First-Order Differential Equations

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First-Order Differential Equations

• CHAPTER 2

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Contents

• 2.1 Solution By Direct Integration
• 2.2 Separable Variables
• 2.3 Linear Equations
• 2.4 Exact Equations
• 2.5 Solutions by Substitutions

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2.1 Solution By Direct Integration

• Consider dy/dx f(x, y) g(x). The DE dy/dx
  g(x) (1)can be solved by direct
  integration. Integrating both sides y ? g(x)
  dx c G(x) c.eg dy/dx 1 e2x, then y
  ? (1 e2x) dx c x ½ e2x c

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2.2 Separable Variables

• Introduction

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• Rewrite the above equation as
  (2)where p(y) 1/h(y).

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• (4)
• Integrating both sides, we have

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Example 2

• Solve
• Solution We also can rewrite the solution
  as x2 y2 c2, where c2 2c1Apply the
  initial condition, 16 9 25 c2See Fig2.18.
  Thus, because y(4)-3.

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Fig2.18
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Losing a Solution

• When r is a zero of h(y), then y r is also a
  solution of dy/dx g(x)h(y). However, this
  solution is not included in the general solution.
  That is a singular solution.

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2.3 Linear Equations

• Introduction Linear DEs are friendly to be
  solved. We can find some smooth methods to deal
  with.

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• Standard FormStandard form of a first-order DE
  can be written as dy/dx P(x)y f(x) (2)

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Solving Procedures

• If (2) is multiplied by (5)then (
  6)or (7)Integrating both sides, we
  get (8)Dividing (8) by
  gives the solution.

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Integrating Factor

• We call as an integrating
  factor and we should only memorize this to solve
  problems.

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Example 1

• Solve dy/dx 3y 6.
• SolutionSince P(x) 3, we have the
  integrating factor is then is the same as
  So e-3xy -2e-3x c, a solution is y -2
  ce3x, -? lt x lt ?.

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Notes

• The DE of example 1 can be written as y
  2 is included in the general solution. The
  general solution of linear first order DE include
  all the solutions.

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Application to Circuits

• See Fig 2.39. (8)

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Fig 2.39
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Example 6

• Refer to Fig 2.39, where E(t) 12 Volt, L ½
  HenryR 10 Ohms. Determine i(t) where i(0) 0.
• SolutionFrom (8), Then Using i(0) 0,
  c -6/5, then i(t) (6/5) (6/5)e-20t.

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Example 6 (2)

• A general solution of (8) is (11)When
  E(t) E0 is a constant, (11) becomes
  (12)where the first term is called a
  steady-state part, and the second term is a
  transient term.

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2.4 Exact Equations

• Introduction

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Differential of a Function of Two Variables

• If z f(x, y), its differential or total
  differential is (1)Now if z f(x, y)
  c,
• (2)eg if x2 5xy y3 c, then (2)
  gives (2x 5y) dx (-5x 3y2) dy 0 (3)
• Q What is the implicit solution of (3)?

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Proof of Necessity for Theorem 2.1

• If M(x, y) dx N(x, y) dy is exact, there
  exists some function f such that for all x in R
  M(x, y) dx N(x, y) dy (?f/?x) dx (?f/?y)
  dyTherefore M(x, y) , N(x, y)
  and
  (why?) The sufficient
  part consists of showing that there is a function
  f for which M(x, y) and N(x,
  y)

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Method of Solution

• Since ?f/?x M(x, y), we have (5)
• Differentiating (5) with respect to y and assume
  ?f/?y N(x, y)Then
• and
• 
  
  (6)
• Which holds if (4) is satisfied.

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• Integrate (6) with respect to y to get g(y), and
  substitute the result into (5) to obtain the
  implicit solution f(x, y) c.

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Example 1

• Solve 2xy dx (x2 1) dy 0.
• SolutionWith M(x, y) 2xy, N(x, y) x2 1,
  we have ?M/?y 2x ?N/?xThus it is exact.
  There exists a function f such that ?f/?x
  2xy, ?f/?y x2 1Then f(x, y) x2y
  g(y) ?f/?y x2 g(y) x2 1 g(y) -1,
  g(y) -yc

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Example 1 (2)

• Hence f(x, y) x2y yc, and the solution
  is x2y y c c, y c/(1 x2)The interval
  of definition is any interval not containing x
  1 and x -1.

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Example 2

• Solve (e2y y cos xy)dx(2xe2y x cos xy
  2y)dy 0.
• SolutionThis DE is exact because ?M/?y 2e2y
  xy sin xy cos xy ?N/?xHence a function f
  exists, and ?f/?y 2xe2y x cos xy 2ythat
  is,

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Example 2 (2)

• Thus h(x) 0, h(x) c. The solution is xe2y
  sin xy y2 c 0

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Example 3

• Solve
• SolutionRewrite the DE in the form (cos x sin
  x xy2) dx y(1 x2) dy 0 Since ?M/?y
  2xy ?N/?x (This DE is exact)Now ?f/?y y(1
  x2) f(x, y) ½y2(1 x2) h(x) ?f/?x
  xy2 h(x) cos x sin x xy2

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Example 3 (2)

• We have h?(x) cos x sin x h(x) -½ cos2
  xcThus ½y2(1 x2) ½ cos2 x c c1
  or y2(1 x2) cos2 x c (7)where c
  2(c1 -c). Now y(0) 2, so c 3.The solution
  is y2(1 x2) cos2 x 3
• Q What is the explicit solution?

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Fig 2.28

• Fig 2.28 shows the family curves of the above
  example and the curve of the specialized IVP is
  drawn in color.

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Integrating Factors

• It is sometimes possible to find an integrating
  factor?(x, y), such that ?(x, y)M(x, y)dx
  ?(x, y)N(x, y)dy 0 (8)is an exact
  differential.Equation (8) is exact if and only
  if (?M)y (?N)x Then ?My ?yM ?Nx
  ?xN, or ?xN ?yM (My Nx) ? (9)

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• Suppose ? is a function of one variable, say x,
  then ?x d? /dx(9) becomes (10)If
  we have (My Nx) / N depends only on x, then
  (10) is a first-order ODE and is separable.
  Similarly, if ? is a function of y only, then
  (11)In this case, if (Nx My) / M is
  a function of y only, then we can solve (11) for
  ?.

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• We summarize the results for M(x, y) dx N(x,
  y) dy 0 (12)If (My Nx) / N depends only on
  x, then (13)If (Nx My) / M depends
  only on y, then (14)

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Example 4

• The nonlinear DE xy dx (2x2 3y2 20) dy 0
  is not exact. With M xy, N 2x2 3y2 20, we
  find My x, Nx 4x. Since depends on both x
  and y. depends only on y.The integrating
  factor is e ? 3dy/y e3lny y3 ?(y)

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Example 4 (2)

• then the resulting equation is xy4 dx (2x2y3
  3y5 20y3) dy 0It is left to you to verify
  the solution is ½ x2y4 ½ y6 5y4 c

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2.5 Solutions by Substitutions

• IntroductionIf we want to transform the
  first-order DE dx/dy f(x, y)by the
  substitution y g(x, u), where u is a function
  of x, then Since dy/dx f(x, y), y g(x,
  u), Solving for du/dx, we have the form du/dx
  F(x, u). If we can get u ?(x), a solution is
  y g(x, ?(x)).

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Bernoullis Equation

• The DE dy/dx P(x)y f(x)yn (4)where n is
  any real number, is called Bernoullis Equation.
• Note for n 0 and n 1, (4) is linear,
  otherwise, let u y1-n to transform (4) into a
  linear equation.

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Example 2

• Solve x dy/dx y x2y2.
• SolutionRewrite the DE as a Bernoullis
  equation with n2 dy/dx (1/x)y xy2For n
  2, then y u-1, and dy/dx -u-2(du/dx) From
  the substitution and simplification, du/dx
  (1/x)u -xThe integrating factor on (0, ?) is

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Example 2 (2)

• Integrating gives x-1u -x c, or u -x2
  cx. Since u y-1, we have y 1/u and the
  general solution of the DE is y 1/(-x2 cx).

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Transformation to Separable DE

• A DE of the form dy/dx f(Ax By
  C) (5)can always be transformed into a
  separable equation by means of substitution u
  Ax By C.

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Example 3

• Solve dy/dx (-2x y)2 7, y(0) 0.
• SolutionLet u -2x y, then du/dx -2
  dy/dx, du/dx 2 u2 7 or du/dx u2
  9This is separable. Using partial fractions,
  or

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Example 3 (2)

• then we have Solving the equation for u
  and the solution is or (6)Applying
  y(0) 0 gives c -1.

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Example 3 (3)

• The graph of the particular solution is shown
  in Fig 2.30 in solid color.

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Fig 2.30

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Thank You !