Problem 32
Question
Verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval. $$ \begin{aligned} &y^{\prime \prime}+y=\sec x \\ &y=c_{1} \cos x+c_{2} \sin x+x \sin x+(\cos x) \ln (\cos x) \\ &(-\pi / 2, \pi / 2) \end{aligned} $$
Step-by-Step Solution
Verified Answer
The given function satisfies the differential equation, confirming it as the general solution.
1Step 1: Identify the General Solution Form
The general solution for the differential equation \( y'' + y = \sec x \) is provided as \( y = c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x) \). Here, \( c_1 \) and \( c_2 \) are arbitrary constants, and other terms represent a particular solution.
2Step 2: Calculate First and Second Derivatives
Find the first derivative \( y' \) of the general solution: \( y' = -c_1 \sin x + c_2 \cos x + (1 + \ln(\cos x)) \cos x + x \cos x \). Then find the second derivative \( y'' \): \( y'' = -c_1 \cos x - c_2 \sin x - (\ln(\cos x) + 2) \sin x \).
3Step 3: Substitute into the Differential Equation
Substitute \( y'' \) and \( y \) into the differential equation \( y'' + y = \sec x \). Add the derivatives: \( (-c_1 \cos x - c_2 \sin x - (\ln(\cos x) + 2) \sin x) + (c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x)) = \sec x \).
4Step 4: Simplify the Expression
Simplify the expression by combining like terms: \( -2 \sin x \ln(\cos x) + x \sin x + \cos x \ln(\cos x) = \sec x \). Using trigonometric identities and simplifications specific to \( \sec x = 1/\cos x \), verify if this equals \( \sec x \).
5Step 5: Verify Particular Solution Contribution
Confirm the expression \( x \sin x + (\cos x) \ln(\cos x) \) satisfies \( y'' + y = \sec x \). When all terms correctly substitute and simplify to satisfy the equation, this verifies the general solution provided works.
Key Concepts
General Solution VerificationTrigonometric IdentitiesParticular Solution Contribution
General Solution Verification
When working with a nonhomogeneous differential equation like \( y'' + y = \sec x \), one key aspect is verifying the general solution. This verification ensures that the solution not only satisfies the given equation but also accounts for all possible solutions within the specified interval.
To do this, you begin by understanding the form of the general solution provided: \( y = c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x) \). Here, \( c_1 \) and \( c_2 \) represent arbitrary constants that account for the homogeneous part, while the rest of the terms form a particular solution that specifically addresses the nonhomogeneous part.
Next comes differentiation. Calculating the first and second derivatives \( y' \) and \( y'' \), you see how these functions change. This differentiation helps evaluate whether substituting these into the original differential equation yields \( \sec x \), as required. If all terms perfectly match after simplification, it confirms that the proposed function is indeed the general solution.
To do this, you begin by understanding the form of the general solution provided: \( y = c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x) \). Here, \( c_1 \) and \( c_2 \) represent arbitrary constants that account for the homogeneous part, while the rest of the terms form a particular solution that specifically addresses the nonhomogeneous part.
Next comes differentiation. Calculating the first and second derivatives \( y' \) and \( y'' \), you see how these functions change. This differentiation helps evaluate whether substituting these into the original differential equation yields \( \sec x \), as required. If all terms perfectly match after simplification, it confirms that the proposed function is indeed the general solution.
Trigonometric Identities
Trigonometric identities play a crucial role in simplifying the expressions involved when verifying differential equations. In this exercise, identifying and applying the correct identities is necessary to bring the equation to the form \( \sec x = 1/\cos x \).
The function \( y = c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x) \) and its derivatives involve terms like \( \sin x \ln(\cos x) \) and \( \cos x \ln(\cos x) \). Here enters the need for identity simplifications. By applying identities such as \( \sec x = 1/\cos x \), we can reframe these functions in a way that becomes manageable and recognizable within the given equation.
Utilizing these mathematical tools appropriately, you can break down complex trigonometric expressions to find the baseline correctness of the generalized solution proposed for the differential equation. This simplification is crucial for understanding the relationships between functions and verifying that the derived solution holds true.
The function \( y = c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x) \) and its derivatives involve terms like \( \sin x \ln(\cos x) \) and \( \cos x \ln(\cos x) \). Here enters the need for identity simplifications. By applying identities such as \( \sec x = 1/\cos x \), we can reframe these functions in a way that becomes manageable and recognizable within the given equation.
Utilizing these mathematical tools appropriately, you can break down complex trigonometric expressions to find the baseline correctness of the generalized solution proposed for the differential equation. This simplification is crucial for understanding the relationships between functions and verifying that the derived solution holds true.
Particular Solution Contribution
The nonhomogeneous differential equation \( y'' + y = \sec x \) includes a particular solution that addresses the \( \sec x \) on the right-hand side. This particular solution is not derived from the homogeneous equation and thus requires special attention.
In the solution \( y = c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x) \), the terms \( x \sin x + (\cos x) \ln(\cos x) \) represent this particular solution. These terms are crafted to counterbalance the \( \sec x \) term from the original differential equation.
To confirm their validity, substitute both the particular solution and its derivatives into the differential equation. If these substitutions, after proper simplification and trigonometric manipulation, result in \( \sec x \), it shows that the particular solution fully contributes to the general solution. This step ensures every part of the general solution is justified, emphasizing the integral role the particular solution plays alongside the arbitrary constants for a complete answer.
In the solution \( y = c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x) \), the terms \( x \sin x + (\cos x) \ln(\cos x) \) represent this particular solution. These terms are crafted to counterbalance the \( \sec x \) term from the original differential equation.
To confirm their validity, substitute both the particular solution and its derivatives into the differential equation. If these substitutions, after proper simplification and trigonometric manipulation, result in \( \sec x \), it shows that the particular solution fully contributes to the general solution. This step ensures every part of the general solution is justified, emphasizing the integral role the particular solution plays alongside the arbitrary constants for a complete answer.
Other exercises in this chapter
Problem 32
$$ \text { Solve the given initial-value problem. } $$ $$ \begin{array}{llll} 4 y^{\prime \prime}-4 y^{\prime}-3 y & 0, y(0) & 1, y^{\prime}(0) & 5 \end{array}
View solution Problem 32
Solve the given boundary-value problem. $$ x^{2} y^{\prime \prime}-3 x y^{\prime}+5 y=0, y(1)=0, y(e)=1 $$
View solution Problem 32
Consider the differential equation \(x^{2} y^{\prime \prime}-\left(x^{2}+2 x\right) y^{\prime}+\) \((x+2) y=x^{3}\). Verify that \(y_{1}=x\) is one solution of
View solution Problem 32
In Problems 31 and 32, solve the given boundary-value problem. $$ x^{2} y^{\prime \prime}-3 x y^{\prime}+5 y=0, y(1)=0, y(e)=1 $$
View solution