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S.l.dr.ing.mat. Alina BogoiDifferential
Equations
POLITEHNICA University of BucharestFaculty of Aerospace Engineering
CHAPTER 2
First-Order Differential
Equations (cont’d)
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Diff_Eq_4_2011 2
Outline• Exact equations
• Integrating Factors
• Homogenous equations
• First Order Ordinary Differential equation
• First Order Non-linear Differential equation
– Bernoulli equation
– Riccati equation
– Clairaut equation
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A first-order DE of the form a1
( x)(dy/dx) +
a0
( x) y = g ( x) (1)
is said to be a linear equation in y.When g ( x) = 0, (1) is said to be homogeneous;
otherwise it is nonhomogeneous.
DEFINITIONLinear Equations
First Order Ordinary Differential
equation
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• we have the general solution:
dx x f eece y y ydx x P dx x P dx x P
pc ∫ ∫ ∫ +∫ =+=−−
)()()()(
• We call is an integrating factor and we should only memorize
this to solve problems.
∫ dx x P
e)(
Integrating Factor
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BERNOULLI’S EQUATION
The differential equation
where n is any real number, is called Bernoulli’sequation. For n = 0 and n = 1, the equation is
linear and can be solved by methods of the lastsection.
,)()( n y x f y x P dx
dy=+
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SOLVING A BERNOULLI
EQUATION
1. Rewrite the equation as
2. Use the substitution w = y1 − n, n ≠ 0, n ≠ 1. NOTE: dw/dx = (1 − n) y−n
dy/dx.
3. This substitution turns the equation in Step 1into a linear equation which can be solved by
the method of the last section.
).()(
1
x f y x P dx
dy
y
nn
=+
−−
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Diff_Eq_4_2011 7
How to transform the Riccati equation to a linear one ?
Somehow we get one solution, , of a Riccati
equation, then the change of variables
transforms the Riccati equation to a linear one.
)()()(2'
x R y xQ y x P y ++=
)( x
z xS y
1)( +=
RICATTI’S EQUATION
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Diff_Eq_4_2011 8
z xQ
z xS x P
z x P z'
z
1)(
1)()(2
1)(
122
++=−
)())()()(2( x P z xQ xS x P z' −=++
[2 ( ) ( ) ( )]
i ntegrating factor
( )P x S x Q x dx
f x e+∫ =
)(dx)()(
)(
1)(
x f
C x f x P
x f x z +−= ∫
Riccati Equation
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Diff_Eq_4_2011 9
CLAIRAUT’S EQUATION
The nonlinear differential equation y = xy′ + F( y′)
is called Clairaut’s equation. Its solution is thefamily of straight lines y = cx + F (c), where c is
an arbitrary constant.
dx
dy p = ( ) y x p F p= +
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Diff_Eq_4_2011 10
PARAMETRIC SOLUTION TO
CLAIRAUT’S EQUATION
2
1 22
1 1 2 1 1
2 1 1 1
( ) 0
0
put into the origional eq. ( )
( ) ( ) general solution
dy dp dF dp dp dF p p x xdx dx dp dx dx dp
dp d y y c x c
dx dxdy
p c c x c c x F cdx
c F c y c x F c
= = + + ⇒ + =
= = ⇒ = +
⇒ = = ⇒ + = +
⇒ = ⇒ = +
0 ( , ) 0 eliminate in the origional ODEdF
x G x p pdp
+ = ⇒ =
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Diff_Eq_4_2011 11
Clairaut’s equation
)( p F px y+
2
2 2
2 2
22
Ex:
2 ( 2 ) 0
(1) ( ) ( ( )) general solution
(2) 2 0
2 2 4
4 0 singular solution4
y px p
dy dp dp dp p x p p x p
dx dx dx dx
F p p y x F c cx c
x x x x p p y
x
y x y
= +
⇒ = + + = ⇒ + =
= ⇒ = + = +
+ = ⇒ = − ⇒ = − +
= − ⇒ + =
'( ) 0 x F p+ =( ) y c x F c= +
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Diff_Eq_4_2011 12
Non-linear 2nd O.D.E.
• p substitution.
•
the second derivative of y is replaced by the first derivative
of p thus eliminating y
• completely and producing a first O.D.E. in p and x.
dx
dy p =
2
2
dx
yd
dx
dp
=
A. Equations where the dependent variables does
not occur explicitly
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Diff_Eq_4_2011 13
Solve axdx
dy xdx
yd =+2
2
Let dx
dy p =
ax xpdx
dp=+
⎟ ⎠
⎞⎜⎝
⎛ 2
2
1exp xintegral factor
Example
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Diff_Eq_4_2011 14
Solve axdx
dy xdx
yd =+2
2
Let dx
dy p = 2
2
dx
yd
dx
dp
=and therefore
ax xpdx
dp=+
⎟ ⎠
⎞⎜⎝
⎛ 2
2
1exp xintegral factor
222
2
1
2
1
2
1 x x x
axe xpeedx
dp=+
22
2
1
2
1
)(
x x
axe pedx
d =
21exp2
ax C x dx⎛ ⎞= + −⎜ ⎟⎝ ⎠∫
Example
dx
dy xC a p =−+= )
2
1exp( 2
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Diff_Eq_4_2011 15
They are solved by differentiation followed by the p
substitution.
When the p substitution is made in this case, the second
derivative of y is replaced as
Let
dx
dy p =
dy
dp p
dx
dy
dy
dp
dx
dp
dx
yd ===∴
2
2
Non-linear 2nd O.D.E.
B. Equations where the independent variables doesnot occur explicitly
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Diff_Eq_4_2011 16
Solve 2
2
2
)(1dx
dy
dx
yd y =+
Let dx
dy p = and therefore
21 p
dy
dp yp =+
Separating the variables
dy
dp
pdx
yd
=2
2
dy y
dp p
p 1
12=
−
)1ln(2
1lnln 2 −=+ pa y
)1( 22 +== yadx
dy p
( )
)sinh(1
)(sinh1
)1(
1
22
caxa
y
baya
x
ya
dy x
+=
+=
+=
−
∫
Example
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S.l.dr.ing.mat. Alina
Bogoi
Differential
Equations
POLITEHNICA University of BucharestFaculty of Aerospace Engineering
CHAPTER 3
Higher Order Linear
Equations
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Diff_Eq_4_2011 18
Outline• Nth-order Linear Homogeneous
Equations with ConstantCoefficients
• Nth-order Linear Nonhomogeneous
Equations:
• Method of Undetermined
Coefficients• The method of variation of
parameters
• Cauchy-Euler Equation
Higher Order Linear Equations
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Diff_Eq_4_2011 19
A set of f 1
( x), f 2
( x), …, f n
( x) is linearly dependent on
an interval I , if there exists constants c1 , c2 , …, cn , not all zero, such that c1
f 1
( x) +
c2
f 2
( x) + …
+
cn
f n
( x) = 0
If not linearly dependent, it is linearly independent.
DEFINITION 1
Linear Dependence and Linear Independence
Higher Order Linear Equations
Homogeneous Equations with Constant Coefficients
Higher Order Linear Equations
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Suppose each of the functions f 1 ( x), f
2 ( x), …, f
n ( x)
possesses at least n – 1 derivatives. The determinant
is called the Wronskian of the functions.
DEFINITION 2Wronskian
)1()1()1(
21
21
1
21
'''),...,(
−−−
=
nn
nn
n
n
n
f f f
f f f
f f f
f f W
L
MMM
L
L
Higher Order Linear EquationsHomogeneous Equations with Constant Coefficients
Higher Order Linear Equations
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Diff_Eq_4_2011 21
Let y1
( x), y2
( x), …, yn
( x) on an interval I . This set of
functions is linearly independent if and only if
W ( y1
, y2
, …, yn
) ≠ 0 for every x in the interval.
THEOREM 3Criterion for Linear Independence
Any set y1
( x), y2
( x), …, yn
( x) of n linearly independent
solutions is said to be a fundamental set of functions.
DEFINITION 3
Fundamental Set
Higher Order Linear EquationsHomogeneous Equations with Constant Coefficients
Higher Order Linear Equations
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Diff_Eq_4_2011 22
• Ini t ial-value Problem An nth-order initial problem is
Solve:
• Subject to n initial conditions :
1
1 1 01[ ] ( ) ( ) ( ) ( ) ( )
n n
n nn n
d y d dy L y a x a x a x a x y g x
dx dx dx
−
− −= + + + + =L
0 0
'
0 0
( 1) 1
0 0
( )( ) , ,
( )
n n
y x y y x y
y x y
− −
=′ =
=
L
Higher Order Linear Equations
Higher Order Linear Equations
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Diff_Eq_4_2011 23
• The following DE
is said to be homogeneous;
• with g(x) not identically zero, isnonhomogeneous .
( )1
1 1 01( ) ( ) ( ) ( ) 0
n n
n nn n
d y d y dy L y a x a x a x a x y
dx dx dx
−
− −= + + + + =L
( )1
1 1 01( ) ( ) ( ) ( ) ( )
n n
n nn nd y d y dy L y a x a x a x a x y g xdx dx dx
−
− −= + + + + =L
Higher Order Linear Equations
L( y ) = g( x )
L( y ) = 0
Higher Order Linear Equations
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Diff_Eq_4_2011 24
1
1 1 01[ ] ( ) ( ) ( ) ( ) ( )
n n
n nn nd y d dy L y a x a x a x a x y g xdx dx dx
−
− −= + + + + =L
10)1(
1000 )(,,)(,)( −− ==′= n
n y x y y x y y x y L
Higher Order Linear Equations
Let an
( x), an-1
( x), …, a0
( x), and g ( x) be continuous on I,
an
( x) ≠
0
for all x on I . If x = x0
is any point in this
interval, then a solution y( x) exists on the intervaland is unique.
THEOREM 1Existence and Uniqueness
Higher Order Linear Equations
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Diff_Eq_4_2011 25
There exists a fundamental set of solutions for DE on an
interval I.
THEOREMExistence of a Fundamental Set
Higher Order Linear EquationsHomogeneous Equations with Constant Coefficients
(A) y = cy1
is also a solution if y1
is a solution.
(B) A homogeneous linear DE always possesses thetrivial solution y = 0.
COROLLARY Corollary
Higher Order Linear Equations
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Diff_Eq_4_2011 26
Let y1
, y2
, …, yk
be a solutions of the homogeneous
nth-order differential equation on an interval I .
Then the linear combination y = c1
y1
( x) + c2
y2
( x) + …+ ck
yk
( x)
where the ci , i = 1, 2, …, k are arbitrary constants, isalso a solution on the interval.
THEOREM
Superposition Principles – Homogeneous Equations
Higher Order Linear EquationsHomogeneous Equations with Constant Coefficients
Higher Order Linear Equations
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Diff_Eq_4_2011 27
Let y1
( x),
y2
( x), …,
yn
( x) be a fundamental set of
solutions of homogeneous DE on an interval I . Thenthe general solution is y
= c1
y1
( x) +
c2
y2
( x) + …
+
cn
yn
( x)
where ci
are arbitrary constants.
THEOREM
General Solution – Homogeneous Equations
Higher Order Linear EquationsHomogeneous Equations with Constant Coefficients
E l
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Diff_Eq_4_2011 28
• The functions y1 = e3x, y2 = e-3x aresolutions of
y″ – 9y = 0 on (-∞
,∞
)Now
for every x.So y = c1e3x + c2e
-3x is the general
solution.
Example
0633),( 33
3333
≠−=−= −
−−
x x
x x x x
ee
ee
eeW
E l
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Diff_Eq_4_2011 29
• The functions y1 = ex, y2 = e2x , y3 = e3x
are solutions of y″′ – 6y″ + 11y′ – 6y = 0
on (-∞
,∞
).Since
for every real value of x.
So y = c1ex + c2 e2x + c3e
3x is the general
solution on (-∞, ∞).
Example
02
94
32),,( 6
32
32
32
32 ≠== x
x x x
x x x
x x x
x x x e
eee
eee
eee
eeeW
Higher Order Linear Equations
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Diff_Eq_4_2011 30
• If y1,…, yn are solutions to ODE, then so
is linear combination
• Every solution can be expressed in thisform, with coefficients determined by
initial conditions, iff we can solve:
1 1 2 2( ) ( ) ( ) ( )n n y x c y x c y x c y x= + + +L
1 1 0 0 0
1 1 0 0 0
( 1) ( 1) ( 1)
1 1 0 0 0
( ) ( )( ) ( )
( ) ( )
n n
n n
n n n
n n
c y x c y x yc y x c y x y
c y x c y x y− − −
+ + =′ ′ ′+ + =
+ + =
LL
M
L
Homogeneous Equations with Constant Coefficients
Higher Order Linear Equations
Higher Order Linear Equations
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Diff_Eq_4_2011 31
( )( )
1 0 2 0 0
1 0 2 0 0
1 2 0
( 1) ( 1) ( 1)
1 0 2 0 0
( ) ( ) ( )
( ) ( ) ( )
, , ,
( ) ( ) ( )
n
n
n
n n n
n
y x y x y x
y x y x y x
W y y y x
x y x y x− − −
′ ′ ′
=
L
L
K M M O M
L
Higher Order Linear Equations
Homogeneous Equations with Constant Coefficients
• The system of equations on the previous slide has aunique solution iff its determinant, or Wronskian, is
nonzero at x0:
COROLLARYCorollar
Higher Order Linear Equations
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Diff_Eq_4_2011 32
•
Solution is:
• The Wronskian of y1 and y2 is
• Since W ≠ 0 for all t, linear combinationsof y1 and y2 can be used to construct
solutions of the IVP for any initial value
t .
1 2
1 2
,
( )
t t
t t
y e y e
y t c e c e
−
−= =
= +
22 0
2121
21
21 −=−=−−=′−′=
′′
= −− eeeee y y y y
y y
y yW t t t t
( ) ( )
0,
0 3, 0 1
y y
y y
′′ − =
′= =EXAMPLE
SOLUTION
g e O de ea quat o s
SOLUTIONS TO HOMOGENEOUS
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Diff_Eq_4_2011 33
SOLUTIONS TO HOMOGENEOUS
LINEAR EQUATIONS WITHCONSTANT COEFFICIENTS
All solutions of the differential equation
are, or are constructed from, exponentialfunctions of the form
y = ert .
NOTE: an
, an−1
, . . . , a1
, a0
are constants.
1
1 1 010
n n
n nn n
d y d y dya a a a y
dt dt dt
−
− −+ + + + =L
Higher Order Linear Equations
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Diff_Eq_4_2011 34
• Consider the nth order linear homogeneous differentialequation with constant, real coefficients:
• By the fundamental theorem of algebra, a polynomial of degree n has n roots r1, r2, …, rn, and hence
[ ] 01
)1(
1
)(
0 =+′+++= −− ya ya ya ya y L nn
nn L
0)( polynomialsticcharacteri
1
1
10 =++++= −
−
4 4 4 4 4 34 4 4 4 4 21 Lr Z
nn
nnrt rt
ar ar ar aee L
)())(()( 210 nr r r r r r ar Z −−−= L
g q
Homogeneous Equations with Constant Coefficients
Higher Order Linear Equations
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Diff_Eq_4_2011 35
• If roots of Z(r) are real and unequal, then
there are n distinct solutions:
• Then general solution of differential
equation is
• The Wronskian can be used to determine
linear independence of solutions.
t r t r t r neee ,,, 21 K
t r
n
t r t r nececect y +++= K21
21
)(
g q
ADistinct Roots
Higher Order Linear Equations
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Diff_Eq_4_2011 36
• Assuming exponential solution leads to
characteristic equation:
• Thus the general solution is
( )( )( )( ) 04321
02414132)( 234
=+−+−⇔
=+−−+⇒=
r r r r
r r r r et y rt
2 3 4
1 2 3 4( ) t t t t y t c e c e c e c e− −= + + +
1)0(,0)0(,1)0(,1)0(
02414132)4(
−=′′′=′′−=′=
=+′−′′−′′′+
y y y y
y y y y y
g q
EXAMPLE
SOLUTION
•Consider the initial value problem
Higher Order Linear Equations
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Diff_Eq_4_2011 37
• The initial conditions
yield
• Solving,
• Hence
164278
01694
1432
1
4321
4321
4321
4321
−=−+−
=+++
−=−+−
=+++
cccc
cccc
cccc
cccc
1)0(,0)0(,1)0(,1)0( −=′′′=′′−=′= y y y y
t t t t
ecececect y4
3
3
3
2
21)(−−
+++=
7
1,70
11,5
4,2
14321 −=−=== cccc
t t t t eeeet y 432
71
7011
54
21)( −− −−+=
g q
SOLUTION
(Cont’d)
Higher Order Linear Equations
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Diff_Eq_4_2011 38
Complex Roots• If Z(r) has complex roots, then they must
occur in conjugate pairs, λ ±
iμ .
• Solutions corresponding to complex roots have
the form
• We use the real-valued solutions
( )
( ) t iet ee
t iet eet t t i
t t t i
μ μ
μ μ λ λ μ λ
λ λ μ λ
sincos
sincos
−=+=
−
+
t et e
t t
μ μ
λ λ
sin,cos
B
Higher Order Linear Equations
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Diff_Eq_4_2011 39
• Consider the equation
• Then
• Now
• Thus the general solution is
( )( ) 01101)(3
=++−⇔=−⇒= r r r r et yrt
0=−′′′ y y
2
1,2
1 1 4 1 3 1 31 0
2 2 2 2
ir r r i
− ± − − ±+ + = ⇔ = = = − ±
2/3sin2/3cos)( 2/3
2/21 t ect ecect y t t t −− ++=
EXAMPLE
SOLUTION
Higher Order Linear Equations
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Diff_Eq_4_2011 40
• Consider the initial value problem
• The roots are 1, -1, i, -i. Thus the generalsolution is
• Using the initial conditions, we obtain
• The graph of solution is given on right.
( )( ) 01101)( 224 =+−⇔=−⇒= r r r et y rt
2)0(,2/5)0(,4)0(,2/7)0(,0)4( −=′′′=′′−=′==− y y y y y y
( ) ( )t ct cecect y t t sincos)(4321
+++= −
( ) ( )t t eet y t t sincos2
130)( −++= −
EXAMPLE
SOLUTION
Higher Order Linear Equations
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Repeated Roots• Suppose a real root rk of characteristic polynomial Z(r) is
a repeated root with multiplicty s.
• Then linearly independent solutions corresponding to this
repeated root have the form
t r st r t r t r k k k k
et et tee
12
,,,,
−
K
C
Higher Order Linear EquationsHomogeneous Equations with Constant Coefficients
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Repeated Roots• If a complex root λ + iμ is repeated s times, then so is its
conjugate λ - iμ .
• There are 2s corresponding linearly independent solns,
derived from real and imaginary parts of
or
1 1
cos , sin ,cos , sin , ,
cos , sin ,k k
t t
t t
r t r t s s t
e t e t te t te t
t e t t e e t
λ λ
λ λ
λ
μ μ μ μ
μ μ
− −
K
( ) ( ) ( ) ( )t iu st iut iut iu
et et tee+−+++ λ λ λ λ 12
,,,, K
C
Homogeneous Equations with Constant Coefficients
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Example 4: Repeated Roots
• Consider the equation
• Then
• The roots are 2i, 2i, -2i, -2i.
• Thus the general solution is
( )( ) 0440168)( 224 =++⇔=++⇒= r r r r et y rt
0168)4( =+′′+ y y y
( ) ( )t t ct t ct ct ct y 2sin2cos2sin2cos)( 4321 +++=
EXAMPLE
SOLUTION
Second Order Differential Equations
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In general, it is not easy to discover particular
solutions to a second-order linear equation.
But it is always possible to do so if the
coefficient functions P , Q and R are constant
functions, that is, if the differential equation hasthe form:
0)()()(2
2
=++ y x R
dx
dy xQ
dx
yd x P
0ay by cy′′ ′+ + =
SOLUTIONS TO THE
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SOLUTIONS TO THE
HOMOGENEOUS LINEAR 2ND
-ORDER DE
The solutions to the homogeneous linear second-
order differential equation depend on the solutionsto the characteristic equation. The solutions to the
characteristic equation fall into three cases:• Distinct (unequal) real roots
• Repeated (equal) real roots
•
Conjugate complex roots
THE CHARACTERISTIC
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THE CHARACTERISTIC
EQUATION
Substituting y = emx
results in the auxiliary
equation or
characteristic equation:
am2
+ bm + c = 0.
From eq am2 + bm + c = 0 the two
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From eq. am2 + bm + c = 0 the two
roots are
(1)b2 – 4ac > 0: two distinct realnumbers.
(2)b
2
– 4ac = 0: two equal realnumbers.
(3)b2 – 4ac < 0: two conjugate complex
numbers.
21 ( 4 ) / 2r b b ac a= − + −
2
2 ( 4 ) / 2r b b ac a= − − −
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If the characteristic equation has two distinct
(unequal) real roots m1
and m2
, the two solutionsto the DE are .
This results in the general solution:
DISTINCT REAL ROOTS
xm xm e ye y 21
21 and ==
xm xm ecec y 21
21 +=
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When the roots of the characteristic equation are
equal (that is, m1
= m2
), then the two solutions are
The general solution is:
REPEATED REAL ROOTS
.and 11
21
xm xm xe ye y ==
.11
21
xm xm
xecec y +=
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CONJUGATE COMPLEX ROOTSIf the roots of the characteristic equation are
conjugate complex roots
m1
= α
+ i β
and m1
= α
−
i β ,
then the two solutions are
y1 = e
α x
cos β x and y2 = e
α x
sin β x,and the general solution is
( ) ( ) ( )α α β β = +1 2e sin e cosx x
y x C x C x
Second Order Differential Equations
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Example 0y y′′ − = CE2 1 0r − = 1 or 1.r r⇒ = = −
1 2e ex xy C C −= +Solution
Example 2 0y y y′′ ′− + = CE2 2 1 0r r− + = 1 (double root).r⇒ =
1 2e ex xy C C x= +Solution
Example 2 5 0y y y′′ ′+ + = CE2 2 5 0r r+ + =
1 2 1 1 2r i⇒ = − ± − = − ±
( ) ( )1 2e sin 2 e cos 2x xy C x C x− −= +
Solution
I i i l l d b d l bl
Second Order Differential Equations
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An initial-value problem for the second-order Equation
consists of finding a solution
y
of the differential equation
that also satisfies initial conditions of the form
where and are given constants. If P, Q, R, and G arecontinuous on an interval and there, then a theorem found in more advanced books guarantees the
existence and uniqueness of a solution to this initial-value problem.
1000 )()( y x y y x y =′=
0 y 1 y0)( ≠ x P
Initial-value and boundary-value problems
Second Order Differential Equations
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A boundary-value problem for Equation 1 consists of finding a solution of the differential equation that also
satisfies boundary conditions of the form
In contrast with the situation for initial-value problems, a
boundary-value problem does not always have asolution.
1100 )()( y x y y x y ==
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• Solve the following DEs:
(a)
(b)
(c)
Example 1
03'5"2 =−− y y y x x ecec y 3
2
2
1 += −
025'10" =+− y y y
x x xecec y 52
51 +=
07'4" =++ y y y
2 1 2( cos 3 sin 3 ) x y e c x c x−
= +
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Example 2Solve
Solut ion:
2)0(',1)0(,017'4"4 =−==++ y y y y y
)2sin432cos(2/ x xe y x+−= −
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