Mathematics in economics
Lecture 11
Mgr. Jiří Mazurek, Ph.D.
Mathematics in Economics/PMAT

Lecture 11
Differential equations - continued




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1.
Now assume that q(x) is not zero:
In such case we use the method called variation of
parameters. We assume the solution of the form:
But C is now a function:

Lecture 11
Differential equations - continued




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1.
Substituting the last formula into                        yields
the solution.
Example:
Solution: we search for a solution of the form                     .
Substituting into the equation:

Lecture 11
Differential equations - continued




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1.
Rearranging of terms yields:
Now we integrate:
Solution of the given equation is:

Lecture 11
Differential equations – Problem 1




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1.
Solve:                        .
Solution:
First, we solve corresponding homogenous equation by
the separation of variables method:
And finally:

Lecture 11
Differential equations – Problem 1 – cont.




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1.
In the second step, we apply the variation of a constant
method:
Substitution:

Lecture 11
Linear differential equations of the second order with constant coefficients




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1.
The last type of differential equation we will address.
It is of the form:                     .
A solution is assumed to be in the form            ,
where lambda is a root of the so called characteristic
equation:

Lecture 11
Linear differential equations of the second order with constant coefficients




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1.
In the aforementioned three cases, we yield following
solutions:
Case 1:
Case 2:
Case 3:
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Lecture 11
Linear differential equations of the second order with constant coefficients




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1.
The characteristic equation is a quadratic equation,
which means we have three cases:
• Two real roots.
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• One real root of the order two.
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• Two imaginary roots.

Lecture 11
Linear differential equations of the second order with constant coefficients  - Problem 1

  Solve:
  Solution: we start with the characteristic equation:
This equation has two real roots: λ1 = 2 and λ2 = 3.
Therefore, th e solution is:


1.

Lecture 11
Linear differential equations of the second order with constant coefficients  - Problem 2

  Solve:
  Solution: we start with the characteristic equation:
This equation has two real roots: λ1 = 3 and λ2 = -1.
Therefore, th e solution is:


1.

Lecture 11
Linear differential equations of the second order with constant coefficients  - Problem 3

  Solve:
  Solution: we start with the characteristic equation:
This equation has two real roots: λ1 = 3 and λ2 = 3.
Therefore, th e solution is:


1.

Lecture 11
Linear differential equations of the second order with constant coefficients  - Problem 4

  Solve:
  Solution: we start with the characteristic equation:
This equation has two real roots: λ1 = -2+i and λ2 = -2-i.
Therefore, th e solution is:


1.

Lecture 11
Linear differential equations of the second order with constant coefficients  - Problem 5

  Solve:
  Solution: we start with the characteristic equation:
This equation has two real roots: λ1 = 1  and λ2 = -1.
Therefore, th e solution is:


1.

Lecture 11
Linear differential equations of the second order with constant coefficients  - Problem 6

  Solve:
  Solution: we start with the characteristic equation:
This equation has two real roots: λ1 = i  and λ2 = -i.
Therefore, the solution is:


1.

Lecture 11
Linear differential equations of the second order with constant coefficients

  Now we will focus on eqautions with non-zero right hand side:

 This type of equation is called non-homogenous.
  Solution of this equation has the following form:


1.

Lecture 11
Linear differential equations of the second order with constant coefficients – cont.

  The solution                   correspond to a homogenous case,
  while           is the so called particular integral, which solves
a nonhomogenous part of an equation.
 A particular integral for the most common functions (polynomials,
exponentials, logarithms, etc.) can be easily “guessed“.
We will illustrate the procedure by several examples.



1.

Lecture 11
Linear differential equations of the second order with constant coefficients – Problem 9

  Solve:

Solution: we begin with the homogenous case and its characteristic polynom:
The roots are λ1 = 2 a λ2 = –1, hence the solution is:
Now we seek a particular integral in the form:

1.

Lecture 11
Linear differential equations of the second order with constant coefficients – Problem 9 – cont.

  Solve:

Solution: we substitute y = P(x) into the given equation:

Which yields: a = -2, b = 2.
Therefore, the general solution to the equation is:

Lecture 11
Linear differential equations of the second order with constant coefficients – Problem 10

  Solve:

Solution: we begin with the homogenous case and its characteristic polynom:
The roots are λ1 = 0 a λ2 = –4, hence the solution is:
Now we seek a particular integral in the form:

1.

Lecture 11
Linear differential equations of the second order with constant coefficients – Problem 10 – cont.

  Solve:

Solution: we substitute y = P(x) into the given equation:

Which yields:
Therefore, the general solution to the equation is:

Lecture 11
Problems to solve - 1

  Solve:

Lecture 11
Problems to solve - 2

  Solve:

Lecture 11
Problems to solve - 3

  Solve:

Lecture 11
Final remarks

•  See the exam dates in STAG. Everybody has 2 attempts.
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• Also, see the older versions of exam tests on my public or Moodle.
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• If you need consultations, write me (or Dr. Stoklasova) an e-mail.
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•Good luck!

Lecture 11
Thank you for your attention!