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Chapter 3: Problem 33

Solve the initial value problems in Problems 31 through \(40 .\) $$ y^{\prime \prime}+9 y=\sin 2 x ; y(0)=1, y^{\prime}(0)=0 $$

Short Answer

Expert verified
The solution is \( y = \cos 3x + \frac{1}{5} \sin 3x - \frac{1}{5} \sin 2x \).

Step by step solution

01

Identify the Type of Equation

The given equation is a second-order linear non-homogeneous differential equation, represented as \( y'' + 9y = \sin 2x \). This type of equation requires finding both the homogeneous solution \( y_h \) and a particular solution \( y_p \).
02

Solve the Homogeneous Equation

To find \( y_h \), solve the associated homogeneous equation: \( y'' + 9y = 0 \). The characteristic equation is \( r^2 + 9 = 0 \), which simplifies to \( r = \pm 3i \). Hence, the general homogeneous solution is \( y_h = C_1 \cos 3x + C_2 \sin 3x \).
03

Find a Particular Solution

Use the method of undetermined coefficients to guess the form of \( y_p \). For \( \sin 2x \), try a solution of the form \( y_p = A \cos 2x + B \sin 2x \). Substitute this into the original equation to determine \( A \) and \( B \).
04

Substitute and Solve for Coefficients

Substitute \( y_p = A \cos 2x + B \sin 2x \) into the differential equation and equate coefficients. After solving, you find: \( A = 0 \) and \( B = -\frac{1}{5} \). Thus, \( y_p = -\frac{1}{5} \sin 2x \).
05

Combine Solutions

The general solution is \( y = y_h + y_p = C_1 \cos 3x + C_2 \sin 3x - \frac{1}{5} \sin 2x \).
06

Apply Initial Conditions

Use the initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \) to find \( C_1 \) and \( C_2 \). Substituting \( x = 0 \) into the general solution yields \( C_1 = 1 \). Differentiating the general solution and applying \( y'(0) = 0 \) gives \( C_2 = \frac{1}{5} \).
07

Formulate the Final Solution

Substitute the values of \( C_1 \) and \( C_2 \) back into the general solution: \( y = \cos 3x + \frac{1}{5} \sin 3x - \frac{1}{5} \sin 2x \).

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Key Concepts

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

Initial Value Problem
An Initial Value Problem (IVP) refers to a differential equation paired with a specified set of initial conditions. These conditions help determine a unique solution for the differential equation. For instance, in our example, we are working with the second-order equation: \( y'' + 9y = \sin 2x \).
The initial conditions given are \( y(0) = 1 \) and \( y'(0) = 0 \).
These initial values enable us to find specific values for the constants that arise in the solution to the homogeneous equation.

Here's why initial values are essential:
  • They ensure a unique solution by removing the arbitrariness caused by constants in the general solution.
  • They provide a starting point that helps validate that the particular and homogeneous solutions combine correctly to solve the equation.
Solving an initial value problem usually involves both solving the differential equation and applying the initial condition constraints.
Non-homogeneous Equation
A non-homogeneous equation, such as our example \( y'' + 9y = \sin 2x \), involves a differential equation with a non-zero term on the right side of the equation.
This term, \sin 2x in our exercise, is what makes it non-homogeneous since it cannot be reduced to just zero.

Here's how you can identify and solve a non-homogeneous equation:
  • Be aware of any non-zero term on the right-hand side.
  • The solution process involves finding two separate components: the homogeneous solution (\(y_h\)) and the particular solution (\(y_p\)).
  • The complete solution is a sum of these two parts: \(y = y_h + y_p\).
The non-homogeneous term often dictates the method needed for finding \(y_p\). Thus, recognizing an equation's structure is crucial for applying the right techniques and simplifying the problem's approach.
Method of Undetermined Coefficients
The Method of Undetermined Coefficients is a strategy used to find a particular solution \(y_p\) for non-homogeneous linear differential equations.
This method relies on the nature of the non-homogeneous term to suggest an appropriate form for \(y_p\).

Here's how this works in our exercise:
  • Since the non-homogeneous term is \(\sin 2x\), we assume a trial solution of the form \(y_p = A \cos 2x + B \sin 2x\).
  • We substitute this trial solution into the original differential equation.
  • By comparing coefficients of similar terms (like \(\sin 2x\) and \(\cos 2x\) on both sides of the equation), we solve for the unknown coefficients \(A\) and \(B\).
This method works well for equations where the non-homogeneous term is a simple polynomial, exponential, or sinusoidal function. Its power lies in transforming the unknowns into solvable algebraic forms.
Homogeneous Solution
A homogeneous solution, \(y_h\), solves the associated homogeneous equation, which in our case is:\( y'' + 9y = 0\).
This part models the natural behavior of the system without external influence.

To find the homogeneous solution:
  • Start by forming the characteristic equation from the homogeneous part: \(r^2 + 9 = 0\).
  • Solve for \(r\), which in this case are imaginary roots: \(r = \pm 3i\).
  • Use these roots to construct the general solution: \(y_h = C_1 \cos 3x + C_2 \sin 3x\).
This solution is crucial because it describes the response of the system to its inherent properties, like damping or oscillation, without any external forces. Therefore, it represents the natural frequency and modes of the system, which when combined with \(y_p\), give the full response of the system.

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Most popular questions from this chapter

In Problems 47 through 56, use the method of variation of parameters to find a particular solution of the given differential equation. $$ y^{\prime \prime}-4 y=\sinh 2 x $$

A front-loading washing machine is mounted on a thick rubber pad that acts like a spring; the weight \(W=m g\) (with \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) ) of the machine depresses the pad exactly \(0.5 \mathrm{~cm}\). When its rotor spins at \(\omega\) radians per second, the rotor exerts a vertical force \(F_{0} \cos \omega t\) newtons on the machine. At what speed (in revolutions per minute) will resonance vibrations occur? Neglect friction.

Make the substitution \(v=\ln x\) of Problem 51 to find general solutions (for \(x>0\) ) of the Euler equations in Problems. $$ x^{2} y^{\prime \prime}+x y^{\prime}+9 y=0 $$

Gives a general solution \(y(x)\) of a homogeneous second-order differential equation \(a y^{\prime \prime}+b y^{\prime}+c y=0\) with constant coefficients. Find such an equation. $$ y(x)=c_{1}+c_{2} x $$

In Problems 47 through 56, use the method of variation of parameters to find a particular solution of the given differential equation. $$ y^{\prime \prime}+9 y=\sin 3 x $$
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