(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)
X Incorrect. Use the transformation u = x-10y, v = 10x + y to find x-10y dA 10x + y R where R is the rectangular region enclosed by the lines X - 10y = 1, x - 10y = 100, 10x + y = 5, 10x + y = 25. (x-104 dA= TS 9999 -In(5) Edit 2 10x + y R
(1 point) Find yy as a function of xx ify′′′+64y′=0,y‴+64y′=0,y(0)=−1, y′(0)=−16, y′′(0)=−192.y(0)=−1, y′(0)=−16, y″(0)=−192.y(x)=
(1 point) Find a particular solution to y" + 7y' +10y = -5te
Entered Answer Preview (9/4)* [e^(4*x)]-x*[e^(4*x)]-(1/4)+4*x The answer above is NOT correct. (1 point) Find y as a function of x if j.(4) – 8y)" + 16y” = 0, y(0) = 2, y(0) = 12, y" (0) = 16, y" (0) = 0. y(x) = (9/4)e^(4x)-xe^(4x)-(1/4)+(4x)
(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) =
(1 point) Solve the boundary-value problem y" – 10y' + 25y = 0, y(0) = 7, y(1) = 0. Answer: y(x) = Note: If there is no solution, type "None".
Problem 3. (1 point) Find y as a function of tif y" + 5y - 14y = 0, y(0) = 5, y(1) = 6, y) = Remark: The initial conditions involve values at two points. Problem 4. (1 point) Find the solution to the linear system of differential equations 8x - 15y 6x-lly satisfying the initial conditions x(0) = -16 and yo) = -10 x(t) = Note: You can earn partial credit on this problem
(1 point) Find y as a function of x if (0)- 3, y' (0)-8, y"00. y(x) -
Suppose the point (2, 4) is on the graph of y = f(x). Find a point on the graph of the given function. The reflection of the graph of y = f(x) across the y-axis O (-2,4) O (2,-4) O (-2,-4) O (2.4)