Solving $\dfrac{\partial{u}}{\partial{t}} + u = \dfrac{\partial^2{u}}{\partial{x}^2} + 4\dfrac{\partial{u}}{\partial{x}}$ Using Separation. 
Proceeding as follows, use the method of separation of variables to solve
$\dfrac{\partial{u}}{\partial{t}} + u = \dfrac{\partial^2{u}}{\partial{x}^2} + 4\dfrac{\partial{u}}{\partial{x}}$
subject to $u(0, t) = 0$ and $u(1, t) = 0$ for $t > 0$ and $u(x, 0) = 1$ for $0 < x < 1$.
(a) Try $u(x, t) = X(x)T(t)$ to get a pair of ordinary differential equations.

There is a discrepancy between my solution and that of the instructor.
The instructor's solution is as follows:

$X(x)T'(t) + X(x)T(t) = X''(x)T(t) + 4X'(x)T(t)$
$\implies \dfrac{X'' + 4X'}{X} = \dfrac{T' + T}{T} = -\lambda$
$\therefore X'' + 4X' + \lambda X = 0$ and $T' + (1 + \lambda)T = 0$.

My solution is as follows:
$X(x)T'(t) + X(x)T(t) = X''(x)T(t) + 4X'(x)T(t)$
$\implies XT' + XT = X''T + 4X'T$
$\implies XT' = X''T + 4X'T - XT$
$\implies XT' = T(X'' 4X' - X)$
$\implies \dfrac{T'}{T} = \dfrac{X'' + 4X' - X}{X}$
$\therefore \dfrac{T'}{T} = - \lambda = \dfrac{X'' + 4X' - X}{X}$
$\therefore$ The two ordinary differential equations are
$T' + \lambda T = 0$ and
$X'' + 4X' - X + \lambda X = 0 \implies X'' + 4X' + X(\lambda - 1) = 0$
I was wondering if people could please take the time to clarify as to whether both solutions are correct or I have made an error?

Your solution is equivalent to the instructor's solution. Only the symbols chosen for the constants $\lambda$ are different, but related : 
$$\lambda_{You}=\lambda_{Instructor}+1$$ 
It doesn't matter since $\lambda$ is arbitrary.
