Chapter 3: Problem 37
In Problems 37 and 38, solve the given initial-value problem. $$ \frac{d^{2} x}{d t^{2}}+4 x=-5 \sin 2 t+3 \cos 2 t, x(0)=-1, x^{\prime}(0)=1 $$
Short Answer
Expert verified
The solution to the initial-value problem is \( x(t) = -\frac{1}{2} \cos 2t + \frac{7}{2} \sin 2t \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \( \frac{d^{2}x}{dt^{2}} + 4x = -5 \sin 2t + 3 \cos 2t \). This is a second-order linear non-homogeneous differential equation.
02
Solve the Homogeneous Equation
The corresponding homogeneous equation is \( \frac{d^{2}x}{dt^{2}} + 4x = 0 \). Solving this gives the characteristic equation \( r^2 + 4 = 0 \). Hence, \( r = \pm 2i \), leading to the complementary solution \( x_c(t) = C_1 \cos 2t + C_2 \sin 2t \).
03
Solve the Particular Equation
We use the method of undetermined coefficients to find a particular solution for the non-homogeneous part. Assume a particular solution of the form \( x_p(t) = A \sin 2t + B \cos 2t \). Substitute into the original equation, differentiate as needed, and equate coefficients to solve for \( A \) and \( B \).
04
Substitute and Solve for Coefficients
Substituting \( x_p(t) = A \sin 2t + B \cos 2t \) into \( \frac{d^{2}x}{dt^{2}} + 4x = -5 \sin 2t + 3 \cos 2t \) gives \( -4A \sin 2t - 4B \cos 2t + 4A \sin 2t + 4B \cos 2t = -5 \sin 2t + 3 \cos 2t \). Solving for coefficients yields \( A = \frac{5}{4} \), \( B = \frac{3}{4} \).
05
Form the General Solution
The general solution is the sum of the complementary and particular solutions: \( x(t) = C_1 \cos 2t + C_2 \sin 2t + \frac{5}{4} \sin 2t + \frac{3}{4} \cos 2t \).
06
Apply the Initial Conditions
Use the initial conditions \( x(0) = -1 \) and \( x'(0) = 1 \) to solve for \( C_1 \) and \( C_2 \). For \( x(0) = -1 \), we have \( C_1 + \frac{3}{4} = -1 \), so \( C_1 = -\frac{7}{4} \). For \( x'(0) = 1 \), differentiate \( x(t) \) and substitute to solve for \( C_2 \). This gives \( C_2 = \frac{9}{4} \).
07
Write the Final Solution
Substituting \( C_1 = -\frac{7}{4} \), \( C_2 = \frac{9}{4} \) into the general solution, we get: \[ x(t) = -\frac{7}{4} \cos 2t + \frac{9}{4} \sin 2t + \frac{5}{4} \sin 2t + \frac{3}{4} \cos 2t \] Simplify to get the final solution: \[ x(t) = -\frac{1}{2} \cos 2t + \frac{7}{2} \sin 2t \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Equations
In mathematics, homogeneous equations are a type of differential equation where the terms involving the dependent variable and its derivatives are directly set to zero. This means these equations have a form something like \( \frac{d^2x}{dt^2} + 4x = 0 \). Here, the left side involves derivatives of \( x \) and \( x \) itself, but all terms sum to zero.
This equation is significant because it helps us find the complementary solution, which is a core component of solving non-homogeneous differential equations.
The importance lies in establishing a basis for the solution. By solving homogeneous equations, you find solutions that form the basis for the solutions to more complex non-homogeneous equations. The characteristic equation, discussed below, is used to find these solutions.
This equation is significant because it helps us find the complementary solution, which is a core component of solving non-homogeneous differential equations.
The importance lies in establishing a basis for the solution. By solving homogeneous equations, you find solutions that form the basis for the solutions to more complex non-homogeneous equations. The characteristic equation, discussed below, is used to find these solutions.
Method of Undetermined Coefficients
The method of undetermined coefficients is a technique used to find a particular solution to non-homogeneous linear differential equations. When an equation involves terms that aren't zero on the right side, as in \( \frac{d^2x}{dt^2} + 4x = -5 \sin 2t + 3 \cos 2t \), we use this method.
Here's how it works:
Here's how it works:
- Assume a particular form for the solution based on the non-homogeneous part. Here, we assume \( x_p(t) = A \sin 2t + B \cos 2t \).
- Substitute this assumed form into the equation.
- Differentiate as necessary and equate coefficients to solve for unknowns, aiming to find the values of \( A \) and \( B \).
Initial-Value Problem
An initial-value problem (IVP) is a differential equation paired with specific conditions known as 'initial conditions.' These allow us to pinpoint a unique solution for the differential equation. In our problem, the initial conditions are \( x(0) = -1 \) and \( x'(0) = 1 \).
These conditions describe the state of a system at the starting point—often, time \( t = 0 \). They guide us in finding the specific constants in our general solution, which makes the solution unique rather than broadly applicable.
These conditions describe the state of a system at the starting point—often, time \( t = 0 \). They guide us in finding the specific constants in our general solution, which makes the solution unique rather than broadly applicable.
- Initial conditions provide specific values for the function and its derivatives at a certain point.
- They are crucial because they help tailor a general solution to fit particular scenarios.
Characteristic Equation
The characteristic equation is a crucial part of solving homogeneous differential equations. It's derived from the differential equation itself and helps find the complementary solution. For the equation \( \frac{d^2x}{dt^2} + 4x = 0 \), the characteristic equation is formed by replacing derivatives with powers of \( r \), giving us \( r^2 + 4 = 0 \).
Solving this equation determines the values of \( r \), which represent the eigenvalues.
Solving this equation determines the values of \( r \), which represent the eigenvalues.
- These values, like \( r = \pm 2i \) in our example, explain the type of solutions: real, complex, or repeated roots.
- These solutions lead to discovering the complementary solution \( C_1 \cos 2t + C_2 \sin 2t \).