prime \prime}+9 y=\sin 2 x ; y(0)=1, y^{\prime}(0)=0 $$

Problems pertain to the solution of differential equations with complex coefficients. Use the quadratic formula to solve the following equations. Note in each case that the roots are not complex conjugates. (a) \(x^{2}+i x+2=0\) (b) \(x^{2}-2 i x+3=0\)
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