Math Problem Statement

(-5+x)y'' + (1+4x)y' +3y = 0, y(0) = 3, y'(0)=4

Assuming the solution can be represented by a power series

y=\sum_{n=0}^{\infty}a_nx^n

y' = \sum_{n=1}^{\infty}na_nx^(n-1)

y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^(n-2)

\sum_{n=2}^{\infty}\left(-5+x\right)n\left(n-1\right)a_nx^{n-2} + \sum_{n=1}^{\infty}na_nx^(n-1)+4na_nx^2 + \sum_{n=0}^{\infty}-3a_nx^n = 0

\sum_{n=0}^{\infty} -5(n+2)(n+1)a_(n+2)+(n+1)^2a_(n+1)+(4n+3)a_n x^n=0

a_(n+2) = ((n+1)^2)/(5(n+2)(n+1)) a_(n+1) + (4n+3)/(5(n+2)(n+1)) a_n

Using the initial values and the recursive formula, determine the first few terms of the series solution

y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...

where

a_0 = ?

a_1 = ?

a_2 =?

a_3=?

a_4=?

Solution

We are tasked with computing the first few terms of the series solution for y(x)y(x) using the initial conditions and the recursive formula.


Given:

The recursive formula is: an+2=(n+1)25(n+2)(n+1)an+1+4n+35(n+2)(n+1)an.a_{n+2} = \frac{(n+1)^2}{5(n+2)(n+1)}a_{n+1} + \frac{4n+3}{5(n+2)(n+1)}a_n.

Initial conditions: a0=3,a1=4.a_0 = 3, \quad a_1 = 4.


Step 1: Compute a2a_2

Using the recursive formula with n=0n = 0: a2=(0+1)25(2)(1)a1+4(0)+35(2)(1)a0.a_2 = \frac{(0+1)^2}{5(2)(1)}a_1 + \frac{4(0) + 3}{5(2)(1)}a_0.

Simplify: a2=110a1+310a0.a_2 = \frac{1}{10}a_1 + \frac{3}{10}a_0.

Substitute a0=3a_0 = 3 and a1=4a_1 = 4: a2=110(4)+310(3)=410+910=1310.a_2 = \frac{1}{10}(4) + \frac{3}{10}(3) = \frac{4}{10} + \frac{9}{10} = \frac{13}{10}.

Thus: a2=1.3.a_2 = 1.3.


Step 2: Compute a3a_3

Using the recursive formula with n=1n = 1: a3=(1+1)25(3)(2)a2+4(1)+35(3)(2)a1.a_3 = \frac{(1+1)^2}{5(3)(2)}a_2 + \frac{4(1) + 3}{5(3)(2)}a_1.

Simplify: a3=2230a2+4(1)+330a1.a_3 = \frac{2^2}{30}a_2 + \frac{4(1) + 3}{30}a_1. a3=430a2+730a1.a_3 = \frac{4}{30}a_2 + \frac{7}{30}a_1.

Substitute a1=4a_1 = 4 and a2=1.3a_2 = 1.3: a3=430(1.3)+730(4).a_3 = \frac{4}{30}(1.3) + \frac{7}{30}(4). a3=5.230+2830=33.230=1.1067.a_3 = \frac{5.2}{30} + \frac{28}{30} = \frac{33.2}{30} = 1.1067.

Thus: a31.107.a_3 \approx 1.107.


Step 3: Compute a4a_4

Using the recursive formula with n=2n = 2: a4=(2+1)25(4)(3)a3+4(2)+35(4)(3)a2.a_4 = \frac{(2+1)^2}{5(4)(3)}a_3 + \frac{4(2) + 3}{5(4)(3)}a_2.

Simplify: a4=3260a3+4(2)+360a2.a_4 = \frac{3^2}{60}a_3 + \frac{4(2) + 3}{60}a_2. a4=960a3+1160a2.a_4 = \frac{9}{60}a_3 + \frac{11}{60}a_2.

Substitute a2=1.3a_2 = 1.3 and a3=1.107a_3 = 1.107: a4=960(1.107)+1160(1.3).a_4 = \frac{9}{60}(1.107) + \frac{11}{60}(1.3). a4=9.96360+14.360=24.26360=0.404.a_4 = \frac{9.963}{60} + \frac{14.3}{60} = \frac{24.263}{60} = 0.404.

Thus: a40.404.a_4 \approx 0.404.


Final Series Solution (First Few Terms):

The solution is: y(x)=3+4x+1.3x2+1.107x3+0.404x4+y(x) = 3 + 4x + 1.3x^2 + 1.107x^3 + 0.404x^4 + \dots


Summary of Coefficients:

a0=3,a1=4,a2=1.3,a31.107,a40.404.a_0 = 3, \quad a_1 = 4, \quad a_2 = 1.3, \quad a_3 \approx 1.107, \quad a_4 \approx 0.404.

Would you like to compute additional terms or analyze the behavior of this series further?


Related Questions:

  1. How does the solution behave for large values of xx?
  2. What is the radius of convergence for this series solution?
  3. Can the solution be approximated by a simpler polynomial for small xx?
  4. How do the initial conditions affect the solution coefficients?
  5. Can we verify the solution by substituting it back into the original differential equation?

Tip:

When working with power series, it's helpful to keep terms unrounded until the very end of your calculations to ensure maximum accuracy in intermediate computations.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Power Series Solutions
Recursive Formulas

Formulas

y = ∑(n=0 to ∞) anx^n
y' = ∑(n=1 to ∞) nanx^(n-1)
y'' = ∑(n=2 to ∞) n(n-1)anx^(n-2)
an+2 = ((n+1)^2 / (5(n+2)(n+1))) an+1 + ((4n+3) / (5(n+2)(n+1))) an

Theorems

Existence and Uniqueness Theorem for Differential Equations
Convergence of Power Series

Suitable Grade Level

Undergraduate Mathematics