Verified answer

Let U(x, t) = X(x) T(t).

So, we obtain from the PDE

XT '' + 2XT ' - 4X''T + XT = 0

==> (T '' + 2T ' + T)/T = 4X''/X = -4λ² for some λ (this gives nonzero solutions).

So, we have two ODE's

X'' + λ²X = 0

T '' + 2T ' + (1 + 4λ²) T = 0.

Next, the BC's Ux(0,t) = Ux(1,t) = 0 transform to X '(0) = X '(1) = 0,

and U(x, 0) = 0 ==> T(0) = 0.

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First, we solve X'' + λ²X = 0 with X '(0) = X '(1) = 0.

The general solution is X = A cos(λx) + B sin(λx)

==> X' = -Aλ sin(λx) + Bλ cos(λx).

X '(0) = 0 ==> B = 0; so X = A cos(λx).

X '(1) = 0 ==> -Aλ sin(λx) = 0 ==> λx = nπ for n = 0, 1, 2, ... (for nonzero solutions).

Hence, X_n = A_n cos(nπx) for n = 0, 1, 2, ... .

(Note that λ = nπ.)

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Next, we solve T '' + 2T ' + (1 + 4λ²) T = 0 with T(0) = 0.

==> T '' + 2T ' + (1 + 4n²π²) T = 0

This has characteristic equation r² + 2r + (1 + 4n²π²) = 0

==> r = -1 ± 2nπi.

So, T_n = e^(-t) [A_n cos(2nπt) + B_n sin(2nπt)].

(Note that the n = 0 case is included in this.)

T(0) = 0 ==> 0 = 1 [A_n + 0] ==> A_n = 0.

Hence, T_n = B_n e^(-t) sin(2nπt).

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Finally, u_n(x, t) = A_n cos(nπx) * B_n e^(-t) sin(2nπt).

==> u_n(x, t) = C_n e^(-t) cos(nπx) sin(2nπt), where C_n = A_n * B_n.

(Note that we have nonzero solutions for n = 1, 2, 3, ... now.)

Since the PDE is linear, we conclude that

u(x, t) = Σ(n = 1 to ∞) C_n e^(-t) cos(nπx) sin(2nπt).

I hope this helps!

If U(x, t) = X(x)T(t), then the PDE becomes

XT '' + 2XT ' - 4X''T + XT = 0

or after division by XT

T ''/ T + 2T '/ T - 4X''/X + 1 = 0

T ''/ T + 2T '/ T + 1 = 4X''/X

The left side is a function of t only, the right is a function of x only, so each side is a constant (yeah, I know you know this).

The BCs give us X '(0) = X '(1) = 0.

Let the constant be -4p² where p is real (and we can take it as positive without loss of generality). Then we have

4X'' + 4p²X = 0

X(x) = c1 cos(px) + c2 sin(px)

X '(x) = -p*c1 sin(px) + p*c2 cos(px)

X '(0) = 0 ⇒ c2 = 0

X '(1) = 0 ⇒ p = n*π for n = 0, 1, 2, ...

k = -4p² = -4n² π²

I'm really pretty rusty on this, but I think this is at least close to correct.

X(0) = 0 ⇒

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