Question

Find the power series solution of the equation: 

x^2frac{d^2y}{dx^2}+4xfrac{dy}{dx}+(x^2+2)y = 0

about its singular point.

21 Mar 2023
Answer :
Word Count : 453

To find the power series solution of the given differential equation, we need to first determine the singular point of the equation. The singular point is the point at which the coefficient of the highest derivative term becomes zero.

Here, the coefficient of the highest derivative term is x^2, so the singular point is x = 0.

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Related Questions

Find the Fourier sine transform of 

f(x)=\left\{\begin{matrix} x, & 0<x<1\\ 2-x,& 1<x<2\\ 0, & x>2 \end{matrix}\right.

Given \frac{dy}{dx}=x-y^2,y(0.02)=(0.02). Find y(0.4)by using modified Euler’s method, correct to two decimal places, taking h = 0.2.

Given

\frac{dy}{dx}=-300y,y(0)=1

Determine the value of h so that the second order Runge-Kutta method applied to be IVP produces stable results. 

State whether the following statements are True or False. Justify your answer with the help of a short proof or a counter-example:

i)  Initial value problem:

\frac{dy}{dx}=\frac{y-1}{x}

y (0)=1 has a unique solution

ii)  The second order Runge-Kutta method when applied to IVP y′ = −100y,  y(0)=1 will produce stable results for  0<h<\frac{1}{50}.

iii) If Fourier cosine transform of  f(x ) is:

F_{c}(n)= \frac{cos(\frac{2n\pi }{3})}{(2n+1)^2}

where  0 ≤ x ≤ then:

f(x)=1+2\sum_{n=1 }^{\infty}\frac{cos(\frac{2n\pi }{3})}{(2n+1)}cosn \pi x.

For the differential equation x_{2}(x-4)^2y''(x)+3xy'(x)-(x-4)y=0,x=0 is a 

regular singular point and x=4,is an irregular singular point.

Using the Crank Nicolson method, integrate upto one time level for the solution of the initial value problem: 

\frac{\partial u}{\partial t}=\frac{\partial ^2u}{\partial x2},0<x<1

with u(x,0)=sin 2\pi x,

          u(0,t)=0=u(1,t)

with h=\frac{1}{3} and \lambda =\frac{1}{6}.

If  f'(x_{k})is approximated by:

f'(x_{k}) = af(x_{k+1})+bf'(x_{k+1}),

find the values of a and b. What is the order of approximation?

 

 

If H_{2} is a Hermite polynomial of degree n, then show that: 

H_{n}^{n}=4n(n=1)H_{n-2}

 

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