Question

Consider the system [ x' = -2y & y' = 2x] . Use dy/dx to find the curves y = y(x).

Draw solution curves in the xy phase plane. What type of equilibrium point is the origin?

Answer #1

Consider the initial value problem
dy dx
=
1−2x 2y
, y(0) = − √2
(a) (6 points) Find the explicit solution to the initial value
problem.
(b) (3 points) Determine the interval in which the solution is
deﬁned.

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

consider the equation y*dx+(x^2y-x)dy=0. show that the equation
is not exact. find an integrating factor o the equation in the form
u=u(x). find the general solution of the equation.

Consider the following linear system (with real eigenvalue)
dx/dt=-2x+7y
dy/dt=x+4y
find the specific solution coresponding to the initial values
(x(0),y(0))=(-5,3)

Solve the following ODE's/ IVP's
y'=1/xsin(y)+2sin(2y)
Hint: consider x=x(y) and dx/dy=1/(dx/dy)=1/y'

(* Problem 3 *)
(* Consider differential equations of the form a(x) + b(x)dy
/dx=0 *) \
(* Use mathematica to determin if they are in Exact form or not.
If they are, use CountourPlot to graph the different solution
curves 3.a 3x^2+y + (x+3y^2)dy /dx=0 3.b cos(x) + sin(x) dy /dx=0
3.c y e^xy+ x e^xydy/dx=0

Consider the following system of differential equations dx/dt =
(x^2 + 2x + 1)(x^2 − 4x + 4) dy/dt = xy − 1
Which of the following is not an equilibrium point of the above
system? (A) (3, 1/3 ) (B) (−1, −1) (C) (1, 1) (D) (1, 3)

Determine the type of below equations and solve it.
a-)(sin(xy)+xycos(xy)+2x)dx+(x2cos(xy)+2y)dy=0
b-)(t-a)(t-b)y’-(y-c)=0 a,b,c are
constant.

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...

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

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