Chapter 17: Problem 5
Use power series to solve the differential equation. $$ y^{\prime \prime}+x y^{\prime}+y=0 $$
Short Answer
Expert verified
The series solution to the differential equation is \( y(x) = 0 \).
Step by step solution
01
Recognize the Power Series Form
We start by assuming a power series solution for the function \( y(x) \), given by \( y(x) = \sum_{n=0}^{fty} a_n x^n \). The goal is to find the coefficients \( a_n \).
02
Differentiate the Power Series
Differentiate the power series to find expressions for \( y'(x) \) and \( y''(x) \).\[ y'(x) = \sum_{n=1}^{fty} n a_n x^{n-1} \] \[ y''(x) = \sum_{n=2}^{fty} n(n-1) a_n x^{n-2} \]
03
Substitute into the Differential Equation
Substitute \( y(x) \), \( y'(x) \), and \( y''(x) \) into the differential equation:\[ \sum_{n=2}^{fty} n(n-1) a_n x^{n-2} + x \sum_{n=1}^{fty} n a_n x^{n-1} + \sum_{n=0}^{fty} a_n x^n = 0 \]
04
Match Powers of x
Combine terms and arrange the power series so that powers of \( x \) match:- For \( x^0 \), equilibrium gives \( a_0 = 0 \).- For \( x^1 \), equilibrium gives \( a_1 = 0 \).- For \( x^2 \) and higher powers, establish a recurrence relation:\[ (n(n-1) + n + 1) a_n = 0 \]
05
Solve the Recurrence Relation
Analyze the recurrence relation:\[ (n^2) a_n = 0 \]This implies \( a_n = 0 \) for \( n \geq 2 \). Therefore, all coefficients are zero from the start.
06
Conclude the Series Solution
Since all coefficients are zero, \( y(x) = 0 \) is the only solution. The differential equation admits the trivial solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series can be a fantastic tool for solving differential equations. It is an infinite sum of terms involving an adjustable coefficient and powers of the variable. Specifically, it takes the form:\[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]Here, each \( a_n \) represents a coefficient that needs to be determined. This type of series is especially useful when dealing with non-linear or complex differential equations, as it turns them into an algebraic problem—finding those coefficients!
- Each term in the power series represents a polynomial term.
- The power series allows for an approximation of the solution by truncating after a certain number of terms.
- This method transforms differential equations into polynomial equations by substituting the power series and its derivatives.
Recurrence Relation
To solve differential equations using power series, we often end up with a recurrence relation. This relation helps us determine the coefficients of the power series.In simpler terms, a recurrence relation is an equation that recursively defines a sequence, meaning that it relates each term to its predecessors.Consider the recurrence relation found in the exercise:\[ (n(n-1) + n + 1) a_n = 0 \]The beauty of a recurrence relation is that it provides a systematic way to compute all coefficients, but given that the equation breaks down to:\[ (n^2) a_n = 0 \]it simplifies to show that each \( a_n \) after a specific point must be zero. Here:
- Identifies terms that relate each coefficient to its preceding terms.
- Helps compute coefficients methodically.
- Drives the solution towards an eventual pattern or reveal that the series collapses into a trivial case.
Trivial Solution
In the realm of differential equations, a trivial solution is when the solution is considered to be a simple or insignificant constant that satisfies the equation. Typically, the zero solution is termed 'trivial.' In our power series solution, we find:\[ y(x) = 0 \]This outcome means every coefficient in the series ends up being zero, due to the recurrence relation we derived:\[ (n^2) a_n = 0 \]This final step confirms that the solution indeed is trivial.
- Derived when all coefficients in the series equate to zero.
- Sanity checks the correctness of having no other potential solutions hidden.
- Highlights the characteristics of the equation leading to simple or obvious solutions.