Author Archives: kumar Pryce

Overview
We are considering methods of solving second order linear equations when the coefficients are functions of the independent variable. We consider the second order linear homogeneous equation
P(x) d2 y/ dx2 + Q(x) dy/ dx + R(x)y = 0 (equation 1)
Since the procedure for the non-homogeneous equation is similar. Many problems in mathematical physics lead to equations of this form having polynomial coefficients; examples include the Bessel equation
X2 y’’ + xy’ + (x2 – a 22) y = 0
Where (a) is a constant, and the Legendre equation
(1 – x 2 ) y’’ – 2xy’ + c(c + 1) y = 0 Where (c) is a constant
Given the equation
P(x) d2 y/ dx2 + Q(x) dy /dx + R(x)y = 0
The equation
d2 y/ dx2 + Q(x) P(x) dy/ dx + R(x) P(x) y = 0 or d2 y/ dx2 + p(x) dy/ dx + q(x)y = 0
Where p(x) = Q(x)/ P(x) and q(x) = R(x) /P(x)
is called the equivalent normalized form of equation. The point (a) is called an ordinary point of equation (1). That is to say that these two quantities have Taylor series around x=x0. We are going to be only dealing with coefficients that are polynomials so this will be equivalent to saying that

for most of the problems. The basic idea to finding a series solution to a differential equation is to assume that we can write the solution as a power series in the form,
Y(x)= ∑_(n=0)^∞▒〖An (x-Xo)^n〗

and then try to determine what the an’s need to be. We will only be able to do this if the point x=x0, is an ordinary point. We will usually say that is a series solution around x=x0.

Example

y”+y=0

Suppose that   has a Taylor series about x=0

y(x) =  xⁿ = a₀+a₁x+a₂x²+a₃x³+a₄x⁴

Substitute into the differential equation and simplify by grouping together terms with similar powers of 

We start with the assumption that

Y(x) = a₀+a₁x+a₂x²+a₃x³+a₄x⁴  + …..

Y’(x) = a₁+2a₂x+3a₃x²+4a₄x³  +……

Y”(x) = 2a₂+6a₃x+12a₄x²+……

Now substitute   and   into the differential equation

Y’’+y= 0:

(2a₂+6a₃x+12a₄x²+……) +( a₀+a₁x+a₂x²+a₃x³+a₄x⁴  + …..) = 0

Get rid of parentheses (don’t forget to distribute the  and the   in front of the second and third sets of parentheses

2a₂+6a₃x+12a₄x²+…+ a₀+a₁x+a₂x²+a₃x³+a₄x⁴  + ….= 0

Now group together by powers of   :

(2a₂₊a₀)+ (6a₃+a₁)x +(12a₄+a₂)x²+(20a₅+a₃)x³+…= 0

Finally, we compare each term on the left with the corresponding term on the right – since the right side is zero, each of the expressions in the parentheses (which give the coefficients of the powers of  ) must also be equal to zero:

• 2a₂+a₀ =0
• 6a₃+a₁ =0
• 12a4+a2=0
• 20a₅+a₃ =0
• ……

Then you find the first 5 terms ( coefficients)

• a₀ =0
• a₁=0
• a₂= a₀/2
• a₃= a₁/6
• a₄= a₀/ 24

 

This gives us the coefficients to determine the first five terms of the Taylor Series, Remembering that Y(x) = a₀+a₁x+a₂x²+a₃x³+a₄x⁴  +…. we substitute in the values of   to obtain

Y=a₀+a₁x+a₀/2 x² +a₁/6 x³ + a₀/24 x⁴

3. In case I wasn’t clear enough with the explanation or steps needed to solve the equation i will add a couple of video hopefully they help.