Differential Equations Practice Exams

Answers


Problem 1. Consider the system

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

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Problem 2. Find the solution of

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with the initial condition

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Problem 3. Find the solution of

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with the initial condition

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Problem 4. Consider the system

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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

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

y-nullclines: We must have

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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

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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

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Answer to Problem 2. The matrix coefficient is

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The characteristic equation is

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

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We may choose

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The general solution is

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where

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satisfies

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This gives tex2html_wrap_inline248 . Hence the vector

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

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

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

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Answer to Problem 3. The matrix coefficient is

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The characteristic equation is

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Its roots are

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

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We may choose

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WE need the real part and imaginary part of

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Easy calculations give

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where

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and

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The general solution is

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

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which yields

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

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

x-nullclines: We must have

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

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y-nullclines: We must have

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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

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

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Therefore, we have

At (0,0)

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

At (0,1)

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

At tex2html_wrap_inline330

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

At tex2html_wrap_inline336

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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.


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