Differential Equations Practice Exams

Answers


Problem 1. Consider the system

displaymath188

Draw the nullclines and find all equilibrium points. Determine the fate of the solutions with initial conditions

displaymath190

Problem 2. Find the solution of

displaymath192

with the initial condition

displaymath194

Problem 3. Find the solution of

displaymath196

with the initial condition

displaymath198

Problem 4. Consider the system

displaymath200

Find the equilibrium points. Find the linearized system at these points and discuss the behavior of the solutions at these points. Determine if the equilibrium points are sinks, saddles, sources, and so on...

Answers.

Answer to Problem 1. Let us find the x-nullclines and y-nullclines.

x-nullclines: We must have

displaymath202

which is equivalent to x = 0 or -4x-y+1 = 0.

y-nullclines: We must have

displaymath208

which is equivalent to y = 0 or tex2html_wrap_inline212 (which is the equation of a circle centered at (0,0)).

The critical points are the intersection between the x-nullclines and y-nullclines. Hence the equilibrium points are

displaymath216

See the figure below for more details about the nullclines and the solutions.

1.
The solution with initial condition (1,1) goes up and left. It dies at the equilibrium point (0,2).

2.
The solution with initial condition (0,1) goes up. Indeed, it is easy to check that there are solutions which lives on the line x=0. Since this initial condition is on this line, the entire solution will stay on the line. When tex2html_wrap_inline226 , the solution dies at the equilibrium point (0,2).

1.
The solution with initial condition (1,-1) goes down and left. It dies at the equilibrium point

displaymath232

Answer to Problem 2. The matrix coefficient is

displaymath234

The characteristic equation is

displaymath236

Its roots are : -3 and -3 (-3 is a double root). An eigenvector associated to -3, is

displaymath238

We may choose

displaymath240

The general solution is

displaymath242

where

displaymath244

satisfies

displaymath246

This gives tex2html_wrap_inline248 . Hence the vector

displaymath250

will do (here we took tex2html_wrap_inline252 ). Hence the general solution is

displaymath254

The initial condition Y(0) = (1,0) gives

displaymath258

This implies tex2html_wrap_inline260 and tex2html_wrap_inline262 . Therefore the solution to the initial condition is

displaymath264

Answer to Problem 3. The matrix coefficient is

displaymath266

The characteristic equation is

displaymath268

Its roots are

displaymath270

(these are complex roots). An eigenvector associated to 4 + 2i, is

displaymath274

We may choose

displaymath276

WE need the real part and imaginary part of

displaymath278

Easy calculations give

displaymath280

where

displaymath282

and

displaymath284

The general solution is

displaymath286

The initial condition Y(0) = (1,1) gives

displaymath290

which yields

displaymath292

This implies tex2html_wrap_inline294 and tex2html_wrap_inline296 . Therefore the solution to the initial condition is

displaymath298

Answer to Problem 4. Let us find the x-nullclines and y-nullclines.

x-nullclines: We must have

displaymath300

which is equivalent to x = 0 or tex2html_wrap_inline304 . This gives

displaymath306

y-nullclines: We must have

displaymath308

which is equivalent to y = 0 or y = 1.

The critical points are the intersection between the x-nullclines and y-nullclines. Hence the equilibrium points are

displaymath314

In order to find the linearized system at these points, we must find the Jacobian of the system which is

displaymath316

Therefore, we have

At (0,0)

displaymath320

The eigenvalues are 1 and -1. Therefore, (0,0) is a saddle.

At (0,1)

displaymath326

The eigenvalues are 1 (which is double). Therefore, (1,1) is a source.

At tex2html_wrap_inline330

displaymath332

The eigenvalues are -1 and -2. Therefore, tex2html_wrap_inline330 is a sink.

At tex2html_wrap_inline336

displaymath338

The eigenvalues are 1 and -2. Therefore, tex2html_wrap_inline336 is a saddle.

See the figure below for more details on the behavior of some solutions.


[Next Exam] [Calculus] [CyberExam]

S.O.S MATH: Home Page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Copyright © 1999-2024 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour