Problem 13
Question
Verify by direct substitution that the given power series is a particular solution of the indicated differential equation. $$ y=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^{n}, \quad(x+1) y^{\prime \prime}+y^{\prime}=0 $$
Step-by-Step Solution
Verified Answer
The given power series satisfies the differential equation by substitution.
1Step 1: Differentiate the Power Series
Given the power series \( y = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^{n} \), first find its first derivative \( y' \). Differentiate term by term: \( y' = \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} \). Notice that the exponent of \( x \) decreases by 1 and the coefficient \( \frac{(-1)^{n+1}}{n} \) becomes \( (-1)^{n+1} \) after differentiating.
2Step 2: Find the Second Derivative
Now find the second derivative, \( y'' \), by differentiating \( y' \) again: \( y'' = \sum_{n=1}^{\infty} (-1)^{n+1} (n-1) x^{n-2} \). Again, the exponent of \( x \) decreases by 1, while the \( n-1 \) factor comes from differentiating the \( x^{n-1} \) term.
3Step 3: Substitute Into the Differential Equation
Substitute \( y' \) and \( y'' \) back into the differential equation \((x+1) y'' + y' = 0\). Rewrite \( y'' \) using the second derivative found: \( (x+1) \sum_{n=1}^{\infty} (-1)^{n+1} (n-1) x^{n-2} + \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} = 0 \).
4Step 4: Simplify the Expression
Distribute \( (x+1) \) across the sums: \( \sum_{n=1}^{\infty} (-1)^{n+1} (n-1) x^{n-1} + \sum_{n=1}^{\infty} (-1)^{n+1} (n-1) x^{n-2} + \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} = 0 \). Simplification shows that the two series cancel out to zero, confirming the given power series is a solution of the differential equation.
Key Concepts
Power SeriesSecond DerivativeSubstitution MethodSolution Verification
Power Series
A power series is an infinite series involving powers of a variable, usually written in the form \( y = \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) represents the coefficients of the series. In the given exercise, the series is \( y = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n \). The power series can converge and represent a function for particular values of \( x \). We usually explore their convergence when solving differential equations through methods such as substitution or by identifying known series representations.
When dealing with differential equations, power series can be used to find solutions in cases where traditional methods aren't suitable or simple. They allow us to express solutions as infinite sums, which can be manipulated through term-by-term differentiation or substitution methods.
When dealing with differential equations, power series can be used to find solutions in cases where traditional methods aren't suitable or simple. They allow us to express solutions as infinite sums, which can be manipulated through term-by-term differentiation or substitution methods.
Second Derivative
Differentiating functions or series multiple times allows us to find higher order derivatives, like the second derivative. In the realm of differential equations, this is particularly useful to solve equations where second derivatives appear.
To find the second derivative of a power series, we start by finding its first derivative by taking the derivative of each term individually. For our series, this first differentiation gives us \( y' = \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} \). Note how the power of \( x \) decreases by one, and the constant terms are derived.
Then, by differentiating this result, we obtain the second derivative \( y'' = \sum_{n=1}^{\infty} (-1)^{n+1} (n-1) x^{n-2} \), where once again, \( x \) decreases in power, and coefficients adjust according to the power rule. This derivative is crucial to understanding how a function changes and is used directly in the substitution step of solving our differential equation.
To find the second derivative of a power series, we start by finding its first derivative by taking the derivative of each term individually. For our series, this first differentiation gives us \( y' = \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} \). Note how the power of \( x \) decreases by one, and the constant terms are derived.
Then, by differentiating this result, we obtain the second derivative \( y'' = \sum_{n=1}^{\infty} (-1)^{n+1} (n-1) x^{n-2} \), where once again, \( x \) decreases in power, and coefficients adjust according to the power rule. This derivative is crucial to understanding how a function changes and is used directly in the substitution step of solving our differential equation.
Substitution Method
The substitution method is a technique used to simplify differential equations by replacing variables, derivatives, or other components with known expressions or series. It works particularly well when validating series solutions.
In the given problem, once we find \( y' \) and \( y'' \), we substitute them back into the original differential equation \((x+1)y'' + y' = 0\). This strategic replacement helps us verify if our series solution is correct.
By performing a direct substitution, we replace each occurrence of \( y'' \) and \( y' \) with our derived series expressions. This allows us to equate the resulting expression to zero, and through simplification, verify that the given power series satisfies the differential equation.
In the given problem, once we find \( y' \) and \( y'' \), we substitute them back into the original differential equation \((x+1)y'' + y' = 0\). This strategic replacement helps us verify if our series solution is correct.
By performing a direct substitution, we replace each occurrence of \( y'' \) and \( y' \) with our derived series expressions. This allows us to equate the resulting expression to zero, and through simplification, verify that the given power series satisfies the differential equation.
Solution Verification
Solution verification is the process of proving that a potential solution satisfies the equation it is supposed to solve. In our differential equation, this means showing that substituting the power series into the equation results in a true statement.
After substituting, distribute any coefficients and combine like terms. In our case, substituting \( (x+1) \) across the series, we get \( \sum_{n=1}^{\infty} (-1)^{n+1} (n-1) x^{n-1} + \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} = 0 \). The verification phase simplifies this expression.
After substituting, distribute any coefficients and combine like terms. In our case, substituting \( (x+1) \) across the series, we get \( \sum_{n=1}^{\infty} (-1)^{n+1} (n-1) x^{n-1} + \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} = 0 \). The verification phase simplifies this expression.
- Combine terms such that like powers of \( x \) are grouped together.
- Check if, after simplification, all terms add up to zero.
Other exercises in this chapter
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