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Chapter 4: Problem 35

\(x^{3} y^{\prime \prime \prime}+10 x^{2} y^{\prime \prime}-20 x y^{\prime}+20 y=0, y(1)=0\), \(y^{\prime}(1)=-1, y^{\prime \prime}(1)=1\)

Short Answer

Expert verified
The series solution can be found using the initial conditions leading to the coefficients of the series.

Step by step solution

01

- Rewrite the Differential Equation

Simplify and rewrite the given differential equation. It is already in a fairly simple form: \[ x^{3} y^{\rm \prime \prime \prime}+10 x^{2} y^{\rm \prime \prime}-20 x y^{\rm \prime}+20 y=0 \]
02

- Assume a Series Solution

Assume a series solution of the form \[ y(x) = \sum_{n=0}^{\infty} a_n x^{n} \]
03

- Compute Derivatives

Compute the first, second, and third derivatives of the series, thus getting: \[ y^{\rm \prime}(x) = \sum_{n=0}^{\infty} (n a_n x^{n-1}) \] \[ y^{\rm \prime \prime}(x) = \sum_{n=0}^{\infty} (n (n-1) a_n x^{n-2}) \] \[ y^{\rm \prime \prime \prime}(x) = \sum_{n=0}^{\infty} (n (n-1) (n-2) a_n x^{n-3}) \]
04

- Substitute Derivatives into Original Equation

Substitute these derivatives back into the original differential equation. \[ x^{3} (\sum_{n=0}^{\infty} n (n-1) (n-2) a_n x^{n-3}) + 10 x^{2} (\sum_{n=0}^{\infty} n (n-1) a_n x^{n-2}) - 20 x (\sum_{n=0}^{\infty} n a_n x^{n-1}) + 20 \sum_{n=0}^{\infty} a_n x^n = 0 \]
05

- Simplify and Combine

Combine and simplify the series terms to form a single summation. Combine like terms and set the coefficient of each power of x to zero to obtain recurrence relations for the coefficients.
06

- Apply Initial Conditions

Apply the initial conditions: \[ y(1) = 0 \implies \sum a_n = 0 \] \[ y^{\rm \prime}(1) = -1 \implies \sum n a_n = -1 \] \[ y^{\rm \prime \prime}(1) = 1 \implies \sum n (n-1) a_n = 1 \]
07

- Solve for Coefficients

Solve the system of equations for the coefficients for values of n. This will allow determining the series coefficients a_n.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are equations that involve an unknown function and its derivatives. These equations appear in many fields, such as physics, engineering, and economics. They help describe the behavior of dynamical systems. For example, Newton's second law of motion is a differential equation. In the given exercise, the equation is \[ x^{3} y^{\rm \tilde \tilderm \tilde \tildeprime}+10 x^{2} y^{\rm \tilde prime -20 x y^{\rm tilde prime} +20 y=0 \] This is a third-order differential equation, meaning it involves the third derivative of y.
Series Solution
A series solution of a differential equation expresses the solution as an infinite sum of terms. Here, we assume that the solution to the differential equation can be written as a power series:

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Most popular questions from this chapter

Solve each of the following initial value problems. Verify that your result satisfies the initial conditions by graphing it on an appropriate interval. (a) \(y^{\prime \prime \prime}+3 y^{\prime \prime}+2 y^{\prime}+6 y=0, y(0)=0, y^{\prime}(0)=\) \(1, y^{\prime \prime}(0)=-1\) (b) \(y^{(4)}-8 y^{\prime \prime \prime}+30 y^{\prime \prime}-56 y^{\prime}+49 y=0\), \(y(0)=1, y^{\prime}(0)=2, y^{\prime \prime}(0)=-1, y^{\prime \prime \prime}(0)=\) \(-1\) (c) \(0.31 y^{\prime \prime \prime}+11.2 y^{\prime \prime}-9.8 y^{\prime}+5.3 y=0\), \(y(0)=-1, y^{\prime}(0)=-1, y^{\prime \prime}(0)=0\)

\(y^{\prime \prime}-y^{\prime}-2 y=e^{-x}\)

\(x^{2} y^{\prime \prime}+x y^{\prime}+y=x^{-2}\)

\(y^{\prime \prime}-11 y^{\prime}+30 y=0\)

\(x^{3} y^{\prime \prime \prime}+6 x^{2} y^{\prime \prime}+7 x y^{\prime}+y=0\)
See all solutions

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