The rate of change of an investment account earning continuous compound interest is given by
dA/dt=kA
where k is a positive constant. The initial account value was $2500. At the end of the third year, the account value was $4200.
Find the particular solution to the differential equation. You may also use A = Pekt.
To find the particular solution to the differential equation, we can use the given information regarding the initial account value and the account value at the end of the third year.
Let's start with the general solution to the differential equation:
dA/dt = kA
Separating the variables, we get:
dA/A = k dt
Integrating both sides, we have:
ln|A| = kt + C
where C is the constant of integration.
Now, let's substitute the initial condition A = $2500 at t = 0:
ln|2500| = 0 + C
Simplifying, we find:
C = ln|2500|
So the particular solution to the differential equation is:
ln|A| = kt + ln|2500|
To find the value of k, we can use the information that the account value was $4200 at the end of the third year (t = 3).
Substituting the values:
ln|4200| = k(3) + ln|2500|
Now, we can solve for k:
ln|4200| - ln|2500| = 3k
ln(4200/2500) = 3k
ln(1.68) = 3k
Now, using the logarithmic property, we get:
1.68 = e^(3k)
From here, we can solve for k:
k = ln(1.68)/3
Now that we know the value of k is approximately 0.2217, we can substitute it back into our particular solution:
ln|A| = 0.2217t + ln|2500|
Therefore, the particular solution to the differential equation is:
A = e^(0.2217t + ln|2500|)
or, simplifying further:
A = e^ln|2500| * e^(0.2217t)
A = 2500 * e^(0.2217t)
So, the particular solution to the differential equation is:
A = 2500e^(0.2217t)
To find the particular solution to the differential equation, we need to substitute the given values into the equation and solve for the constant k.
We are given that the initial account value (A) is $2500 and the end value after 3 years is $4200.
Using the formula A = Pekt, we can substitute these values into the equation:
$4200 = $2500e^3k
To isolate k, divide both sides of the equation by $2500:
($4200/$2500) = e^3k
1.68 = e^3k
To solve for k, take the natural logarithm (ln) of both sides of the equation:
ln(1.68) = ln(e^3k)
Using the property of logarithms, ln(e^a) = a, we can simplify the equation:
ln(1.68) = 3k
Now, divide both sides of the equation by 3 to solve for k:
k = ln(1.68)/3
Using a calculator, we can find the approximate value of k:
k ≈ 0.246
Now that we have determined the value of k, we can substitute it back into the differential equation to find the particular solution:
dA/dt = kA
dA/dt = 0.246A
To solve this separable differential equation, we can rearrange it as follows:
dA/A = 0.246dt
Now, integrate both sides of the equation:
∫ dA/A = ∫ 0.246dt
ln|A| = 0.246t + C
where C is the constant of integration.
To determine the value of C, we can use the initial condition that the initial account value is $2500. Substituting this value into the equation, we get:
ln|2500| = 0.246(0) + C
ln|2500| = C
Using a calculator, the natural logarithm of 2500 is approximately 7.8241, so:
C ≈ 7.8241
Now, substituting the value of C back into the equation, we have:
ln|A| = 0.246t + 7.8241
To find the particular solution, we need to remove the absolute value by exponentiating both sides of the equation:
|A| = e^(0.246t + 7.8241)
Since e^(0.246t + 7.8241) is always positive, we can remove the absolute value signs, and the particular solution to the differential equation is:
A = e^(0.246t + 7.8241)