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Linear ODEs with Constant Coefficients

The document discusses homogeneous and non-homogeneous linear ordinary differential equations with constant coefficients. It provides the general solutions for different cases based on the nature of the roots of the auxiliary equation. It also introduces the method of undetermined coefficients for finding particular solutions to non-homogeneous equations.
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0% found this document useful (0 votes)
43 views5 pages

Linear ODEs with Constant Coefficients

The document discusses homogeneous and non-homogeneous linear ordinary differential equations with constant coefficients. It provides the general solutions for different cases based on the nature of the roots of the auxiliary equation. It also introduces the method of undetermined coefficients for finding particular solutions to non-homogeneous equations.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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DE 730 MW

Homogeneous Linear ODE with Constant Coefficients

Any homogeneous linear ordinary differential equation with constant coefficients,


dn y dn1y dy
ao n  a1 n1  ...  an1  an y  0
dx dx dx

may be written in the form of f(D) = 0, where f(D) is a linear differential operator.

Based on the given differential equation, an auxiliary equation is obtained, that is, f(m) = 0, where
m is any root of f(m) = 0.

Then the general solution is determined based on the type of roots obtained from the auxiliary
equation.

Nature of Roots Roots General Solution


1. Distinct roots m = a, -b
y  c1eax  c 2ebx

2. Equal roots m = a, a, a
y  c1eax  c 2 xeax  c 3 x 2eax

3. Imaginary roots m = a + bi, a – bi


y  c1eax cosbx  c 2eax sinbx

Discriminant

is the most important part of quadratic functions when we talk about the nature of its roots. Given
a quadratic equation: y  ax2  bx  c , discriminant (D) can be defined mathematically as
D  b2  4ac .

Discriminant Nature of Roots

D=0 Equal Roots

D>0 Real and Distinct Roots

D<0 Imaginary Roots

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DE 730 MW
Exercises

1. (D2 + 2D – 3)y = 0

2. (D2 + D – 6)y = 0

3. (D3 + 3D2 – 4D)y = 0

4. (4D3 – 7D + 3)y = 0
d3 y d2 y dy
5. 3
 2 2 0
dx dx dx

6. (D2 – 6D + 9)y = 0

7. (4D3 + 4D2 + D)y = 0

8. (D4 + 6D3 + 9D2)y = 0

9. (D5 – D3)y = 0

10. (D2 – 2D + 5)y = 0

11. (D2 + 6D +13)y = 0

12. (D4 + 2D3 + 10D2)y = 0

d2 y
 2

D  2D  3 y  0
dx 2
2
dy
dx
 3y  0 y " 2y ' 3y  0

d d2
D D2  2
dx dx
dy d2 y
Dy  D2 y  2
dx dx

D 2

 2D  3 y  0 HLDE
m2  2m  3  0 auxiliary eqn
m  3 m  1  0
m  3, m  1 roots
3x
y  c1e  c 2e x
G.S.

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DE 730 MW
Nonhomogeneous Linear ODE with Constant Coefficients

The Method of Undetermined Coefficients


If the nonhomogeneous term d( x) in the general second‐order nonhomogeneous differential
equation

a  x  y" b  x  y' c  x  y  d  x 

is of a certain special type, then the method of undetermined coefficients can be used to obtain a
particular solution

Example
Find a complete solution of the differential equation

y " 3y ' 10y  5x 2

the family of d = 5 x 2 is { x2, x, 1}; therefore, the most general linear combination of the functions
in the family is y = Ax 2 + Bx + C (where A, B, and C are the undetermined coefficients).
Substituting this into the given differential equation gives

yh  general solution of the homogeneous equation


Auxiliary Equation
m2  3m  10  0
m  5 m  2   0
m  5, 2
yh  c1e5x  c 2e2x

y  Ax 2  Bx  C
y '  2Ax  B
y "  2A


2A  3  2Ax  B   10 Ax 2  Bx  C  5x 2 
1
x2 :  10A  5 A
2
3
x: 6A  10B  0 B
10
19
c: 2A  3B  10C  0 C
100

1 3 19
yp   x 2  x
2 10 100

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DE 730 MW

y  complete solution of nonhomogeneous linear differential equation


y  yh  yp
1 2 3 19
y  c1e5x  c 2e2x  x  x
2 10 100

Exercises

Find the complete solution of the following differential equation

1 7x
1. y " 6y ' 25y  8e 7x y  e 3x  c1 cos 4x  c 2 sin 4x   e
4

2. y " y ' 6y  10e3x y  c1e3x  c 2e 2x  2xe3x

1 1
3. y " y ' 6y  e x  12x y  c 1 e 3 x  c 2 e 2 x  e x –2 x 
6 3

4. y' 8y  16; y 0  5 y  5e8x  2

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DE 730 MW

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