- #1
jezznar
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1. Homework Statement
Find two linearly independent solutions valid for x>0
xy''+y=02. Homework Equations
well first of all, all the solutions that i get to see in the forums is interms of the euler formula.. but in the course that I am taking(advanced engineering mathematics), it is expressed using a power formula... can anyone please teach me how to do this?..3. The Attempt at a Solution
y = Summation of (a_n)(x^n) n=0 -> infinity
y'' = Summation of (n)(n-1)(a_n)(x^[n-2]) n=0 -> infinity // so yeah just differentiation
*substituting it to the equation
= [ Summation of (n)(n-1)(a_n)(x^[n-1]) n=0 -> infinity ] + [Summation of (a_n)(x^n) n=0 -> infinity]
*shifting the y equation to achieve same degree of x
= [ Summation of (n)(n-1)(a_n)(x^[n-1]) n=0 -> infinity ] + [Summation of (a_[n-1] )(x^[n-1]) n=1 -> infinity]
*getting the recurrence releationship
a_2k = [-1]^k a_1 / 2k! [2k-1]!
a_[2k+1] =[-1]^k a_1 / [2k+1]![2k]!
*now substituting it to a power series formula
y= a_0 + [ SUmmation of [a_2k] x^2k k=1 -> infinity ] + a_1 x + [ Summation of a_[2k+1] x^[2k+1] k=1 -> infinity ] +...
*now after that I really don't know how to obtain the two linearly independent solutions.. can someone enlighten me?.. did I even do this right? =[
any help will be very much appreciated...
Find two linearly independent solutions valid for x>0
xy''+y=02. Homework Equations
well first of all, all the solutions that i get to see in the forums is interms of the euler formula.. but in the course that I am taking(advanced engineering mathematics), it is expressed using a power formula... can anyone please teach me how to do this?..3. The Attempt at a Solution
y = Summation of (a_n)(x^n) n=0 -> infinity
y'' = Summation of (n)(n-1)(a_n)(x^[n-2]) n=0 -> infinity // so yeah just differentiation
*substituting it to the equation
= [ Summation of (n)(n-1)(a_n)(x^[n-1]) n=0 -> infinity ] + [Summation of (a_n)(x^n) n=0 -> infinity]
*shifting the y equation to achieve same degree of x
= [ Summation of (n)(n-1)(a_n)(x^[n-1]) n=0 -> infinity ] + [Summation of (a_[n-1] )(x^[n-1]) n=1 -> infinity]
*getting the recurrence releationship
a_2k = [-1]^k a_1 / 2k! [2k-1]!
a_[2k+1] =[-1]^k a_1 / [2k+1]![2k]!
*now substituting it to a power series formula
y= a_0 + [ SUmmation of [a_2k] x^2k k=1 -> infinity ] + a_1 x + [ Summation of a_[2k+1] x^[2k+1] k=1 -> infinity ] +...
*now after that I really don't know how to obtain the two linearly independent solutions.. can someone enlighten me?.. did I even do this right? =[
any help will be very much appreciated...
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