Question

1. Using the Frobenius method solve the following ODE:

x²y" + 4xy' + (x² + 2)y = 0

2. Determine the first and second order partial derivatives for the function:

u(x,t) = Ce^(1-n²π²)tsin(nπx)

3. Obtain the Fourier series for the following function:

0 -π < x < 0

f(x) =

sin x 0 < x < π

Answer 1

Let the solution of above differential equation be

$y=a_ox^m+a_1x^{m+1}+a_2x^{m+2}+......................+a_nx^{n+m}$

$\dfrac{dy}{dx}=ma_ox^{m-1}+(m+1)a_1x^m+(m+2)a_2x^{m+1}+.........................$

$\dfrac{d^2y}{dx^2}=m(m-1)a_ox^{m-2}+(m+1)ma_1x^{m-1}+(m+2)(m+1)a_2x^{m}+..................$

$x^2\dfrac{d^2y}{dx^2}+4x\dfrac{dy}{dx}+(x^2+2)y=0$

$x^2[m(m-1)a_ox^{m-2}+(m+1)ma_1x^{m-1}+(m+2)(m+1)a_2x^{m}+..................]+4x[ma_ox^{m-1}+(m+1)a_1x^m+(m+2)a_2x^{m+1}+.........................]+(x^2+2)[a_ox^m+a_1x^{m+1}+a_2x^{m+2}+......................+a_nx^{n+m}]=0$

Equating the coefficient of lowest degree in x (which is x^m) equal to zero

$m(m-1)a_o+4ma_o+2a_o=0$

$m^2+3m+2=0$

$m=-2,-1$

For further powers of x,

$x^{m+1}$:

$m(m+1)a_1+4(m+1)a_1+2a_1=0$

$m^2+m+4m+4+2=0$

$m^2+5m+6=0$

$m=-2,-3$

Since the equation has three roots it cannot be solved by frobenius method

The given differential equation refers to Sturm-Liouville equation.

2.

$u(x,t) = Ce^{(1-n²π²)t}sin(nπx)$

$u_x=n\pi Ce^{(1-n²π²)t}cos(nπx)$

$u_t= C(1-n²π²) e^{(1-n²π²)t}sin(nπx)$

$u_{xx}=-n^2\pi^2 Ce^{(1-n²π²)t}sin(nπx)$

$u_{tt}= C(1-n²π²)^2 e^{(1-n²π²)t}sin(nπx)$

$u_{xt}= n\pi C(1-n²π²) e^{(1-n²π²)t}cos(nπx)$

3.

$f(x)=a_0+\displaystyle{\sum_{n=1}^{\infin}}a_ncos(n\pi x/L)+\displaystyle{\sum_{n=1}^{\infin}}b_nsin(n\pi x/L)$

we have: $L=\pi$

$a_0=\frac{1}{2L}\int^L_{-L}f(x)dx=\frac{1}{2\pi}\int^{\pi}_{0}sinxdx=\frac{1}{\pi}$

$a_n=\frac{1}{L}\int^L_{-L}f(x)cos(n\pi x/L)dx=\frac{1}{\pi}\int^{\pi}_{0}sinxcos(nx)dx=$

$=\frac{nsinxsin(nx)+cosxcos(nx)}{\pi (n^2-1)}|^{\pi}_0=-\frac{cos(n\pi)+1}{\pi (n^2-1)}$

$b_n=\frac{1}{L}\int^L_{-L}f(x)sin(n\pi x/L)dx=\frac{1}{\pi}\int^{\pi}_{0}sinxsin(nx)dx=$

$=\frac{cosxsin(nx)-nsinxcos(nx)}{\pi (n^2-1)}|^{\pi}_0=0$

$f(x)=\frac{1}{\pi}-\displaystyle{\sum_{n=1}^{\infin}}\frac{cos(n\pi)+1}{\pi (n^2-1)}cos(n x)$