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Magazine Chapter 9: Problem 1
Solve. \(y^{\prime \prime}+y=7\)
Short Answer
Expert verified
\( y(t) = C_1 \text{cos}(t) + C_2 \text{sin}(t) + 7 \)
Step by step solution
01
Write the homogeneous equation
First, consider the homogeneous version of the differential equation, which is obtained by setting the non-homogeneous term to zero. This gives: \[ y^{\text{''}} + y = 0 \]
02
Find the characteristic equation
To solve the homogeneous equation, form the characteristic equation by substituting a trial solution of the form \( y = e^{rt} \). The characteristic equation becomes: \[ r^2 + 1 = 0 \]
03
Solve the characteristic equation
Solve for \( r \) in the characteristic equation: \[ r^2 = -1 \] This gives: \[ r = \text{卤}i \]
04
Write the general solution of the homogeneous equation
Using the results from the characteristic equation, the general solution to the homogeneous equation is: \[ y_h(t) = C_1 \text{cos}(t) + C_2 \text{sin}(t) \]
05
Find a particular solution to the non-homogeneous equation
Since the non-homogeneous term is a constant (7), we guess a particular solution of the form \( y_p = A \), where \( A \) is a constant. Substitute \( y_p \) into the original differential equation: \[ y^{\text{''}} + y = 7 \] Giving: \[ 0 + A = 7 \] Thus: \[ A = 7 \] Therefore, the particular solution is: \[ y_p = 7 \]
06
Write the general solution of the non-homogeneous equation
Combine the homogeneous and particular solutions to form the general solution of the non-homogeneous differential equation: \[ y(t) = y_h(t) + y_p(t) = C_1 \text{cos}(t) + C_2 \text{sin}(t) + 7 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Differential Equation
A homogeneous differential equation is one where the right-hand side is zero. This means the equation is of the form: \[ y'' + P(t)y' + Q(t)y = 0 \] In the given exercise, the homogeneous version of the equation is: \[ y'' + y = 0 \] Such equations can often be solved using standard methods like finding the characteristic equation. For our specific problem, transforming the non-homogeneous equation \[ y'' + y = 7\] into its homogeneous counterpart helps to find the overall solution.
Characteristic Equation
The characteristic equation is derived from a homogeneous differential equation by assuming a solution of exponential form, typically given by \[ y = e^{rt} \]. This simplification allows us to transform the differential equation into an algebraic equation. For our exercise, the homogeneous equation \[ y'' + y = 0 \] leads to the characteristic equation: \[ r^2 + 1 = 0 \]. Solving this provides roots in the form of complex numbers. Here, \[ r = 卤i \] indicates oscillatory behavior characteristic of solutions involving sine and cosine functions.
Non-Homogeneous Differential Equation
A non-homogeneous differential equation includes a non-zero term on the right-hand side. This form is expressed as: \[ y'' + P(t)y' + Q(t)y = G(t) \]. In our example, the equation \[ y'' + y = 7 \] is non-homogeneous because of the constant term '7'. To solve this, we need a particular solution, which in our case is a constant guess, leading to the solution \[ y_p = 7 \]. This part of the solution addresses the 'forcing' term contributing to the non-homogeneous nature.
General Solution
The general solution of a non-homogeneous differential equation combines the solutions of its homogeneous counterpart with a particular solution to the non-homogeneous part:
- Homogeneous solution, derived from the characteristic equation
- Particular solution, addressing the non-homogeneous term
