A-Level Differential equations: worked solution

Question

In a chemical reaction, compounds X and Y combine to form a product. Let x and y be the concentrations (mol/kL) of X and Y at time t minutes, with initial values x₀ and y₀ mol/kL. When Y is in large excess, the reaction behaves as pseudo-first-order, giving dx/dt=−ax for a positive constant a. Solve this equation, giving x in terms of t, x₀ and a.

Method: dx/dt = −ax is a SEPARABLE differential equation. Get every x onto one side with the dx, and every t onto the other side with the dt, then integrate both sides. Step 1 - separate the variables. Divide both sides by x (allowed because a concentration x is positive, so it is never 0) and multiply by dt: dx/dt = −ax (1/x) dx = −a dt Step 2 - integrate both sides. Remember that the integral of 1/x is ln|x|, and a is a constant so it integrates to −at. ∫ (1/x) dx = ∫ −a dt ln|x| = −at + C, where C is an arbitrary constant Step 3 - undo the logarithm by taking exponentials of both sides. |x| = e^(−at + C) = e^C × e^(−at) Since x is a concentration it is positive, and e^C is just some positive constant. Write A = e^C: x = A e^(−at) Step 4 - find A from the initial condition. At t = 0 the concentration of X is x₀: x₀ = A e^(−a×0) = A e⁰ = A so A = x₀. Step 5 - write the solution. x = x₀ e^(−at) Check: at t = 0 this gives x₀, and dx/dt = −a x₀ e^(−at) = −ax, as required.