Solve the following initial value problem for u as a function ...

Question

Solve the following initial value problem for u as a function of t:

du/dt + (k/m) u = 0 (k and m positive constants), u(0) = u₀

a. as a first-order linear equation.

Accepted Answer

Welcome back everyone. A chemical reactant in a perfectly mixed batch reactor decays at a rate proportional to its current concentration. Let b oft denote the chemical's concentration in moles per liter at time t in minutes. The proportionality constant is alpha greater than 0, and initially b of 0 equals b subscript 0. Solve the following equation for bt by considering it as a first order linear differential equation. DB divided by dt plus alpha multiplied by b equals 0. So for this problem we can apply the concept of the integrating factor since we have an equation in the form of b plus alpha multiplied by b equals 0. So immediately we can identify the integrating factor mu equals e to the power of the integral of alpha. With respect to see you, right? And this is going to be equal to e to the power of alpha multiplied by c. So we have our integrating factor. And now let's remember that we multiply both sides of our equation by the integrating factor. So we get e to the power of alpha c multiplied by b. Derivative plus. E to the power of alpha c multiplied by alpha multiplied by b equals 0. And now what we have to notice is that on the left hand side we have the derivative of, let's write it down, d divided by dt of e to the power of alphat multiplied by b, right? We are basically applying the reverse product rule of differentiation. This is equal to 0. So now what we're going to do is simply find the integral of each side. In particular, we can show that the integral of the left-hand side with respect to T. is going to be integral of the right hand side with respect to t. So on the left hand side we can cancel out dt, and we're basically finding the integral of the differential. We're going to get e to the power of alphat multiplied by b. On the right hand side we have the integral of 0, which is 0 plus a constant of integration c. What we're going to do from here is sol for b. B is going to be equal to c divided by e to the power of alphat, or basically c multiplied by e to the power of negative alphat. And now we can find the constant c, knowing that b of 0 equals 0. In other words, when c is equal to 0, b becomes 0. So we're going to say 0 equals. C multiplied by e to the power of 0, which is basically equal to. One, right? So now, I'm sorry. So instead of 0, we have B subscript 0. So let's modify this part. So, B subscript 0 on the left-hand side. is going to be equal to c multiplied by e to the power of 0, which is equal to c. So we have identified the constant c which is equal to b subscript 0, right? And now we can write our solution. So b oft is equal to our constant c, which is b subscript 0. Multiplied by e to the power of negative alpha c, and this corresponds to our final answer for this problem. Thank you for watching.

Solve the following initial value problem for u as a function of t:du/dt + (k/m) u = 0 (k and m positive constants), u(0) = u₀a. as a first-order linear equation.

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Solve the following initial value problem for u as a function of t:

du/dt + (k/m) u = 0 (k and m positive constants), u(0) = u₀

a. as a first-order linear equation.