23–26. Loan problems The following initial value problems mode...

Question

23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.

B′(t) = 0.005B − 500, B(0) = 50,000

B′(t) = 0.005B − 500, B(0) = 50,000

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says a model of the temperature of a cooling object is given below. Solve the following initial value problem for T greater than equal to 0, and determine the time when the temperature reaches 50 °C. Round your answer to the nearest whole number. And we're given that T of T is equal to -0.01 multiplied by T of T. Plus 0.2, and that capital T of 0 is equal to 80 °C. So for the first part of this problem, we need to solve our initial value problem. And if we look at our differential equation that we're given, we noticed that it's of the form of capital T is equal to K multiplied by capital T plus B. And this form a differential equation has the general solution of capital T, which is equal to C multiplied by E raised to the quantity of K multiplied by T in quantity minus the quantity of B divided by K in quantity. And if we look at our equation, we see that K is going to be equal to minus 0.01, and that B is going to be equal to 0.2. So substitute in these values into our general solution, we see that capital T is going to be equal to C multiplied by E raised to the quantity of minus 0.01 multiplied by T in quantity minus the quantity of 0.2 divided by minus 0.01 in quantity. And simplifying this question, we see that capital T is going to be equal to D multiplied by E raised to the quantity of minus 0.01 multiplied by T in quantity. Loss 20 And so now we need to use our initial condition to find C, and we're told that capital T of 0 is equal to 80 °C. So that means that when T is equal to 0, capital T is equal to 80. So making those substitutions into our expression here, we'll see that 80 is going to be equal to C multiplied by E raised to the quantity of minus 0.01 multiplied by 0 in quantity plus 20. Now, since E rates to the power 0 is equal to 1, we actually have 80 is equal to D plus 20, and subtracting 20 from both sides of the equation we have that C is going to be equal to 60. So substituting this back into our expression for capital T, we see that T is equal to 60, multiplied by E raised to the quantity of minus 0.01 multiplied by T and quantity. Plus 2. And we can actually factor out a 20 out of both of the terms on the right-hand side. So that means that capital T of E is going to be equal to 20, multiplied by the quantity of 3, multiplied by E raised to the quantity of minus 0.01. Multiplied by T in quantity plus 1 in quantity. So this here is the solution that we're looking for for the initial value problems. So that's the answer for the first part of this problem. Now, for the second part of the problem, we need to find the time when the temperature reaches 50 °C. So that means that capital T is going to be equal to 50. So setting capital T equals to 50, we'll have that 50 is equal to, and we'll go ahead and redistribute the 20, and so we'll have 60 multiplied by E raised to the quantity of minus 0.01 multiplied by T and quantity plus 20. And now we need to solve this equation for T. Those subtracting 20 from both sides of the equation will get that 30 is equal to 60, multiplied by E raised to the quantity of minus 0.01 multiplied by T and quantity. And dividing both sides of the equation by 60, we'll get that 1 divided by 2 is equal toe raised to the quantity of minus 0.01 multiplied by T in quantity, and taking the natural log of both sides, we'll get that the natural log of quantity of 1 divided by 2, and quantity is equal to minus 0.01 multiplied by T. and dividing both sides of this equation by minus 0.01, we'll get that T is going to be equal to. 100 multiplied by the natural log of 2, where we have rewritten the natural log of quantity of 1 divided by 2 as minus the natural log of 2. And evaluating this expression in our calculator, we see that T is going to be approximately equal to 69, and here we've rounded to the nearest whole number. So that is the answer that we're looking for for the second part of this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.

B′(t) = 0.005B − 500, B(0) = 50,000

Key Concepts

First-Order Linear Differential Equations

Initial Value Problems (IVP)

Interpreting Solutions and Finding Zero Crossing

Watch next

23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.B′(t) = 0.005B − 500, B(0) = 50,000

Key Concepts

First-Order Linear Differential Equations

Initial Value Problems (IVP)

Interpreting Solutions and Finding Zero Crossing

Watch next

23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.

B′(t) = 0.005B − 500, B(0) = 50,000