Question

Please the steps clearly and simply. This is a differential equations question

Answer 1

The
`\(b_n\)'s` are all equal to 1.

The given diffusion equation is:

`\[ 8u(x, t) = \sum_(n=1)^(\infty) b_n \sin\left((n\pi x)/(L)\right) e^{-D\left((n^2\pi^2)/(L^2)\right)t} \]`

With the boundary conditions
`\(u(0, t) = 0\)` and
`\(u(L, t) = 0\)`, and using the fact that L = 3, we have:

`\[ 8u(x, t) = \sum_(n=1)^(\infty) b_n \sin\left((n\pi x)/(3)\right) e^{-D\left((n^2\pi^2)/(9)\right)t} \]`

To find the
`\(b_n\)'s`, we can use the orthogonality of sine functions over the interval [0, L]:

`\[ \int_(0)^(L) \sin\left((m\pi x)/(L)\right) \sin\left((n\pi x)/(L)\right) \,dx = (L)/(2) \delta_(mn) \]`

where
`\(\delta_(mn)\)` is the Kronecker delta. Applying this to our problem:

`\[ \int_(0)^(3) \sin\left((m\pi x)/(3)\right) \sin\left((n\pi x)/(3)\right) \,dx = (3)/(2) \delta_(mn) \]`

Now, multiply both sides by
`\(\sin\left((k\pi x)/(3)\right)\)` and integrate from 0 to 3:

`\[ \int_(0)^(3) \sin\left((k\pi x)/(3)\right) \left[ \sum_(n=1)^(\infty) b_n \sin\left((n\pi x)/(3)\right) \right] \,dx = (3)/(2) \delta_(mk) \]`

This simplifies to:

`\[ \sum_(n=1)^(\infty) b_n \int_(0)^(3) \sin\left((k\pi x)/(3)\right) \sin\left((n\pi x)/(3)\right) \,dx = (3)/(2) \delta_(mk) \]`

Now, evaluate the integral:

`\[ \int_(0)^(3) \sin\left((k\pi x)/(3)\right) \sin\left((n\pi x)/(3)\right) \,dx = (3)/(2) \delta_(mk) \]`

This integral will be non-zero only when k = n, so the sum becomes:

`\[ b_n \int_(0)^(3) \sin^2\left((n\pi x)/(3)\right) \,dx = (3)/(2) \]`

Now, evaluate the integral:

`\[ b_n \int_(0)^(3) (1 - \cos\left((2n\pi x)/(3)\right))/(2) \,dx = (3)/(2) \]`

`\[ b_n \left[(x)/(2) - (3\sin\left((2n\pi x)/(3)\right))/(4n\pi)\right]_0^3 = (3)/(2) \]`

`\[ b_n \left[(3)/(2) - (3\sin\left(2n\pi\right))/(4n\pi)\right] = (3)/(2) \]`

`\[ b_n \left[(3)/(2) - (3\sin(0))/(4n\pi)\right] = (3)/(2) \]`

`\[ b_n \left[(3)/(2)\right] = (3)/(2) \]`

`\[ b_n = 1 \]`

Therefore, the
`\(b_n\)'s` are all equal to 1.