2. Find the real-valued solution to the initial value problem: y"-2y' + 17y 0 y(0) -2,...
(5 points) Find the general solution to the differential equation y" – 2y + 17y=0. In your answer, use Cį and C2 to denote arbitrary constants and t the independent variable. Enter Cų as C1 and C2 as С2. y(t) = help (formulas) Find the unique solution that satisfies the initial conditions: y(0) = -1, y'(0) = 7. y(t) =
Find the solution of the given initial value problem: y(4) + 2y" + y y(3) (0) y, (0) 0, y', (0) llt + 2; y (0) 1 Enclose arguments of functions in parentheses. For example, sin (2x) Find the solution of the given initial value problem: y(4) + 2y" + y y(3) (0) y, (0) 0, y', (0) llt + 2; y (0) 1 Enclose arguments of functions in parentheses. For example, sin (2x)
dy Find solution for initial valued problem:-3y + 2 3t ,y(0)--1 dt dy Find solution for initial valued problem:-3y + 2 3t ,y(0)--1 dt
Find the solution of the given initial value problem. y" + 2y' +2y = cost+8(-5); y(0) = 3, y'(0) = 5 ° 20) = us20e" sin + + cost ( +ş) + sint (36+}) x() ==««n6e8cose + cost (3e* +) + sint (80* + }) 20 = usz beé" sin + sing (54* +5.) +cos (34++}) ° 40 = =uaz(Dei* cost + cost ("* + 5 ) + sint (3*+ }) 209 = 192(e“ cose + cost (* +) +sint(****+})
Consider the initial value problem y'' – 2y' – 8y = 0, y(0) = a, y'(0) = 6 Find the value of a so that the solution to the initial value problem approaches zero as t + oo Q = a =
21. Solve the initial value problem y" - y-2y= 0, y(0) = a , y ( 0) the solution approaches zero as t 0o. 2. Then find a so that
Consider the initial value problem y'' + 2y' – 15y = 0, y0) = a, y'0 = 1 Find the value of a so that the solution to the initial value problem approaches zero as t → a= Preview Get help: Video Points possible: 2 Unlimited attempts.
Solve the initial value problem: 4y" +12y + 17y= 0, y(1/2) = 1, y(7/2) = 1. Give your answer as y=... . Use x as the independent variable. Answer:
Find the solution of the given initial value problem in explicit form. y′=(9x)/(y+x^2y), y(0)=−3 Enclose arguments of functions in parentheses. For example, sin(2x).
The only solution of the initial value problem ay' + by' + 2y = 4, y(0) = 2, y'(0) = 0 where a,and b are positive constants is y(x) = 2. True False