Question

dX(t) = bX(t)dt + cX(t)dW(t) for contant values of X(0), b and c

(a) Find E[X(t)] (hint: look at e ^(−bt)X(t))

(b) The Variance of X(t)

Answer #1

Let w(x,y,z) = x^2+y^2+z^2 where x=sin(8t), y=cos(8t) , z=
e^t
Calculate dw/dt by first finding dx/dt, dy/dt, and dz/dt and using
the chain rule
dx/dt =
dy/dt=
dz/dt=
now using the chain rule calculate
dw/dt 0=

Find the matrix operator T: P3 --> P2 where T [= T(a + bx +
cx^2 + dx^3) = b + 2cx + 3dx^2 with respect to bases B = {1, x,
x^2, x^3} and C = {1, x, 2x^2, -1}.

dx/dt = 1 - (b+1) x + a x^2 y
dy/dt = bx - a x^2 y
find all the fixed point and classify them with Jacobian.

Let f(x)= a -bx^c + dx^e where a, b,c,d,e >0 and c<e.
Suppose that f(x0)= 0 and f '(x0)=0 for some x0>0. Prove that
f(x) greater than or equal to 0 for x greater than or equal to
0

a). Find dy/dx for the following integral.
y=Integral from 0 to cosine(x) dt/√1+ t^2 ,
0<x<pi
b). Find dy/dx for tthe following integral
y=Integral from 0 to sine^-1 (x) cosine t dt

If x1(t) and x2(t) are solutions to the differential
equation
x"+bx'+cx = 0
1. Is x= x1+x2+c for a constant c always a solution?
(I think No, except for the case of c=0)
2. Is tx1 a solution? (t is a constant)
I have to show all works of the whole process, please
help me!

solve the given initial value problem
dx/dt=7x+y x(0)=1
dt/dt=-6x+2y y(0)=0
the solution is x(t)= and y(t)=

1.express dw/dt as a function of t, both by using the Chain Rule
and by expressing w in terms of t and differentiating directly with
respect to t. Then evaluate dw/dt at the given value of t.
a)w= -6x^2-10x^2 , x=cos t,y=sint, t=pi/4
b)w=4x^2y-4y^2x, x=cost y=sint, --> express n terms of t
2.Find the linearization L(x,y) of the function (x,y)=e^x
cos(9y) at points (0,0) and (0,pi/2)

dy/dt=x , dx/dt=6x-8y, x=1 and y=-1 when t=0

Use the Chain Rule to find dw/dt.
w = ln
x2 + y2 + z2
, x = 9 sin(t), y
= 4 cos(t), z = 5 tan(t)
dw
dt
=

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