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Find the general solution to the differential equation. $$y^{\prime \prime}+9 y=3 \sin (3 t)$$

Short Answer

Expert verified
The general solution is \(y(t) = c_1 \cos(3t) + (c_2 - \frac{1}{6}t) \sin(3t)\).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \(y'' + 9y = 3\sin(3t)\). This is a second-order linear non-homogeneous differential equation. The general solution will be the sum of the homogeneous solution and a particular solution.
02

Solve the Homogeneous Equation

First, solve the homogeneous equation \(y'' + 9y = 0\). To solve it, assume a solution of the form \(y = e^{rt}\), which gives the characteristic equation \(r^2 + 9 = 0\). The roots of this equation are \(r = \pm 3i\), leading to the homogeneous solution \(y_h = c_1 \cos(3t) + c_2 \sin(3t)\).
03

Solve the Non-homogeneous Equation

To find a particular solution to the non-homogeneous equation, consider a solution of the form \(y_p = A t \cos(3t) + B t \sin(3t)\) due to the form of the right-hand side. After substituting into the original differential equation and equating coefficients, solve for constants \(A\) and \(B\). For this equation, you will find that \(A = 0\) and \(B = -\frac{1}{6}\). Thus, a particular solution is \(y_p = -\frac{1}{6}t \sin(3t)\).
04

Construct the General Solution

The general solution is the sum of the homogeneous and particular solutions: \(y = y_h + y_p = c_1 \cos(3t) + c_2 \sin(3t) - \frac{1}{6}t \sin(3t)\).
05

State the General Solution

Thus, the general solution to the differential equation is \(y(t) = c_1 \cos(3t) + (c_2 - \frac{1}{6}t) \sin(3t)\).

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Key Concepts

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

Understanding Second-Order Linear Differential Equations
A second-order linear differential equation is an equation that involves the second derivative of a function, typically denoted as \( y'' \), and it is of the form:
\[ a y'' + b y' + c y = g(t) \]This means the highest derivative is the second derivative (hence second-order), and linear because the function and its derivatives appear to the first power.
For example, the given equation \( y'' + 9y = 3\sin(3t) \) is second-order because it involves \( y'' \), and linear because both \( y'' \) and \( y \) are linear terms. The non-homogeneous part here is represented by \( 3\sin(3t) \). Understanding the order of a differential equation helps us determine the complexity of the solution and the methods required.
To solve it, you always need to find both the homogeneous solution and a particular solution, adding them together to get the general solution.
Exploring Homogeneous Solutions
The homogeneous solution deals with solving the equation when the non-homogeneous part or function \( g(t) \) is zero. For the given differential equation, it involves solving:
\[ y'' + 9y = 0 \]
To find the homogeneous solution, we use an approach that involves assuming a solution of the form \( y = e^{rt} \), leading to the characteristic equation:
  • \( r^2 + 9 = 0 \)
After solving, we find the roots as \( r = \pm 3i \), indicating complex roots. Therefore, the general solution to this homogeneous equation is expressed using trigonometric functions:
  • \( y_h = c_1 \cos(3t) + c_2 \sin(3t) \)
The \( c_1 \) and \( c_2 \) are constants determined by initial conditions, if given. This solution describes a system that only depends on the characteristic equation, without any influence from external forces.
Understanding Non-Homogeneous Differential Equations
A non-homogeneous differential equation includes a non-zero function on the right side, such as:
\[ y'' + 9y = 3 \sin(3t) \]
This equation has a part \( g(t) = 3\sin(3t) \), making it non-homogeneous. To solve it, you first find the homogeneous solution, solving the equation as if \( g(t) \) was zero.
Then, locate a particular solution \( y_p \) that fits the form of \( g(t) \). A common technique is the method of undetermined coefficients, where you assume a form for \( y_p \):
  • Example for \( 3\sin(3t) \): \( y_p = A t \cos(3t) + B t \sin(3t) \)
Substitute into the original equation and solve for the coefficients \( A \) and \( B \). In this case:
  • \( A = 0 \), \( B = -\frac{1}{6} \)
This yields a particular solution of \( y_p = -\frac{1}{6}t \sin(3t) \), which addresses the influence of \( g(t) \).
Combining to Find the General Solution
The general solution of a second-order linear differential equation combines both the homogeneous and particular solutions. It provides a complete description of all potential solutions
For such problems, the general solution is formulated as:
\[ y = y_h + y_p \]This means you add the homogeneous solution \( y_h \) and the particular solution \( y_p \).
  • Homogeneous solution: \( c_1 \cos(3t) + c_2 \sin(3t) \)
  • Particular solution: \( -\frac{1}{6}t \sin(3t) \)
Combining these, the comprehensive solution becomes \( y(t) = c_1 \cos(3t) + (c_2 - \frac{1}{6}t) \sin(3t) \).
The constants \( c_1 \) and \( c_2 \) are determined by specific initial values or boundary conditions, providing solutions adapted to particular scenarios or physical problems.

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