I need help in solving an initial-value problem and a few series problems (Especially on #45 & #46). I don't really understand how to do the series problems...majority of the time. An explanation would be great as well. Thank you for your time.
#20) Solve the initial-value problem.
(9 + e^x)dy/dx + e^xy = sin x, y(0) = 1/10
#45) Determine whether the given series converges or diverges.
(∞ on top of summation symbol) (n = 1 under the summation symbol) ∑ [cos(4/n) - cos(4/(n+1)]
If it converges, find its sum.
#39) Consider the following series.
(∞ on top of summation symbol) (n = 1 under the summation symbol) ∑ (1+9^n)/8^n
(a) Determine whether the series is convergent or divergent.
(b) If it is convergent, find its sum.
#46) Determine whether the given series converges or diverges.
(∞ on top of summation symbol) (n = 1 under the summation symbol) ∑ ln(n/n+1)
If it converges, find its sum.
Sure! I'd be happy to help you out with these problems. Let's start with the initial-value problem.
#20) Solve the initial-value problem.
(9 + e^x)dy/dx + e^xy = sin x, y(0) = 1/10
To solve this initial-value problem, we can use an integrating factor. The integrating factor is given by the formula:
IF = e^(∫P(x)dx)
where P(x) is the coefficient of dy/dx (in this case, 9 + e^x). So, let's find the integrating factor:
IF = e^(∫(9 + e^x)dx)
= e^(9x + e^x)
Now, multiply both sides of the differential equation by the integrating factor:
e^(9x + e^x) * [(9 + e^x)dy/dx + e^xy] = e^(9x + e^x) * sin x
Simplify the left side by applying the product rule for differentiation and the chain rule:
d/dx [e^(9x + e^x) * y] = e^(9x + e^x) * sin x
Integrate both sides with respect to x:
∫ [d/dx [e^(9x + e^x) * y]] dx= ∫ [e^(9x + e^x) * sin x] dx
The left side becomes:
e^(9x + e^x) * y = ∫ [e^(9x + e^x) * sin x] dx
Now, evaluate the integral on the right side. This requires using integration by parts. Once you have the integral evaluated, rearrange the equation to solve for y:
y = [∫ [e^(9x + e^x) * sin x] dx] / e^(9x + e^x)
You can use numerical approximation methods, such as Simpson's rule or the trapezoidal rule, to evaluate the integral.
Let's move on to the series problems.
#45) Determine whether the given series converges or diverges.
(∞ under the summation symbol) (n = 1 on top of the summation symbol) ∑ [cos(4/n) - cos(4/(n+1))]
To determine whether this series converges or diverges, we can use the limit comparison test. Let's compute the limit as n approaches infinity of the absolute value of the nth term divided by the corresponding term of a known convergent series. If the limit is a finite positive number, then the given series converges; if the limit is zero or infinite, it diverges.
In this case, let's compare the series to the series ∑ (4/n). The limit we need to compute is:
lim (n→∞) [(cos(4/n) - cos(4/(n+1))) / (4/n)]
Simplify the expression inside the limit:
lim (n→∞) [n * (cos(4/n) - cos(4/(n+1))) / 4]
Note that as n approaches infinity, cos(4/n) approaches 1 and cos(4/(n+1)) approaches 1. Therefore, the numerator in the above expression goes to zero, and the denominator goes to infinity.
So, the limit becomes:
lim (n→∞) [0 / ∞] = 0
Since the limit is zero, the given series converges.
Let's move on to the next series problem.
#39) Consider the following series.
(∞ on top of the summation symbol) (n = 1 under the summation symbol) ∑ (1+9^n)/8^n
(a) Determine whether the series is convergent or divergent.
To determine whether this series converges or diverges, we can use the ratio test. Let's compute the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term. If the limit is less than 1, the series converges; if the limit is greater than 1, it diverges; if the limit is equal to 1, the test is inconclusive.
In this case, let's compute the limit:
lim (n→∞) [(1 + 9^(n+1)) / 8^(n+1)] / [(1 + 9^n) / 8^n]
Simplify the expression inside the limit:
lim (n→∞) [(1 + 9^(n+1)) * 8^n] / [(1 + 9^n) * 8^(n+1)]
Cancel out the common terms:
lim (n→∞) [(1 + 9^(n+1))] / [(1 + 9^n)] * [(8^n) / (8^(n+1))]
= lim (n→∞) [(1 + 9^(n+1))] / [(1 + 9^n)] * [1 / 8]
Now, let's simplify the expression inside the limit:
lim (n→∞) [(1 + 9^(n+1))] / [(1 + 9^n)] * [1 / 8]
= [(9^n * (1 + 9))] / (9^n * (1 + 1/9)) * (1/8)
= (10/9) * (1/8)
= 10/72
= 5/36
Since the limit is less than 1, the series converges.
(b) If it is convergent, find its sum.
To find the sum of this convergent series, we can use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
where a is the first term and r is the common ratio. In this case, the first term a is (1+9^1)/8^1 = 10/8 = 5/4, and the common ratio r is 9/8.
So, the sum of the series is:
S = (5/4) / (1 - 9/8)
= (5/4) / (-1/8)
= -10
Therefore, the sum of the series is -10.
Finally, let's move on to the last series problem.
#46) Determine whether the given series converges or diverges.
(∞ on top of the summation symbol) (n = 1 under the summation symbol) ∑ ln(n/n+1)
To determine whether this series converges or diverges, we can use the integral test. Let's compute the integral of the function f(x) = ln(x/x+1) from 1 to infinity. If the integral converges, the series converges; if the integral diverges, the series diverges.
Let's calculate the integral:
∫ (ln(x/x+1)) dx
= ∫ ln(x) - ln(x+1) dx
= xln(x) - x - ln(x+1) + C
Now, evaluate the integral from 1 to infinity:
∫ (ln(x/x+1)) dx = lim (t→∞) [(tln(t) - t - ln(t+1))] - [(1ln(1) - 1 - ln(1+1))]
= lim (t→∞) [tln(t) - t - ln(t+1) - 0]
= ∞
Since the integral of the function diverges, the series also diverges.
I hope this explanation helps! If you have any further questions, please feel free to ask.