Solve the following differential Equation using Galerkin Method.

$\frac{d^2y}{dx^2}+3x \frac{dy}{dx}-6y = 0 \hspace{0.6cm} 0 \lt x \lt 1$

Boundary conditions are : y (0) = 1 , y'(1) = 0.1

Find y (0.2) and compare with exact solution.

Galerkin Method:

order of d.e = 2

Degree of polynomial = 3

Let approximate solution be y

$y = C_0 + C_1x + C_2x^2 + C_3x^3\\ y' = C_1x + 2 C_2x + 3C_3x^2$

From boundary condition 1: y(0) = 1

$C_0 = 1$

From boundary condition 2 : y'(1) = 0.1

$0.1 = C_1 + 2C_2 + 3C_3\\ '\therefore C_1 = 0.1 - 2C_2 - 3C_3$

Approximate solution be,

$y = 1 + (0.1 - 2C_2 - 3C_3x) + C_2x^2 + C_3x^3\\ \therefore y = 1 + 0.1x + C_2 (x^2 - 2x) + C_3(x^3 - 3x)\\ \frac{dy}{dx} = 0.1 + C_2(2x - 2) + C_3(3x^2 - 3)\\ \frac{d^2y}{dx^2} = 2C_2 + 6x C_3$

Residue,

$R = \frac{d^2y}{dx^2} + 3x \frac{dy}{dx}-6y\\ = 2C_2 + 6xC_3+3x [0.1 + C_2(2x + 2)+C_3(3x^2 - 3)] - 6 [1 + 0.1x + C_2(x^2 - 2x)+C_3(x^3 - 3x)]\\ = C_2[2 + 3x(2x - 2)+ 12x - 6x^2] + C_3[6x + 3x(sx^2 - 3)+18x - 6x^3] + 0.3x - 6 - -.6x\\ C_2(6x+2) + C_3(3x^2 + 15x) - 0.3x - 6$

Weighted Integral form

$\int\limits_0^1 WiR dx = 0 \hspace{1cm} ------ (1)$

For Galerkin method,

wi = coefficients of $C_i$ in y

$W_1 = (x^2 - 2x)$ and $w_2 = (x^3 - 3x)$

Equation 1, i=1

$\int\limits_0^1(x^2 - 2x)[C_2 (6x+2)+ C_3(3x^3 + 15x) - 0.3x - 6]dx = 0\\ - 3.883C_2 - 6.95C_3 + 4.125 = 0\\ 3.883C_2 + 6.95C_3 = 4.125 \hspace{1cm} ------ (2)$

Equation 1: i = 2

$\int\limits_0^1(x^3 - 3x) [C_2 (6x + 2) + C_3 (3x^3 + 15x) - 0.3x - 6]dx = 0\\ -7.3C_2 - 13.371C_3 + 7.74 = 0\\ 7.3C_2 + 13.371C_3 = 7.74 \hspace{1cm} ------ (3)$

Solving equations (2) and (3),

$C_2 = 2.64$ and $C_3 = -0.862\\ \therefore y = 1 + 0.1x + 2.64 (x^2 - 2x) - 0.862 (x^3 - 3x)\\ y(0.2) = 0.58$