Solve following differential equation

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

BCs : y (0) = 0 and y' (1) = 0.1; Find y(0.2) using variational method and Compare with exact solution

Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis

Marks: 12M

Year: Dec 2015

$y = C_0 + C_1x + C_2 x^2 + C_3 x^3\\ y' = C_1x + 2C_2x + 3C_3x^2$

Apply B.C (1) : y(0) = 1

$\therefore C_0 = 1$

Apply B.C. (2) :y'(1) = 0.1

$0.1 = C_1 + 2C_2 +3C_3\\ \therefore C_1 = 0.1 - 2C_2 - 3C_3\\ \therefore y = 1 + (0.1 - 2C_2 - 3C_3)x + C_2x^2 + C_3x^3\\ y = 1 + 0.1x + C_2(x^2 - 2x)+ C_3(x^3 - 3x)\\ \frac{dy}{dx} = 0.1 + C_2(2x - 2) + C_3(3x^2 - 3)\\ \frac{d^2y}{dx^2} = 2C_2 + 6xC_3$

Residue,

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

Weighted integral form,

$\int\limits_0^1 W_i R dx = 0\\ \int\limits_0^1 W_i [C_2(6x+2)+C_3(3x^3 + 15 x) - 0.3x - 6]dx =0$

Galerkin Method:

$W_1 = (x^2 - 2x)$ and $W_2 = (x^3 - 3x)$

For i = 1

$\int\limits_0^1(x^2 - 2x)[C-2(6x + 2)+ C_3(3x^3 + 15x) - 0.3x - 6]dx = 0\\ - 3.833C_2 - 6.95C_3 + 4.125 = 0\\ 3.833C_2 + 6.95 C_3 = 4.125 \hspace{2cm} ------ (1)$

For i = 2

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

From (1) and (2)

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