Solve the following differential equation

$\frac{d^2y}{dx^2} - 10x^2= 5; \hspace{2cm} 0 \leq x \leq 1$

Bcs: y(0) = y(1) = 0. Using Rayleigh Ritz Method mapped over entire domain using one parameter method.

Given the differential equation, $ \frac{d^2y}{dx^2}−10x^2 = 5; \hspace{0.5cm} 0≤x≤1 $

y(0) = y(1) = 0

Since only one parameter is to be used, so the polynomial will be of order 2.

i) Approximate solution is given by,

$ y = C_0 + C_1x + C_2x^2 $ ....(1)

Applying boundary conditions,

a) At x = 0; y = 0, equation (1) becomes,

0 = C$_2$ + 0 + 0

C$_2$ = 0

b) At x = 1; y = 0, equation (1) becomes,

0 = C$_0$ + C$_1$ + C$_2$

C$_1$ = -C$_2$ [$\because C_0 = 0 $]

Substitute the value of C$_0$ and C$_1$ in equation (1), we get,

$ y = - C_2x + C_2x^2 \\ y = (x^2-x)C_2 $

ii) To find Residue, Equate general D.E. to residue R

$ \frac{d^2y}{dx^2}−10x^2 - 5 = R $

Weighted integral form,

$ \int_0^1 w_i R \,\, dx = 0 \\ \int_0^1 w_i [\frac{d^2y}{dx^2}−10x^2 - 5] \,\, dx = 0 \\ \int_0^1 w \frac{d^2y}{dx^2} \,\, dx − 10 \int_0^1 w x^2 \,\, dx - 5 \int_0^1 w \,\, dx = 0 \hspace{0.5cm} .....(2) $

Now, w = coefficient of C$_2$ = x$^2$ - x

Now consider the integral $ \int_0^1 w \frac{d^2y}{dx^2} \,\, dx $,

Integrating by parts,

$ \int_0^1 w \frac{d^2y}{dx^2} \,\, dx = w [\frac{dy}{dx}]_0^1 - \int_0^1 \frac{dw}{dx} \frac{dy}{dx} \,\, dx $ ....(3)

We know,

$ w = x^2 - x \hspace{0.25cm} \therefore \frac{dw}{dx} = 2x -1 \\ y = C_2(x^2-x) \hspace{0.25cm} \therefore \frac{dy}{dx} = C_2(2x-1) \\ [w \frac{dy}{dx}]_0^1 = [(x^2-x)(2x-1)C_2]_0^1 = 0 $

Equation (3) becomes,

$ \int_0^1 w \frac{d^2y}{dx^2} \,\, dx = 0 - \int_0^1 (2x -1)C_2(2x-1) \,\, dx $

Substitute above value in equation (2),

$ - \int_0^1 (2x -1)C_2(2x-1) \,\, dx - 10 \int_0^1 (x^2 - x) x^2 \,\, dx - 5 \int_0^1 (x^2 - x) \,\, dx = 0 \\ -0.33C_2 + 0.5 + 0.833 = 0 \\ -0.33C_2 = -1.333 \\ C_2 = 4 $

Therefore, Approximate solution is given by, y = 4(x$^2$ - x)