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# Series Solution of Linear Ordinary Differential Equations

Aim: To study methods for determining series expansions for solutions to linear ODE with variable coefficients.
In particular, we shall obtain
• the form of the series expansion,
• a recurrence relation for determining the coefficients

Given the differential equation $$\frac{d^2y}{dx^2} + P(x)\frac{dy}{dx} + Q(x)y = 0$$, substitute$$y=\sum_{n=0}^{\infty}{a_n(x − x_0)^n }$$ and solve for the $$a_n$$ to find a power series solution centered at $$x_0$$.

Power Series
A series of the form $$\sum_{n=0}^{\infty}{a_n(x − x_0)^n } =a_0 + a_1(x − x_0) + a_2(x − x_0)^2 + · · ·$$ is called a power series about the point $$x_0$$. Here, x is a variable and $$a_n$$’s are constants.

Analytic function
A function $$f(x)$$ defined on an interval containing the point $$x_0$$ is called analytic at $$x_0$$ if its Taylor series about $$x_0$$, i.e., $$\sum_{n=0}^{\infty}{\frac{f^n(x_0)}{n!}(x − x_0)^n }$$ exists and converge to $$f(x)$$ for all x in some nbd of $$x_0$$.
For example, $$e^x, sinx, cosx, sinhx$$ are analytic everywhere, where as the function $$\frac{x-2}{x-3}$$ is analytic except $$x=3$$.

## Power Series Solutions to D.E.s at Ordinary Point

Let us consider a differential equation in the standard form :$$\frac{d^2y}{dx^2} + P(x)\frac{dy}{dx} + Q(x)y = 0$$

Definition

A point $$x_0$$ is an ordinary point if both P(x) and Q(x) are analytic at $$x_0$$. If a point in not ordinary it is a singular point.

#### Solved Problems on Power series method about an ordinary point

Prob.(1): Find the ordinary and singular points of the D.E. $$(1-x^2)\frac{d^2y}{dx^2} -2x\frac{dy}{dx} + 2y = 0$$

Prob.(2): Consider the differential equation $$(x-1)\frac{d^2y}{dx^2} +x\frac{dy}{dx} + \frac{1}{x} y = 0$$
Then
1. x=1 is the only singular point
2. x=0 is the only singular point
3. both x=0 and x=1 are singular points
4. neither x=0 nor x=1 are singular
points
[NET 2017(D)]

Prob.(3): Find the power series solution of  the D.E. $$\frac{d^2y}{dx^2} +3x\frac{dy}{dx} + 3y = 0$$ about x=0
[C.H. 2006] [C.H. 2009] [C.H. 2012] [C.H. 2015]

#### EXERCISE

1. Find the power series solution of  the D.E. $$\frac{d^2y}{dx^2} +3x\frac{dy}{dx} + 3y = 0$$ near the ordinary point x=0
[C.H. 2006] [C.H. 2009] [C.H. 2012] [C.H. 2015]
2. Find the power series solution of  the D.E. $$(1-x^2)\frac{d^2y}{dx^2} +2x\frac{dy}{dx} -y = 0$$ about x=0
[C.H. 2011] [C.H. 2013
3. Find the power series solution of  the D.E. $$\frac{d^2y}{dx^2} – y = x$$ about x=0
[C.H. 2007] [C.H. 2009
4. Find the power series solution of  the D.E. $$\frac{d^2y}{dx^2} +x\frac{dy}{dx} +x^2 y = 0$$ about x=0
[C.H. 2013] [C.H. 2014]
5. Find the power series solution in power of x:$$\frac{d^2y}{dx^2} + y = 0$$ near the ordinary point x=0
[C.H. 2007
6. Find the power series solution of  the D.E. $$(x^2+1)\frac{d^2y}{dx^2} +x\frac{dy}{dx} -x y = 0$$ about x=0
[C.H. 2014
7. Find the power series solution of  the D.E. $$(1-x^2)\frac{d^2y}{dx^2} +2x\frac{dy}{dx} – y = 0$$ about x=0
8. Find the power series solution of  the D.E. $$(x^3-1)\frac{d^2y}{dx^2} +x^2\frac{dy}{dx} +xy = 0$$ about x=0
9. Find the power series solution of  the D.E. $$\frac{d^2y}{dx^2} -4(x-1)y = 0$$ near the ordinary point x=1
[C.H. 2008
10. Find the power series solution of  the D.E. $$\frac{d^2y}{dx^2} +(x-1)^2\frac{dy}{dx} -4(x-1) y = 0$$ near the ordinary point x=1
[C.H. 2012
11. Find the power series solution of the following initial-value problems:
(i) $$\frac{d^2y}{dx^2} -4y = 0$$ satisfying $$y(0)=1, y^{/} (0)=-1$$
[C.H. 2008] [C.H. 2011]
(ii) $$\frac{d^2y}{dx^2} -x\frac{dy}{dx}-y = 0$$ satisfying $$y(0)=1 , y^{/} (0)=0$$
[C.H. 2009]
(iii) $$\frac{d^2y}{dx^2} +x\frac{dy}{dx}-2y = 0$$ satisfying $$y(0)=1 , y^{/} (0)=1$$
[C.H. 2009]
(iv)$$\frac{d^2y}{dx^2} +x\frac{dy}{dx}+2y = 0$$ satisfying $$y(0)=1 , y^{/} (0)=1$$
[C.H. 2010]
(v) $$\frac{d^2y}{dx^2} +8x\frac{dy}{dx}-4y = 0$$ satisfying $$y(0)=1 , y^{/} (0)=0$$
[C.H. 2010]
12. Find the power series solutions $$x^2\frac{d^2y}{dx^2} +3x\frac{dy}{dx}-y = 0$$
1. $$y(x)=a_0 (1-\frac{3}{2} x^2+\frac{9}{8} x^4-\dots)+a_1 (x-x^3+\frac{3}{5} x^5-\dots)$$
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3. $$y(x)=a_0 (1+\frac{1}{2} x^2+\frac{1}{24} x^4+⋯)+a_1 (x+\frac{1}{6} x^3+\frac{1}{120} x^5+⋯)+\\(\frac{1}{6} x^3+\frac{1}{120} x^5+⋯)$$
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