Question

(dx/dt) = x+4z

(dy/dt) = 2y

(dz/dt) = 3x+y-3z

Answer #1

find dx/dx and dz/dy
z^3 y^4 - x^2 cos(2y-4z)=4z

dx
dt
= y − 1
dy
dt
= −3x + 2y
x(0) = 0, y(0) = 0

(x+1)y'=y-1 2) dx+(x/y+(e)?)dy=0 3)ty'+2y=sint 4) y"-4y=-3x²e3x
5) y"-y-2y=1/sinx 6)2x2y"+xy'-2y=0 ea)y=x' b) x=0
2dx/dt-2dy/dt-3x=t; 2

dx/dt=y, dy/dt=2x-2y

Solve the system of equations by method of the Laplace
transform:
3 dx/dt + 3x +2y = e^t
4x - 3 dy/dt +3y = 3t
x(0)= 1, y(0)= -1

Use the Laplace transform to solve the given system of
differential equations. dx/dt=x-2y dy/dt=5x-y x(0) = -1, y(0) =
6

Solve the following system of differential equations:
dx/dt =x+2y
dy/dt =−x+3y

dx/dt - 3(dy/dt) = -x+2
dx/dt + dy/dt = y+t
Solve the system by obtaining a high order linear differential
equation for the unknown function of x (t).

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=

Initial value problem : Differential equations:
dx/dt = x + 2y
dy/dt = 2x + y
Initial conditions:
x(0) = 0
y(0) = 2
a) Find the solution to this initial value problem
(yes, I know, the text says that the solutions are
x(t)= e^3t - e^-t and y(x) = e^3t + e^-t
and but I want you to derive these solutions yourself using one
of the methods we studied in chapter 4) Work this part out on paper
to...

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