solve the initial values:
if Y(3)-4Y"+20Y'=51e^3x
Y"(0)=41, Y'(0)= 11. Y(0)= 7 > solution is Y(x)= e^3x+2 e^2x
sin(4x)+6
so, what is the solution for:
Y(3)-8Y"+17Y'=12e^3x
Y"(0)=26, Y'(0)= 7. Y(0)= 6
Y(x)=???
Given:
f(x,y) = 5 - 3x - y for 0 < x,y < 1 and x + y < 1, 0
otherwise
1) find the covariance of x and y
2) find the marginal probability density function for x
c) find the probability that x >= 0.6 given that y <=
0.2