If fY(y) = 3y^2e^(-y^3) for y>0 what are the principles for finding E[Y] ?
(ie what to ask Wolfram Alpha for integration)
If fY(y) = 3y^2e^(-y^3) for y>0 what are the principles for finding E[Y] ? (ie what...
Solve the IVP. y' – 3y = t + 2e, y(0) = 0
JO SUUS. 7.12. Solve the BVP y" = -2e-3y + 4(1+x)-3, 0<x<1, subject to y(0) = 0, y' (O) = 1, y(1) = In 2. Compare to the exact solution, y(x) = ln(1+x).
Solve the initial value problem y" + 3y' + 2y = 8(t – 3), y(0) = 2, y'(0) = -2. Answer: y = u3(t) e-(-3) - u3(t)e-2(1-3) + 2e-, y(t) ={ 2e-, t<3, -e-24+6 +2e-l, t>3. 5. [18pt] b) Solve the initial value problem y' (t) = cost + Laplace transforms. +5° 867). cos (t – 7)ds, y(0) – 1 by means of Answer:
(10 points) Let Y have probability density function (pdf) 3y?, for ( <y<1 fy(y) = 10, otherwise (a) Compute the probability density function (pdf) of 1/Y. (b) Compute the probability density function (pdf) of Y1 +Y2, if Yį and Y2 are inde- pendent random variables with the same pdf as Y. (You can use a computer to help with the integration).
Question 27 DE y"- 3y'- 4y = 0 has a general solution y = C1 e4x + C2e* Find the particular solution, if exists, with the initial conditions: y(O) = 0 and y(1) = 2. Yp 2e e5--1 (241 – e-") None of them 2e Ур 2e 5-1 etr + e 1- Both of them
Solve the initial value problem y" – 3y' + 2y = e3r, y(0) = 2, y'(0) = -1. (a) y(x) = 40-1 – 4e2+ 2e 32 (b) y(x) = 1 e?' – 4e-2x + £230 (c) y(x) = 40-1 – 4e-2x + 3e3x (d) y(x) = 40" – 4e2x + e3r Select one: a с b d
(1 point) Given the third order homogeneous constant coefficient equation y" + 3y" + 3y + y = 0 1) the auxiliary equation is ar3 + br2 + cr + d = ^3+3r^2+3r+1 - = 0. 2) The roots of the auxiliary equation are -1,-1,-1 (enter answers as a comma separated list). 3) A fundamental set of solutions is e^(-x),xe^(-x),x^2e^(-x) (enter answers as a comma separated list). 4) Given the initial conditions y(0) = -1, ý (0) = 2 and...
Solve the initial value problem. y'" – 3y" - y' + 3y = 0; y(0)=5, y'0) = -3. y'(0)=5 The solution is y(t) =
Suppose FY (y) = y3 for 0 ≤ y < 1/2, and FY (y) = 1 − (1-y)3 for 1/2 ≤ y ≤ 1. Compute each of the following. (a) P(1/3 < Y < 3/4) (b) P(Y = 1/3) (c) P(Y = 1/2)
Solve: y' – 4y' + 3y = 9t – 3 y(0) = 3, y'(0) = 13 y(t) = Preview