Chapter 4: Problem 27
The indicated functions are known linearly independent solutions of the associated homogeneous differential equation on \((0, \infty)\). Find the general solution of the given non-homogeneous equation. $$\begin{aligned} &x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{4}\right) y=x^{3 / 2}\\\ &y_{1}=x^{-1 / 2} \cos x, y_{2}=x^{-1 / 2} \sin x \end{aligned}$$
Short Answer
Step by step solution
Verify Linear Independence of Solutions
Write the General Solution for Homogeneous Equation
Find the Particular Solution Using Variation of Parameters
Solve for \( u_1 \) and \( u_2 \)
Compute Integrals to Find \( u_1 \) and \( u_2 \)
Write the Particular Solution
Formulate the General Solution for the Non-Homogeneous Equation
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Independence
To verify if two functions, such as the solutions to a homogeneous differential equation, are linearly independent, we use the Wronskian determinant. A key takeaway is that if the Wronskian is not zero for some point in the interval of interest, then the functions are linearly independent.
- Helps in identifying unique solutions.
- Ensures the solvability of the differential equation.
- Essential for constructing the general solution.
Wronskian Determinant
If you already know the solutions are linearly independent, computing the Wronskian confirms this by yielding a non-zero result. It's a great external check to ensure mathematical rigor and accuracy. For instance, in the given exercise, the Wronskian turns out to be \( \frac{1}{x} \), which is non-zero for \( x > 0 \).
Variation of Parameters
Here's how it works:
- Assume a solution in the form: \( y_p = u_1(x)y_1(x) + u_2(x)y_2(x) \), where \( u_1 \) and \( u_2 \) are functions to be determined.
- To find \( u_1' \) and \( u_2' \), we use:\[u_1' = \frac{-y_2 g(x)}{W(y_1, y_2)}\] \[u_2' = \frac{y_1 g(x)}{W(y_1, y_2)}\], where \( g(x) \) is the non-homogeneous term, and \( W \) is the Wronskian.
- The results are integrated to find \( u_1 \) and \( u_2 \).
General Solution
For a second order linear differential equation, if \( y_1 \) and \( y_2 \) are linearly independent solutions, the general solution to the homogeneous equation is:
\[ y_h = C_1 y_1 + C_2 y_2 \], where \( C_1 \) and \( C_2 \) are arbitrary constants.
To find the general solution of the non-homogeneous equation, we add the particular solution \( y_p \) to the homogeneous solution:
- \( y = y_h + y_p \)
- This represents all solutions that satisfy the original differential equation.