(1 point) Find as a function of if
(1 point) Find y as a function of x if y′′′+64y′=0, y(0)=−1, y′(0)=−16, y′′(0)=−192. y(x)=
(1 point) Find y as a function of t if 64y"-81y-0 with O) 3, (0)7.
(1 point) We know that y(x) = ** is a solution to the differential equation y - 12y - 64y = 0 for x € (-0,00) Use the method of reduction of order to find the second solution to y - 12y - 64 y = 0 for x € (-0, 0). (a) After you reduce the second order equation by making the substitution w = C', you get a first order equation of the form w = f(x, w)...
(1 point) Find y as a function of x if y" – 7y" + 10y' = 12et, y(0) = 10, y(0) = 29, y' (0) = 10. y(x) = (21/2)+(41/2)^(2x)-3e^(5x)+3e^(x) 000 (1 point) Find a particular solution to y" + 36y = –24 sin(6t). yp = 16-3e^(-3t)-8cos(3t)
(1 point) Find y as a function of x if (0)- 3, y' (0)-8, y"00. y(x) -
(1 point) Find y as a function of x if y(4) – 10y" + 254" = -392e-27, = 16. y(0) = 4, y(0) = 24, y" (O) = 17, y" (0) y(x) =
(16 points) Find y as a function of x if y'" + 25y' = 0, y(0) = -7, y' (O) = -15, y" (0) = 100. y(x) =
(1 point) The joint probability density function of X and Y is given by f(x, y) = cx – 16 c”, - <x< 0 < b < co alt 0 < y < 0 Find c and the expected value of X: c = E(X) =
(16 points) Find y as a function of a if y" + 16y' = 0, y(0) = -7, y(0) = 12, 7(0) = 48. y(x) =
(1 point) Find y as a function of lif y" - 11y +24y = 0 y(0) - S WI) = 4 W = Remark: The initial conditions involve values at two points. Problem 4. (1 point) Find the solution to the linear system of differential equations 59x +84 -42x - 607 satisfying the initial conditions (0) = 10 and y(0) -7. = X(t) = y = Note: You can earn partial credit on this problem.
13. Find the length of the part of the curve y=3/16e^(2x)+1/3e^(-2x) for 0<x<ln2 16 13. Find the length of the part of the curve y-で2" +-e-2x for 0-rS In 2. 29 64 13 16 16 16 13. Find the length of the part of the curve y-で2" +-e-2x for 0-rS In 2. 29 64 13 16 16