Engineering Mathematics Class

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Engineering Mathematics Class 3First-Order ODE
(Part2)

• Sheng-Fang Huang

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1.2 Geometric Meaning of y (x, y).
Direction Fields
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Introduction

• Given a first-order ODE
• (1) y f(x, y)
• Geometrically, the derivative y(x) denotes the
  slope of y(x). Hence, for a given point (x0, y0)
• y(x0) f(x0, y0)
• denotes the slope on (x0, y0).
• We can indicate the solution curves by drawing
  short straight-line segments that describe slopes
  (called direction fields) to fit the solution
  curve.

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• Consider
• (2) y xy
• Using Isoclines to Find Direction Fields
• For (2), these are the hyperbolas (x, y) xy
  k const in Fig. 7b.
• By (1), these are the curves along which the
  derivative y is constant.
• Note
• These are not yet solution curves. Along each
  isocline draw many parallel line elements of the
  corresponding slope k. This gives the direction
  field, into which you can now graph approximate
  solution curves.

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Fig. 7. Direction field of y xy
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Fig. 8. An ODE that do need direction field
direction field of

• (3) y 0.1(1 x2)
• It is related to the van der Pol equation of
  electronics, a circle and two spirals approaching
  it from inside and outside.

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1.3 Separable ODEs. Modeling
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Introduction

• Many practically useful ODEs can be reduced to
  the form
• (1) g(y)y (x)
• Then we can integrate on both sides with respect
  to x, obtaining
• (2) ?g(y)y dx ?(x) dx
  c.
• By calculus, y dx dy, so that
• (3) ?g(y) dy ?(x) dx
  c.

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Example 1 A Separable ODE

• The ODE y 1 y2 is separable because it can
  be written
• By integration, arctan y x c or y tan
  (x c).
• Note
• Remember to introduce the constant immediately
  when the integration is performed.
• If we wrote arctan y x, then y tan x, and
  then introduced c, we would have obtained y tan
  x c, which is not a solution (when c ? 0).
  Verify this.

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Example 2 Radiocarbon Dating

• Physical Information
• In the atmosphere and in living organisms, the
  ratio of radioactive carbon 6C14 to ordinary
  carbon 6C12 is constant.
• When an organism dies, its absorption of 6C14 by
  breathing and eating terminates. Hence one can
  estimate the age of a fossil by comparing the
  radioactive carbon ratio in the fossil with that
  in the atmosphere. To do this, one needs to know
  the half-life of 6C14, which is 5715 years.

continued
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Example 2 Radiocarbon Dating

• In 1991, the famous Iceman (Oetzi), a mummy from
  the Neolithic period of the Stone Age was found
  in the ice of the Oetztal Alps in Southern
  Tyrolia near the AustrianItalian border.
• When did Oetzi approximately live and die if the
  ratio of carbon 6C14 to carbon 6C12 in this mummy
  is 52.5 of that of a living organism?

continued
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Solution
continued
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Example 3 Mixing Problem

• The tank in Fig. 9 contains 1000 gal of water in
  which initially 100 lb of salt is dissolved.
• Brine runs in at a rate of 10 gal/min, and each
  gallon contains 5 lb of dissoved salt.
• The mixture in the tank is kept uniform by
  stirring. Brine runs out at 10 gal/min.
• Find the amount of salt in the tank at any time t.

continued
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Fig. 9. Mixing problem in Example 3
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Solution
continued
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Example 5 Leaking Tank

• This problem concerns the outflow of water from a
  cylindrical tank with a hole at the bottom (Fig.
  11).
• If the tank has diameter 2 m, the hole has
  diameter 1 cm, and the initial height of the
  water when the hole is opened is 2.25 m.
• When will the tank be empty?

continued
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Physical information

• Under the influence of gravity the outflowing
  water has velocity
• (7)
  (Torricellis law),
• where h(t) is the height of the water above
  the hole at time t, and g 980 cm/sec2 is the
  acceleration of gravity at the surface of the
  earth.

continued
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Fig. 11. Outflow from a cylindrical tank.
Torricellis law
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Solution
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Extended Method

• Reduction to Separable Form
• Certain nonseparable ODEs can be made separable
  by transformations that introduce for y a new
  unknown function. For equations
• (8)
• Here, is any (differentiable) function of
  y/x, such as sin (y/x), (y/x)4, and so on.

continued
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Extended Method

• The form of such an ODE suggests that we set
  y/x u thus,
• Substitution into y (y/x) then gives ux
  u (u) or ux (u) u. We see that this
  can be separated

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Example 6 Reduction to Separable Form

• Solve 2xyy y2 x2.
• Solution.

continued
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1.4 Exact ODEs. Integrating Factors
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Introduction to Exact ODEs

• For those equations that are not separable,
  verify if they are exact.
• A first-order ODE M(x, y) N(x, y)y 0,
  written as
• (1) M(x, y) dx N(x, y) dy 0
• is called an exact differential equation if
  the differential form M(x, y) dx N(x, y) dy is
  exact, that is, this form is the differential
• (2)

continued
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Introduction to Exact ODEs

• of some function u(x, y). Then (1) can be
  written
• du 0.
• By integration we immediately obtain the
  general solution of (1) in the form
• (3) u(x, y) c.
• The equation (1) is an exact differential
  equation if there is some function u(x, y) such
  that
• (4)
• Thus,

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Introduction to Exact ODEs

• (5)
• This condition is not only necessary but also
  sufficient for (1) to be an exact differential
  equation.

continued
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Introduction to Exact ODEs

• If (1) is exact, from (4a) we can obtain u(x,
  y) have by integration with respect to x
• (6) u ?M dx k(y)
• in this integration, y is to be regarded as a
  constant, and k(y) plays the role of a constant
  of integration.
• To determine k(y), we derive ?u/?y from (6), use
  (4b) to get dk/dy, and integrate dk/dy to get k.

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Example 1 An Exact ODE

• Solve
• (7) cos (x y) dx (3y2 2y cos (x
  y)) dy 0.
• Solution.
• Step 1. Test for exactness. Our equation is of
  the form (1) with
• M cos (x y),
• N 3y2 2y cos (x y).
• Thus
• From this and (5) we see that (7) is exact.

continued
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• Step 2. Implicit general solution. From (6) we
  obtain by integration
• (8)
• To find k(y), we differentiate this formula
  with respect to y and use formula (4b), obtaining
• Hence dk/dy 3y2 2y. By integration, k
  y3 y2 c. Inserting this result into (8) and
  observing (3), we obtain the answer
• u(x, y) sin (x y) y3 y2
  c.

continued
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• Step 3. Checking an implicit solution.
• Check by differentiating the implicit solution
  u(x, y)c and see whether this leads to the given
  ODE (7)
• (9)
• This completes the check.

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Example 2 An Initial Value Problem

• Solve the initial value problem
• (10) (cos y sinh x 1) dx sin y cosh x
  dy 0, y(1) 2.
• Solution. You may verify that the given ODE is
  exact.

continued
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Fig. 14. Particular solutions in Example 2
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Example 3 WARNING!
Breakdown in the Case

• The equation y dx x dy 0 is not exact
  because M y and N x, so that in (5), ?M/?y
  1 but ?N/?x 1.
• Let us show that in such a case the present
  method does not work. From (6),
• Now, ?u/?y should equal N x, by (4b). However,
  this is impossible because k(y) can depend only
  on y. Try (6) it will also fail.

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Reduction to Exact Form. Integrating Factors

• We multiply a given nonexact equation
• (12) P(x, y) dx Q(x, y) dy
  0,
• by a function F that, in general, will be a
  function of both x and y. We want the result to
  be a exact equation
• (13) FP dx FQ dy 0
• So we can solve it as just discussed. Such a
  function F(x, y) is then called an integrating
  factor of (12).

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Example 4 Integrating Factor

• The integrating factor in (11) is F 1/x2. Hence
  in this case the exact equation (13) is
• These are straight lines y cx through the
  origin.
• We can readily find other integrating factors for
  the equation y dx x dy 0, namely, 1/y2,
  1/(xy), and 1/(x2 y2), because

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How to Find Integrating Factors

• We let
• (16)

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Example 5 Application of Theorems 1 and 2.

• Initial Value Problem
• Using Theorem 1 or 2, find an integrating factor
  and solve the initial value problem
• (exy yey) dx (xey 1) dy 0, y(0) 1
• Solution.
• Step 1. Nonexactness. The exactness check fails

continued
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• Step 2. Integrating factor. General solution.
  Theorem 1 fails because R the right side of
  (16) depends on both x and y,

continued
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• Step 3. Particular solution.
• Step 4. Checking.

continued