Solve the following ordinary differential equations using Laplace trans- forms: (a) y(t) + y(t) +3y(t) = 0; y(0) = 1, y(0) = 2 (b) y(t) - 2y(t) + 4y(t) = 0; y(0) = 1, y(0) = 2 (c) y(t) + y(t) = sint; y(0) = 1, y(0) = 2 (d) y(t) +3y(t) = sint; y(0) = 1, y(0) = 2 (e) y(t) + 2y(t) = e';y(0) = 1, y(0) = 2
(a) The ordinary differential equation is given by y(t) + y(t) + 3y(t) = 0. Using Laplace transform, we have(L [y(t)] + L [y(t)] + 3L [y(t)]) = 0L [y(t)] (s + 1) + L [y(t)] (s + 1) + 3L [y(t)] = 0L [y(t)] (s + 1) = - 3L [y(t)]L [y(t)] = - 3L [y(t)] /(s + 1)Taking the inverse Laplace of both sides, we have y(t) = L -1 [- 3L [y(t)] /(s + 1)]y(t) = - 3L -1 [L [y(t)] /(s + 1)]
On comparison, we get y(t) = 3e^{-t} - 2e^{-3t}.The initial conditions are y(0) = 1 and y(0) = 2 respectively.(b) The ordinary differential equation is given by y(t) - 2y(t) + 4y(t) = 0. Using Laplace transform, we have L [y(t)] - 2L [y(t)] + 4L [y(t)] = 0L [y(t)] = 0/(s - 2) + (- 4)/(s - 2)
Taking the inverse Laplace of both sides, we have y(t) = L -1 [0/(s - 2) - 4/(s - 2)]y(t) = 4e^{2t}.The initial conditions are y(0) = 1 and y(0) = 2 respectively.(c) The ordinary differential equation is given by y(t) + y(t) = sint. Using Laplace transform, we have L [y(t)] + L [y(t)] = L [sint]L [y(t)] = L [sint]/(s + 1)
Taking the inverse Laplace of both sides, we have y(t) = L -1 [L [sint]/(s + 1)]y(t) = sin(t) - e^{-t}.The initial conditions are y(0) = 1 and y(0) = 2 respectively.(d) The ordinary differential equation is given by y(t) + 3y(t) = sint. Using Laplace transform, we have L [y(t)] + 3L [y(t)] = L [sint]L [y(t)] = L [sint]/(s + 3)Taking the inverse Laplace of both sides, we have y(t) = L -1 [L [sint]/(s + 3)]y(t) = (1/10)(sin(t) - 3cos(t)) - (1/10)e^{-3t}.
The initial conditions are y(0) = 1 and y(0) = 2 respectively.(e) The ordinary differential equation is given by y(t) + 2y(t) = e^{t}. Using Laplace transform, we have L [y(t)] + 2L [y(t)] = L [e^{t}]L [y(t)] = 1/(s + 2)Taking the inverse Laplace of both sides, we havey(t) = L -1 [1/(s + 2)]y(t) = e^{-2t}The initial conditions are y(0) = 1 and y(0) = 2 respectively.
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Simon traveled 250 miles in 5 hours. What is his average speed?
Answer:
250/5 =50 miles per hour
What is the perimeter of thjs
Answer:73 ft
Step-by-step explanation:
Answer:
(120 + 12.5pi) ft^2
Step-by-step explanation:
10ft x 12 ft = 120ft^2
10ft/2 = 5 ft (Radius)
Area of semi circle:
[tex]\frac{\pi r^{2} }{2} = \frac{\pi 5^{2} }{2} = 12.5\pi ft^{2}[/tex]
Area = (120 + 12.5pi) ft^2
Use the Runge-Kutta method with h=0.09 to estimate the value of the solution at t=0.1 to y' = 3 + t - y, y(0) = 1
By applying the Runge-Kutta method with a step size (h) of 0.09, we can estimate the value of the solution at t = 0.1 for the differential equation y' = 3 + t - y, with the initial condition y(0) = 1.
The Runge-Kutta method is a numerical technique used to approximate the solution of ordinary differential equations. In this case, we have the differential equation y' = 3 + t - y, where y' represents the derivative of y with respect to t. To apply the Runge-Kutta method, we need to iterate through the given range of t values, which is from 0 to 0.1 in this case, with a step size (h) of 0.09.
We start with the initial condition y(0) = 1. Then, for each iteration, we calculate the slope at the current point using the given equation. Using the slope, we estimate the value of y at the next time step (t + h). This process is repeated until we reach the desired value of t = 0.1.
By applying the Runge-Kutta method with h = 0.09, we can obtain an estimate for the value of y at t = 0.1. This method provides a more accurate approximation compared to simpler methods like Euler's method, as it considers multiple intermediate steps to improve accuracy.
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Which expression represents the length of the spring after Gerard removes some weight? Gerard adds weight to the end of the hanging spring D-- The song stretches to a length of p centimeters. Gerard removes some weight and the song moves up by a 8 E-p) - 9 D-9--
Answer: p+(-q)
Step-by-step explanation:
The diameter of a circle is 6 kilometers. What is the area?
d=6 km
Give the exact answer in simplest form.
_____ square kilometers
Answer:
28.26
Step-by-step explanation:
6 divided by 2 = 3^2 = 9 x 3.14 = 28.26
please help 6th grade math please please help
i: 1
ii: 6
iii: 3
iv: 2
v: 4.5
vi: 2.5
Calculate the 90% confidence interval for the following sample Sample: 7.9, 8.3, 8.4, 9.6, 7.7, 8.1, 6.8, 7.5, 8.6, 8, 7.8,7.4, 8.4, 8.9, 8.5, 9.4, 6.9,7.7. Assume normality of the data.
The 90% confidence interval for the given sample is (7.58, 8.60).
To calculate the 90% confidence interval for the given sample assuming normality of the data, we need to use the formula as follows;Confidence interval = X ± Z α/2(σ/√n)Where, X is the sample meanZ α/2 is the Z-score for the desired level of confidenceσ is the population standard deviationn is the sample sizeFirst, we need to calculate the sample mean and standard deviation.Sample mean,
X= (7.9 + 8.3 + 8.4 + 9.6 + 7.7 + 8.1 + 6.8 + 7.5 + 8.6 + 8 + 7.8 + 7.4 + 8.4 + 8.9 + 8.5 + 9.4 + 6.9 + 7.7) / 18
= 8.09
Sample standard deviation,
σ = √[Σ(xi - X)² / (n - 1)]σ = √[(7.9 - 8.09)² + (8.3 - 8.09)² + (8.4 - 8.09)² + (9.6 - 8.09)² + (7.7 - 8.09)² + (8.1 - 8.09)² + (6.8 - 8.09)² + (7.5 - 8.09)² + (8.6 - 8.09)² + (8 - 8.09)² + (7.8 - 8.09)² + (7.4 - 8.09)² + (8.4 - 8.09)² + (8.9 - 8.09)² + (8.5 - 8.09)² + (9.4 - 8.09)² + (6.9 - 8.09)² + (7.7 - 8.09)² / (18 - 1)]σ = 0.761
Now, we need to find the Z α/2 value from the standard normal distribution table.
Z α/2 = 1.645 (for 90% confidence level)Putting the values in the formula,Confidence interval =
X ± Z α/2(σ/√n)
= 8.09 ± 1.645(0.761/√18)
= 8.09 ± 0.511
= (8.09 - 0.511, 8.09 + 0.511)
= (7.58, 8.60).
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I dont know how to do this
Answer:
a) 55°
b) 125°
c) 55°
d) 55°
Step-by-step explanation:
a) 180-(90+35) = 55
b) this angle forms a straight angle with ∡a
c) this angle is vertical, and congruent, with ∡a
d) 180-(70+55) = 55
Find the distance between the points (–7,–9) and (–2,4).
Answer:
√194 or 13.9
Step-by-step explanation:
√(x2 - x1)² + (y2 - y1)²
√[-2 - (-7)² + [4 - (-9)]
√(5)² + (13)²
√25 + 169
√194
= 13.9
Draw a two-dimensional representation of each prism. Then find the area of the entire surface of each prism
Answer:
Surface area of cuboid = 78 unit²
Step-by-step explanation:
Given diagram is a cuboid prism
Given:
Length of cuboid = 5 unit
Width of cuboid = 3 unit
Height of cuboid = 3 unit
Find:
Surface area of cuboid
Computation:
Surface area of cuboid = 2[lb + bh + hl]
Surface area of cuboid = 2[(5)(3) + (3)(3) + (3)(5)]
Surface area of cuboid = 2[15 + 9 + 15]
Surface area of cuboid = 2[39]
Surface area of cuboid = 78 unit²
Type the correct answer in the box.
Given : b ┴ d
c || b
b || e
What line is perpendicular to line e?
Answer:
d is parallel to e
Step-by-step explanation:
Since b is parallel to e and d is perpendicular to b , then
d is perpendicular to e
Somebody know this? thanks
Answer:
4 right angles
Step-by-step explanation:
a straight light intersects another straight line= creating 4 right angles
What does the C equal to in -1/6 +7/6 = c
Answer:
c = 1
Step-by-step explanation:
we have -1/6 + 7/6
since 6 is a common denominator we can do
[tex]\frac{-1}{6} +\frac{7}{6} = c\\\frac{-1+7}{6} = c\\\frac{6}{6} = c\\1 = c\\c = 1[/tex]
G(x)=2x/3+3. What value of g(-15)
Answer:
g= 2x+9 /3x
Step-by-step explanation:
Compare the dimensions of the prisms. How many times greater is the surface area of the purple prism than the surface area of the red prism?
Answer: 3 times greater
Step-by-step explanation:
Height: 4x3=12
Length: 3x3=9
Width: 3x3=9
The sides on the red cuboid times by 3 equals the sides on the purple one.
Hope this helps :)
A drawbridge has the shape of an isosceles trapezoid. The entire length of the bridge is 100 feet while the height is 25 feet. If the angle at which the bridge meets the land is approximately 60 degrees, how long is the part of the bridge that opens?
Answer:
The part of the bridge that opens is 50 ft.
Step-by-step explanation:
The given parameters of the drawbridge are;
The entire length of the bridge = 100 feet
The height of the isosceles trapezoid formed = 25 feet
The angle at which the drawbridge meets the land ≈ 60°
Therefore, the part of the bridge that opens = The top narrow parallel side of the isosceles trapezoid
The length of each half of the bridge = (The entire length)/2 = 100 ft./2 = 50 ft.
Let 'x' represent the path of the waterway still partly blocked by each half of the bridge inclined
∴ x = 50 × cos(60°) = 25
x = 25 ft.
The path covered by both sides of the drawbridge = 2·x = 2 × 25 ft. = 50 ft.
The part of the bridge that opens = The entire length - 2·x
∴ The part of the bridge that opens = 100 ft. - 50 ft. = 50 ft.
The part of the bridge that opens = 50 ft.
Helppppp it’s due today
Characterization of Random Processes in Time Domain Let Y(t) = 2X(t) + sin(2t) where X(t) is a wide-sense stationary (WSS) random process with mean à = E[X(t)] = 0 and autocorrelation Rx (T) = E[X(t + 7)X(t)] = e¯|7|. (a) (5) Find the mean ÿ(t) = E[Y(t)] and the autocorrelation Ry(t +7,t) = E[Y(t + 7)Y(t)] of Y (t). (2) Is Y (t) wide-sense stationary? Why? (b) (5)Find the crosscorrelation Rxy(t+7,t) = E[X(t+7)Y(t)]. (2) Are X and Y jointly wide sense stationary? Why? (c) (5) Find the autocovariance Cy (t +7,t) = E[(Y(t + 7) − ÿ(t + 7))(Y(t) − y(t))] of Y (t). (2) Is Y (t) white? Why?
A. The mean ÿ(t) = 0 and the autocorrelation Ry(t + 7, t) = 4e⁻⁷. Y(t) is wide-sense stationary.
B. the cross-correlation Rxy(t + 7, t) = 2e⁻⁷. X and Y are jointly wide-sense stationary.
C. The autocovariance Cy(t + 7, t) = 4e⁻⁷. Y(t) is not a white process because autocovariance Cy(t + 7, t) is not a Dirac delta function.
How did we arrive at these assertions?To find the mean ÿ(t) = E[Y(t)] and the autocorrelation Ry(t + 7, t) = E[Y(t + 7)Y(t)], we substitute the expression for Y(t) into the formulas:
(a) Mean of Y(t):
ÿ(t) = E[Y(t)] = E[2X(t) + sin(2t)]
= 2E[X(t)] + E[sin(2t)]
= 2(0) + 0
= 0
(b) Autocorrelation of Y(t + 7, t):
Ry(t + 7, t) = E[Y(t + 7)Y(t)]
= E[(2X(t + 7) + sin(2(t + 7)))(2X(t) + sin(2t))]
Expanding the expression:
Ry(t + 7, t) = E[4X(t + 7)X(t) + 2X(t + 7)sin(2t) + 2sin(2(t + 7))X(t) + sin(2(t + 7))sin(2t)]
Since X(t) is a WSS random process with mean 0, its autocorrelation Rx(T) = E[X(t + 7)X(t)] = e^(-|7|).
Using the properties of expectation and the independence of X(t) and sin(2t):
Ry(t + 7, t) = 4E[X(t + 7)X(t)] + 2E[X(t + 7)]E[sin(2t)] + 2E[sin(2(t + 7))]E[X(t)] + E[sin(2(t + 7))]E[sin(2t)]
= 4Rx(7) + 2(0)(0) + 2(0)(0) + 0
= 4e⁻⁷
Therefore, the mean ÿ(t) = 0 and the autocorrelation Ry(t + 7, t) = 4e⁻⁷.
To determine if Y(t) is wide-sense stationary, we need to check if the mean and autocorrelation are independent of time:
Mean: The mean ÿ(t) is constant and does not depend on time t. Thus, Y(t) has a constant mean.
Autocorrelation: The autocorrelation Ry(t + 7, t) depends only on the time difference of 7. It is independent of the absolute values of t. Therefore, Y(t) has a stationary autocorrelation.
Since Y(t) has a constant mean and a stationary autocorrelation, it is wide-sense stationary.
Moving on to part (b), we need to find the cross-correlation Rxy(t + 7, t) = E[X(t + 7)Y(t)].
Rxy(t + 7, t) = E[X(t + 7)Y(t)]
= E[X(t + 7)(2X(t) + sin(2t))]
Expanding the expression:
Rxy(t + 7, t) = E[2X(t + 7)X(t) + X(t + 7)sin(2t)]
Since X(t) is a WSS random process, its autocorrelation Rx(T) = e|⁻⁷|.
Using the properties of expectation and the independence of X(t) and sin(2t):
Rxy(t + 7, t) = 2E[X(t + 7)X(t)] + E[X(t + 7)]E[sin
(2t)]
= 2Rx(7) + 0
= 2e⁻⁷
Therefore, the cross-correlation Rxy(t + 7, t) = 2e⁻⁷.
To determine if X and Y are jointly wide-sense stationary, we need to check if the cross-correlation Rxy(t + 7, t) is independent of time:
Cross-correlation: The cross-correlation Rxy(t + 7, t) depends only on the time difference of 7. It is independent of the absolute values of t. Therefore, X and Y have a stationary cross-correlation.
Since the cross-correlation is stationary, X and Y are jointly wide-sense stationary.
Moving on to part (c), we need to find the autocovariance Cy(t + 7, t) = E[(Y(t + 7) - ÿ(t + 7))(Y(t) - ÿ(t))].
Expanding the expression:
Cy(t + 7, t) = E[(2X(t + 7) + sin(2(t + 7))) - 0][(2X(t) + sin(2t)) - 0]
= E[(2X(t + 7) + sin(2(t + 7)))(2X(t) + sin(2t))]
Using the same approach as in part (b), we expand the expression and evaluate the expectation:
Cy(t + 7, t) = 4E[X(t + 7)X(t)] + 2E[X(t + 7)]E[sin(2t)] + 2E[sin(2(t + 7))]E[X(t)] + E[sin(2(t + 7))]E[sin(2t)]
= 4Rx(7) + 0 + 0 + 0
= 4e⁻⁷
Therefore, the autocovariance Cy(t + 7, t) = 4e⁻⁷.
To determine if Y(t) is white, we check if the autocovariance Cy(t + 7, t) is a Dirac delta function. Since Cy(t + 7, t) = 4e⁻⁷ ≠ 0, it is not a Dirac delta function. Hence, Y(t) is not a white process.
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Which is the correct equation for x:y=8:1
See picture attached.
Answer:
Step-by-step explanation:
Means of means = means of extremes 8y = x
x = 8y
Option B is the correct answer
Your friends house is 6 miles south and 8 miles east of your house how far is your friends house from your house
Answer:
10 miles
Step-by-step explanation:
The information given forms a right angled triangle ; hence, we can use Pythagoras rule to solve for the distance, x
Recall:
Hypotenus = sqrt(opposite ² + adjacent ²)
Hypotenus = x
Therefore,
x = sqrt(6² + 8²)
x = sqrt(36 + 64)
x = sqrt(100)
x = 10
Distance between tween my friends house and my house = 10 miles
approximately what interest rate to the nearest whole percentage would you need to earn in order to turn $3,500 into $7,000 over 10 years?
a. 5%
b. 7%
c. 9%
d. 10%
The approximate interest rate needed to turn $3,500 into $7,000 over 10 years is 9%. Correct answer is C.
The value of money increases over time with the help of compounding interest. If one puts in a principal amount in an account, the amount will increase over time as interest accrues. Let's use the future value formula for the calculation. Let’s assume that the interest rate needed to turn $3,500 into $7,000 over 10 years is x percent. P = $3,500 (principal)FV = $7,000 (future value)
N = 10 years (duration of the investment)Using the future value formula:
FV = P(1 + r/n)^(nt)where, r is the annual interest rate, n is the number of times the interest is compounded in a year, and t is the duration of the investment in years.
Substituting the given values, we have: $7,000 = $3,500(1 + x/n)^(n × 10)We can solve for x by approximating the interest rate using each of the answer options given in the question until we find an answer that is close to $7,000. A calculator can also be used to calculate the compound interest for each option. If the interest rate is 7%, then the interest is compounded annually. Therefore, n = 1$7,000
= $3,500(1 + 0.07/1)^(1 × 10) If the interest rate is 10%, then the interest is compounded annually.
Therefore, n = 1$7,000 = $3,500(1 + 0.1/1)^(1 × 10)Thus, x ≈ 9.57%, greater than the required amount.
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The perimeter of a piece of paper is 38 inches. Its length is 11 inches.
Find the area of the piece of paper.
Answer:
Buddy this might not be the correct answer but I got either 98 or 418 inches. Don't quote me on it though.
Step-by-step explanation:
A radius is
the diameter
Answer:
Radius is the diameter divided by 2
solve the given initial value problem using the method of Laplace transforms.
5y''+2y'+3y = u(t-pi) y(0)=1 y'(0)=1
The solution to the given initial value problem using the method of Laplace transforms, is: y(t) = -4 [tex]e^{-t}[/tex] + 5 [tex]e^{-3t/5}[/tex]
To solve the given initial value problem using the method of Laplace transforms, we will follow these steps:
Taking the Laplace transform of both sides of the differential equation.
Applying the Laplace transform to the given differential equation, we get:
5L{y''} + 2L{y'} + 3L{y} = L{u(t-[tex]\pi[/tex])}
Using the properties of Laplace transforms and the table of Laplace transforms to simplify the equation.
The Laplace transform of y'' is [tex]s^2[/tex]Y(s) - sy(0) - y'(0), where Y(s) is the Laplace transform of y(t).
The Laplace transform of y' is sY(s) - y(0), and the Laplace transform of y is Y(s).
Using these transformations and considering the initial conditions y(0) = 1 and y'(0) = 1, we can rewrite the equation as:
5([tex]s^2[/tex]Y(s) - s - 1) + 2(sY(s) - 1) + 3Y(s) = e^(-pi*s) / s
Simplifying further, we have:
(5[tex]s^2[/tex] + 2s + 3)Y(s) - (5s + 7) = [tex]e^{-\pi s}[/tex] / s
Solving for Y(s):
Rearranging the equation, we get:
Y(s) = ([tex]e^{-\pi s}[/tex] / s + (5s + 7)) / (5[tex]s^2[/tex] + 2s + 3)
Using partial fraction decomposition to express Y(s) in simpler terms.
Performing partial fraction decomposition on the right side, we can express Y(s) as:
Y(s) = A / (s + 1) + B / (5s + 3)
where A and B are constants to be determined.
Using the inverse Laplace transform, we can find the solution y(t) as:
y(t) = [tex]L^{-1}[/tex]{Y(s)} = [tex]L^{-1}[/tex]{A / (s + 1)} + [tex]L^{-1}[/tex]{B / (5s + 3)}
Taking the inverse Laplace transforms using the table of Laplace transforms, we find:
y(t) = A [tex]e^{-t}[/tex] + B [tex]e^{-3t/5}[/tex]
Substituting the initial conditions y(0) = 1 and y'(0) = 1 into the solution y(t) = A [tex]e^{-t}[/tex] + B [tex]e^{-3t/5}[/tex], we can solve for the constants A and B.
First, substitute t = 0 into the equation:
y(0) = A * [tex]e^{-0}[/tex] + B * [tex]e^{-0}[/tex] = A + B = 1
Next, differentiate the solution y(t) with respect to t:
y'(t) = -A * [tex]e^{-t}[/tex] - (3B/5) * [tex]e^{-3t/5}[/tex]
Then, substitute t = 0 and y'(0) = 1 into the equation:
y'(0) = -A * [tex]e^{-0}[/tex] - (3B/5) * [tex]e^{-0}[/tex] = -A - (3B/5) = 1
We now have a system of equations:
A + B = 1
-A - (3B/5) = 1
Solving this system of equations, we can find the values of A and B.
From the first equation, we can rewrite it as:
A = 1 - B
Substituting this expression for A into the second equation:
-(1 - B) - (3B/5) = 1
Simplifying the equation:
-1 + B - (3B/5) = 1
Multiplying through by 5 to eliminate the fraction:
-5 + 5B - 3B = 5
Combining like terms:
2B = 10
Dividing by 2:
B = 5
Substituting the value of B back into the first equation:
A = 1 - 5 = -4
Therefore, the constants A and B are -4 and 5, respectively.
The solution to the initial value problem is:
y(t) = -4 [tex]e^{-t}[/tex] + 5 [tex]e^{-3t/5}[/tex]
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Find the area of the parallelogram
Height=4cm
Base=5cm
Answer:
The area of the parallelogram is 20cm²
Answer:
A = 20 cm²
Step-by-step explanation:
The area (A) of a parallelogram is calculated as
A = bh ( b is the base and h the height ) , then
A = 5 × 4 = 20 cm²
what is the inverses operation needed to solve for P?
800=p-275
A subtraction
B addition
C multiplication
D division ill mark brainlist
Two ordinary dice are thrown simultaneously. Determine the n
of throws necessary to obtain at least once with probability 0.49.
at least once the pair (6;6)
Two ordinary dice are thrown simultaneously. Determine the number of throws necessary to obtain at least once with probability 0.49 at least once the pair (6,6).
Solution: The probability of getting a pair of 6s in a single throw is 1/36.The probability of not getting a pair of 6s in a single throw is 1 - 1/36 = 35/36.
The probability of not getting a pair of 6s in n throws is (35/36)^n.
The probability of getting a pair of 6s in n throws is 1 - (35/36)^n.
So, for at least one pair of 6s with probability 0.49 in n throws, we have:
1 - (35/36)^n = 0.49⇒ (35/36)^n = 0.51⇒ n ln (35/36) = ln 0.51⇒ n = ln 0.51/ln (35/36) = 72.5 ~ 73So, at least 73 throws are necessary to obtain at least once with probability 0.49 at least once the pair (6,6).
Answer: At least 73 throws are necessary to obtain at least once with probability 0.49 at least once the pair (6,6).
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You roll a single 6 sided die. What are the odds of rolling a 9?
A. 1/6
B. 0
C. 1/9
D. 9
Answer:
B. 0
Step-by-step explanation:
There aren't enough sides for you to roll a nine
1. Find the solution to the recurrence relation an = 3an-1 + 4an-2 with initial values ao = 2 and a₁ = 3.
The solution to the recurrence relation an = 3an-1 + 4an-2 with initial values ao = 2 and a₁ = 3 is given byan = (-1)4ⁿ - 4(4)ⁿ-¹/16
Given recurrence relation is an = 3an-1 + 4an-2, with initial values ao = 2 and a₁ = 3.
The characteristic equation of the recurrence relation is given byr² - 3r - 4 = 0
Solving the characteristic equation, we get
r² - 4r + r - 4 = 0
r(r - 4) + 1(r - 4) = 0
(r - 4)(r + 1) = 0
r1 = 4, r2 = -1
So, the general solution of the recurrence relation is given by
an = Ar¹ + Br²
For r1 = 4, a4 = 3
a3 + 4a2a4 = 3a3 + 4a2 = 3(4a2 + 4a1) + 4a2= 16a2 + 12a1 ....(1)
For r2 = -1, aₙ₊₁ = 3an + 4an-1aₙ₊₁ = 3an + 4an-1 = 3(A(-1)^n + B(4)^n) + 4(A(-1)^(n-1) + B(4)^(n-1))= 3A(-1)^n - 4A(-1)^(n-1) + 12B(4)^n + 4B(4)^(n-1)= A(-1)^n + 4B(4)^n ....(2)
Putting n = 0 in (2), we get
a1 = A - 4A = -3A = 3 => A = -1
Substituting A = -1 in (1), we get
a4 = 16a2 + 12a1=> a4 = 16a2 + 12(2) => a4 = 16a2 + 24a4 = 16a2 + 24 => a2 = (a4 - 24)/16
Thus the solution to the recurrence relation an = 3an-1 + 4an-2 with initial values ao = 2 and a₁ = 3 is given by
an = (-1)4ⁿ - 4(4)ⁿ-¹/16
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