continous compounding you have $1000 wiht which to open an account to which you plan also to add $ 1000 per year. all funds in the account will earn 10% annual interest compounded continously. if the added deposits are also credited to your account continously, the number of dollars x in your account at time t(years) will satisfy th einitial calue problem.
differentital equation
dx/dt = 1000+0.10x
initial condition
x(0) = 1000
A) solve the initial value problem for x as a function of t
B) about how many years will it take for the amount in your account to reach $ 100,000
[I know how to plug in for part , but can anyone help me on the part A to find t]
Use the method of separation of variables to solve the differential equation, and the initial condition x(0) = 1000 to get the cosnstant term. The resulting x(t) will have an exponential form.
Separation of variables will involve the integration of
dx/(1000+0.10x) = dt
The resulting equation will be an approximation that is valid only when t is an integral number of years. During the interim, the balance will be a stepwise function until the next deposit and interest payment is made.
To solve the initial value problem for x as a function of t, we can use the technique of separation of variables. Here's how you can do it:
A) Solve the initial value problem for x as a function of t:
The given differential equation is:
dx/dt = 1000 + 0.10x
To solve this equation, we can separate the variables by moving all terms involving x to one side and the terms involving t to the other side:
dx / (1000 + 0.10x) = dt
Now, we can integrate both sides with respect to their respective variables:
∫ (1 / (1000 + 0.10x)) dx = ∫ dt
The integration on the left side is a little more involved. We can use a substitution to simplify it. Let's substitute u = 1000 + 0.10x, then du = 0.10 dx.
∫ (1 / u) du = ∫ dt
ln|u| = t + C1
ln|1000 + 0.10x| = t + C1
Now, we can exponentiate both sides to get rid of the natural logarithm:
|1000 + 0.10x| = e^(t + C1)
Since e^(t + C1) is always positive, we can remove the absolute value signs:
1000 + 0.10x = e^(t + C1) or 1000 + 0.10x = -e^(t + C1)
Now, let's solve each equation separately:
1) 1000 + 0.10x = e^(t + C1)
Subtracting 1000 from both sides:
0.10x = e^(t + C1) - 1000
Dividing both sides by 0.10:
x = 10(e^(t + C1) - 1000)
2) 1000 + 0.10x = -e^(t + C1)
Subtracting 1000 from both sides:
0.10x = -e^(t + C1) - 1000
Dividing both sides by 0.10:
x = -10(e^(t + C1) + 1000)
So, the general solution to the differential equation is given by both equations: x as a function of t.
B) To find out how many years it will take for the amount in your account to reach $100,000, we can use the solution we obtained in part A. We need to find the value of t when x = 100,000.
Let's use the equation we derived in step 1):
x = 10(e^(t + C1) - 1000)
Set x = 100,000 and solve for t:
100,000 = 10(e^(t + C1) - 1000)
Dividing both sides by 10:
10,000 = e^(t + C1) - 1000
Adding 1000 to both sides:
11,000 = e^(t + C1)
Taking the natural logarithm of both sides:
ln(11,000) = t + C1
Subtracting C1 from both sides:
ln(11,000) - C1 = t
Now, you can use a calculator to find the approximate value of ln(11,000) and subtract C1 if it is given. This will give you an estimate of the number of years it will take for the amount in your account to reach $100,000.