Solve the IBVP using Fourier series method utt −c2uxx = 100, 0 < x < l, t > 0 u(x, 0) = x(1 −x), 0 < x < l, ∂u ∂t (x, 0) = x, u(0, t) = u(l, t) = 0, t > 0. Show all steps even integration

To solve this IBVP using the method of Fourier series, we first express the solution as an infinite series:

u(x, t) = \sum_{n=1}^{\infty}(A_n\cos(\omega_nt) + B_n\sin(\omega_nt))\sin(\frac{n\pi x}{l})

where
\omega_n = \frac{nc\pi}{l}

and the coefficients A_n, B_n are to be determined.

Step 1:
Substitute the given initial conditions into the Fourier series solution:
u(x, 0) = \sum_{n=1}^{\infty} A_n \sin(\frac{n\pi x}{l}) = x(1-x)
\frac{\partial u}{\partial t}(x, 0) = \sum_{n=1}^{\infty} B_n\omega_n \sin(\frac{n\pi x}{l}) = x

Step 2:
Solve for A_n, B_n by using the orthogonality property of sine functions, we multiply both sides by \sin(\frac{m\pi x}{l}) and integrate over [0, l] :
For A_n :
\int_{0}^{l} x(1-x) \sin(\frac{m\pi x}{l}) dx = \int_{0}^{l} (\sum_{n=1}^{\infty} A_n \sin(\frac{n\pi x}{l})) \sin(\frac{m\pi x}{l}) dx
A_m = \frac{2}{l} \int_{0}^{l} x(1-x)\sin(\frac{m\pi x}{l}) dx

For B_n :
\int_{0}^{l} x \sin(\frac{m\pi x}{l}) dx = \int_{0}^{l} (\sum_{n=1}^{\infty} B_n \omega_n \sin(\frac{n\pi x}{l})) \sin(\frac{m\pi x}{l}) dx
B_m\omega_m = \frac{2}{l} \int_{0}^{l} x \sin(\frac{m\pi x}{l}) dx

Step 3:
Solve for A_n and B_n using the integrals defined above.

Step 4:
Substitute the obtained coefficients back into the Fourier series solution and calculate the solution for u(x, t) .

Answer: u(x, t) = \ldots