Answer to Question #146751 in Differential Equations for Nikhil Singh

Question

Using the method of product solution,solve ∂u/∂x= 2∂u/∂x+u where u(x,0)=6e^-3x

Answer 1

$\displaystyle u(x, t) = X(x) T(t)\\ u_x(x, t) = \frac{\partial(u(x, t))}{\partial x} =X'(x)T(t),\\ u_t(x, t) = \frac{\partial(u(x, t))}{\partial t} = X(x)T'(t)\\ u_x(x, t) = 2u_t(x, t) + X(x) T(t)\\ X'(x)T(t) = 2X(x)T'(t) + X(x) T(t)\\ (X'(x) - X(x))T(t) = 2X(x)T'(t)\\ \frac{X'(x) - X(x)}{X(x)} = 2\frac{T'(t)}{T(t)}\\ \frac{X'(x)}{X(x)} - 1 = 2\frac{T'(t)}{T(t)} = k\\ \frac{X'(x)}{X(x)} - 1 = k, \frac{X'(x)}{X(x)} = k + 1\\ \int \frac{X'(x)}{X(x)}\,\mathrm{d}x = \int k + 1 \,\mathrm{d}x\\ \log(X(x)) = (k + 1)x + c_1\\ X(x) = c_2e^{(k + 1)x}\\ \int\frac{T'(t)}{T(t)}\,\mathrm{d}t = \frac{k}{2}\,\mathrm{d}t\\ \log(T(t)) = \frac{kt}{2} + c_2\\ T(t) = c_3e^{\frac{kt}{2}}\\ \implies u(x, t) = X(x) T(t) = c_2e^{(k + 1)x} \times c_3e^{\frac{kt}{2}}= ce^{k\left(x + \frac{t}{2}\right) + x}\\ u(x, t) = ce^{k\left(x + \frac{t}{2}\right) + x}, u(x, 0) = ce^{kx + x} = ce^{(k + 1)x} = 6e^{-3x}\\ c = 6, k + 1 = -3, k = -4\\ \therefore u(x, t) = 6e^{-4\left(x + \frac{t}{2}\right) + x} = 6e^{-(3x + 2t)}$