-2s+9s2-8s+18 Find L-1 s(s²+4)(s+1) O (2- 5e +3 sin 2t) O (3 - 5e-t + 3 sin 2t) O (1-3e-t +3 sin 2t) O (4 5e +2 sin 3t)
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- Find all solutions to cos(7x)−cos(x)=sin(4x)cos(7x)-cos(x)=sin(4x) on 0≤x<2π30≤x<2π31. (2x4-x3-2x+3) dx 2. (3-2x)2 dx 3. 3(x-1)(2-3x) dx 4. (5-5x+x5) dx5. 1/3 ex dx 6. (x2-cos x) dx 7. (2 cos x- 3 sin x) dxFor the new solution problem at the bottom, how did you get (4^n/2^2n) = 1/4^n. Plus, why does (theta^(4)/4^n) times (1/4!) = 0?
- 1. Express as a sum or difference a.) sin 7t sin t b.) cos 6u cos (-4u) c.) cos 2t sin 6t d.) 2 sin 5θ cos 3θ e.) 2 sin 7θ sin 5θ f.) 3 cos x sin 2x g.) 5 cos 4u cos 5uMy particular form solution looks like At^2+Bt+C+Dsin(4t)+Ecos(4t)+Fe^(2t), I know this is correct.I keep getting sin(4t)*(-16D-16E+4D)+cos(4t)*9-16E+16D+4E) which reduces to sin(4t)*(-12D-16E)+cos(4t)*9-12E+16D) which gets me D=(-4/3)E and E=-47/18 which is wrong, I even plugged it into an online diffeq solver which shows D=2/5 and E=-3/10 respectively. So I know there's something wrong with my application of method of undetermined coefficients.Here are the choices:- square root of (x-1)- 126 units- x-1- integral of square root of (1+squareroot of (x-1)) dx with the limits from 9 to 36- 189 units- integral of square root of (1+squareroot of (x-1) squared) dx with the limits from 9 to 36P.S no need for solutions
- Find the solution in the interval from 0 ≤ θ < 2π 2sin2 x - cosx -1 = 0Solve sin(3x)cos(5x)−cos(3x)sin(5x)=−0.25sin(3x)cos(5x)-cos(3x)sin(5x)=-0.25 for the smallest positive solution.x =Show your solution. Given the differential equation: x(x+3)y'' - 9y' - 6y = 0 Question: Using the suggested approach, the recurrence relation for n ≥ 1 is: