5 Linear Differential Equation of Higher Order

5.1 Linear Dependence and Linear Independence of Functions

Def. Given f_i 2 C^{n 1}(I), i = 1; 2; ; n. If a linear relation Xn

i=1

c_if_i = 0

holds only for c_i = 0; 8i & 8x 2 I^{, then} f_i; i = 1; 2; n are said "linearly independent" in I: Otherwise, f₁;f₂; f_n are said

"linearly dependent" in I:

Theorem 6 Given f_i 2 C^{n 1}(I), i = 1; 2; ; n:8x 2 I^{. Then} f₁; f₂; f_n are linearly indep. iff.

f₁ ; f₂ ; : : : f_n f₁⁰ ; f₂⁰ ; : : : f_n⁰ ... ... ... ...

f₁^{[n 1]} ; f₂^{[n 1]} ; f_n^{[n 1]}

6= 0 pf.

Xn i=1

c_if_i = 0 (1)

Differential Equations 62

Differential Equations 63 5.2 De nition and Properties of Linear D. E. of n th Order Def.

P₀(x)y^[n] + P₁(x)y^{[n 1]} + + P_n(x)y = f (x) (1) s. t.

P_i; f 2 C⁰; i = 0; 1; n; P₀(x) 6= 0:

Then (1) is called a linear D. E. of n th order.

(1) can also be written as

y^[n] + eP₁(x)y^{[n 1]} + + fP_n(x)y = Q(x) (2) Consider a linear transformation

T ( ₁y₁ + + _ny_n) =

Xn i=1

iT (y_i)

A linear operator L is also a linear transformation. We may de ne L[y] =

Xn i=0

P_i(x)y^{[n i]}; then (1) becomes L[y] = f (x) (1)⁰.

Def. L[y] = 0 is called the homogeneous D. E. of L[y] = f (x).

Properties :

(i) Let y_i; i = 1; r be P. S. of L[y] = 0. Then y = Pr

i=1 c_iy_i is also a particular solution of L[y] = 0.

Differential Equations 64

Differential Equations 65 n solutions that are L. I.

Explanation :

Since L [ y ] = 0 is of order n;

y =

Xn i=1

c_iy_i

is the G. S. of L[y] = 0: And clearly, y₁; y_n are L. I. P. S.s of L[y] = 0.

Properties :

(i) Let y₁; y_n be n L. I. P. S.s of L[y] = 0. Then y = Pn

i=1 c_iy_i is the G. S. of L[y] = 0.

(ii) Let y₁; y_n be n L. I. P. S. of L[y] = 0 and y^ be a P. S.

of L[y] = f (x). Then

Xn i=1

c_iy_i + yb is the G. S. of L[y] = f (x).

Figure.

Ex. y⁰⁰ y = x² + 2 (1)

Differential Equations 66 Solution: Let

L[y] = y⁰⁰ y;

and f (x) = x² + 2;

where e^x and e ^x are 2 P. S.s of L[y] = 0 & ^y = x² is a P. S. of L[y] = f (x).

) c¹e^x + c₂e ^x : G. S. of L[y] = 0;

and the G. S. of L[y] = f (x) is

c₁e^x + c₂e ^x + x²: # (iii) Let y₁; y₂ be 2 P. S. of L[y] = f (x).

Then y₁ y₂ is a P. S. of L[y] = 0. pf.

* L[y¹] = f (x); L[y₂] = f (x);

) L[y¹ y₂] = L[y₁] L[y₂] = f (x) f (x);

) L[y¹ y₂] = 0:

i. e. y₁ y₂ : P. S. of L[y] = 0. #

(iv) Let y₁; y_n be n linearly independent P. S.s of an n th order linear D. E. L[y] = 0.

If w(y₁; y_n)(x₀) 6= 0; ^{for some} x₀ 2 I; ^then w(y₁; y_n)(x) 6= 0; 8x 2 I; ^where

Differential Equations 67

Differential Equations 68

Differential Equations 69 Let

g(x) =

Z x x₀

(p₁

p₀)dx; then w(x) = w(x₀)e^g(x):

* w(x⁰) 6= 0 & e^g(x) 6= 0; 8x 2 I;

) w(x) 6= 0; 8x 2 I: #

Differential Equations 70 5.3 Linear D. E. of n-th Order with Constant Coef cients 5.3.1 D-Operator

L[y] =

Xn i=0

a_iy^{[n i]} = f (x) (1) where a_i : constants, i = 0; 1; 2; n.

(1) can also be written as L[y] =

Xn i=0

a_iD^{n i}y = f (x) (1)⁰ where

Xn i=0

a_iD^{n i} = F (D):

Hence

(1)⁰ ) F (D)y = f(x) (2):

The polynomial F of D has the following properties:

i)

(aD^r + bD^s)y = (bD^s + aD^r)y :commutative of +

ii)

(aD^r) (bD^s)y = (bD^s) (aD^r)y :commutative of

Differential Equations 71 iii)

[(aD^r + bD^s) + cD^t]y = [aD^r + (bD^s + cD^t)]y :associative of +

iv)

[(aD^r bD^s) cD^t]y = [aD^r (bD^s cD^t)]y :associative of

v)

[aD^r (bD^s + cD^t)]y = [(aD^r) (bD^s)]y

+[(aD^r) (cD^t)]y :distributive law.

Formulas:

(i)

F (D)e^ax = F (a)e^ax pf.

* F(D)e^ax =

Xn i=0

a_i(a^{n i})e^ax

= e^ax

Xn i=0

a_ia^{n i} = e^axF (a):

Note:

Dⁱe^kx = kⁱe^kx:

Differential Equations 72

Differential Equations 73 Ex.

(D⁴ 2D² + 1) cos 2x =?

Solution:

((D²)² 2D² + 1) cos 2x

= (( 4)² 2( 4) + 1) cos 2x

= 25 cos 2x: # (iii)

F (D²) cosh ax

sinh ax = F (a²) cosh ax sinh ax : Pf. *

cosh ax = 1

2(e^ax + e ^ax);

sinh ax = 1

2(e^ax e ^ax);

) F(D²) cosh ax

sinh ax = F (a²) cosh ax sinh ax : (iv)

F (D)(e^ax (x)) = e^axF (D + a) (x)

Differential Equations 74

Differential Equations 75 Ex.

(D 1)(D + 2)(e^xx³) =?

Solution: ( In fact, (_dx^d2₂ + _dx^d 2)(e^xx³))

* F(D) = (D 1)(D + 2)

) F(D)(e^xx³) = e^xF (D + 1)x³

= e^x(D + 1 1)(D + 1 + 2)x³

= e^xD(D + 3)x³ = e^x(6x + 9x²): # (v)

(D a)^re^ax (x) = e^axD^r (x):

In particular, (x) = x^k; k = 0; 1; 2; 3; ; then we have the following conditions:

(D a)^r(e^axx^k) = (e^axD^rx^k)

=

( 0; if r > k

e^axk(k 1) (k r + 1)x^{k r}; if k r:

i. e.

(D a)^r(e^axx^k) = 0; k = 0; 1; 2; r 1 Hence for

(D a)^ry = 0; (1) we obtain

e^ax; xe^ax; x²e^ax; ; x^{r 1}e^ax

Differential Equations 76 being L. I. , and

y = C₁e^ax + C₂xe^ax + + C_rx^{r 1}e^ax

:a G. S. of (1), where C₁; C₂; C₃; C_r are arbitrary constants.

Differential Equations 77 5.3.2 G. S. of a D. E. F (D)y = 0:

Let F & G be two polynomials of operators, and let (x) be a solution of G(D)y = 0. Then (x) is also a solution of F (D)G(D)y = 0:

(* F(D)G(D)y = F(D)(G(D)y) = F(D)0 = 0^.)

Ex. Note that e^x is a P. S. of (D 1)y = 0, then e^x is also a P. S. of

(D + 2)(D 1)y = 0:

Discussion: Consider (D a)^ry = 0 has the G. S.

y_G = (C₁ + C₂x + + C_rx^{r 1})e^ax: Let

F (D) = a₀(D ₁)^r1 (D _k)^rk; where r₁ + r₂ + + r_k = n; and for

(D ₁)^r1y = 0; y = (C₁ + C₂x + + C_r1x^{r1 1})e ^1x; (D ₂)^r2y = 0; y = (C₁ + C₂x + + C_r2x^{r2 1})e ^2x;

... ...

(D _k)^rky = 0; y = (fC₁ + fC₂x + + fC_rkx^{rk 1})e ^kx: i. e.

(C₁ + C₂x + + C_r1x^{r1 1})e ^1x +

+(fC₁ + fC₂x + + fC_rkx^{rk 1})e ^kx is the G. S. of F (D)y = 0: #

Differential Equations 78 Ex.

(D 1)²(D + 1)y = 0:

Solve y:

Solution:

i) (D + 1)y = 0 has G. S.

y = C₁e ^x;

or y⁰ + y = 0, a 1st order linear D. E. with solution

y = e ^x

Z x

0 e^Rudtdu + c = ce ^x: ii) (D 1)²y = 0 has G. S.

y = (C₂ + C₃x)e^x So combine i) and ii),

y = C₁e ^x + (C₂ + C₃x)e^x : G. S. of y: # Ex.

y^[4] y = 0:

Solve y(x). Solution:

i. e.

Differential Equations 79 Then

F (D) = (D 1)(D + 1)(D² + 1)

) (D 1)(D + 1)(D i)(D + i)y = 0:

) G: S: of y = C¹e^x + C₂e ^x + C₃e^ix + C₄e ^ix:

Since e^ix = cos x + i sin x

e ^ix = cos x i sin x ^{, we have}

y = C₁e^x + C₂e ^x + fC₃cos x + fC₄sin x:

Ex. (D⁶ 1)y = 0: Solve for y:

Solution:

(D⁶ 1) = (D³ 1)(D³ + 1)

= (D 1)(D + 1)(D² + D + 1)(D² D + 1):

) D⁶ 1 = 0 has roots 1; 1 p

3i

2 ; and 1 p 3i 2 : Hence the G. S.

Differential Equations 80

y = C₁e^x + C₂e ^x + C₃e ¹⁺

p3i

2 x + C₄e ¹

p3i

2 x

+C₅e¹⁺

p3i

2 x + C₆e¹

p3i

2 x

= C₁e^x + C₂e ^x + e ^2x(C₃⁰ cos(

p3

2 x) + C₄⁰ sin(

p3 2 x)) +e^x2(C₅⁰ cos(

p3

2 x) + C₆⁰ sin(

p3

2 x)): # Note: Given

F (D)y = f (x) ( ) The polynomial equation

F (u) = 0 is called the "auxillary equation" of ( ).

Differential Equations 81 5.3.3 Particular Solution of F (D)y = f (x) : Inverse

D Operator

If we can nd a P. S. y_p of

F (D)y = f (x) and solve the homogeneous D. E.

F (D)y = 0 to nd the G. S. y_c of F (D)y = 0. Then

y_c + y_p is the G. S. of F (D)y = f (x).

De nitions

i) De ne the P. S. of (A) by _{F (D)}¹ f (x).

ii) We regard 2 P. S.s y₁ & y₂ of (A) are equivalent if y₂ y₁ = y_p

with F (D)y_p = 0:

explanation:

F (D)y₁ = f (x) = F (D)y₂

) F (D)(y² y₁) = 0: # iii)

F (D) 1

F (D)f (x) = F (D) 1

F (D) f (x) = f (x); 8f 2 C¹:

Differential Equations 82 (i. e.F (D)_{F (D)}¹ = 1:)

iv)

a 1

F (D) + b 1

G(D) f (x) = a 1

F (D)f (x) + b 1

G(D)f (x):

v)

1

F (D) (G(D)f (x)) = 1

F (D)G(D) f (x):

Propositions (i)

F (D) 1

F (D) = 1 (By Def.)

(ii)

1

F (D)F (D) = 1:

pf. * f(x) is a P. S. of F (D)y = F (D)f (x); by Def., f (x) = 1

F (D)F (D)f (x):

i. e.

1

F (D)F (D) = 1: #

Differential Equations 83 (iii)

1

F (D)(af (x) + bg(x)) = a 1

F (D)f (x) + b 1

F (D)g(x):

pf.

1

F (D)f (x) is a P. S. of

F (D)y = f;

and

1

F (D)g(x) is a P. S. of

F (D)y = g(x):

Let

y₁ = 1

F (D)f (x)& y₂ = 1

F (D)g(x):

) F(D)(ay¹ + by₂) = aF (D)y₁ + bF (D)y₂

= af (x) + bg(x):

) 1

F (D)(af (x) + bg(x)) is a P. S. of

F (D)y = af (x) + bg(x);

Differential Equations 84

Differential Equations 85 pf. Let

y₁ = 1

G(D)f (x); y₂ = 1

F (D)f (x):

i. e.

G(D)y₁ = f (x); F (D)y₂ = f (x):

Let

y₃ = 1

F (D)G(D)f (x);

thenF (D)G(D)y₃ = f (x):

) F(D)G(D) 1

F (D)G(D)f (x) = f (x)

, G(D)F (D) 1

F (D)G(D)f (x) = f (x) , G(D) F (D) 1

F (D)

1

G(D)f (x) = f (x)

) G(D) 1

G(D)f (x) = f (x): # Ex.

1

(D a)^re^ax =?

Solution:

Differential Equations 86

Differential Equations 87

(iii) Hence we may do this again and again, and we obtain 1

Differential Equations 88

To solve the problem, consider

F (D) = D² + D 2 = (D + 2)(D 1):

Differential Equations 89

Differential Equations 90 Then (N) ^becomes

1 = X

By using the above notations, we nd that F (D)X

Differential Equations 91

Differential Equations 92

k is a constant.

pf. Let y₁ be the P. S. of F (D)y = f (x), i. e.

Differential Equations 93

Differential Equations 94

Differential Equations 95

Differential Equations 96

Differential Equations 97

Differential Equations 98 iii).

1

D² + D + 2 sin 2x = Im( 1

D² + D + 2e^2ix)

= Im( 1

4 + 2 + 2ie^2ix)

= 1

2Im e^2ix i 1

= 1

2Imi + 1

2 (cos 2x + i sin 2x)

= 1

4Im[(cos 2x sin 2x) +i(sin 2x + cos 2x)]

= 1

4(sin 2x + cos 2x):

Homework. Find

1

D³ + D² 1 cos 3x:

vi).

F ¹(D)(e^ax (x)) = e^axF ¹(D + a) (x):

pf. Consider the P. S. of

F (D)y = e^ax (x) being y_p. In particular, set

y_p = e^ax (x):

Differential Equations 99 Plug y_p into F (D)y = e^ax (x), we nd

F (D)(e^ax (x)) = e^axF (D + a) (x)

= e^ax (x):

) F (D + a) (x) = (x) Then (x) is a P. S. of F (D + a)y = (x); i. e.

(x) = 1

F (D + a) (x):

So

y_p = 1

F (D)(e^ax (x))

= e^ax (x)

= e^ax( 1

F (D + a) (x)):

Ex.

1

(D a)^re^ax =?

Solution: We have solved _{(D a)}¹ re^ax previously as ^x_r!^re^ax. Now

Differential Equations 100 consider the following:

1

(D a)^re^ax = 1

(D a)^re^ax 1

= e^ax 1 D^r 1

= e^ax x^r r!:

* _D^1r1 is a P. S. of D^ry = 1,

) y = Z Z

| {z }

r times

dx = x^r r!

(without arbitrary constant.) i. e.

1

D^r 1 = x^r r!

Homework. Find

1

D⁴ 163 cos 2x:

vii) Let f (x) be a polynomial of degree . For

F ¹(x) = a₀ + a₁x + + a_rx^r + x ⁺¹ (x) F (x) : Then

F ¹(D)f (x) = (a₀ + a₁D + + a D )f (x);

Differential Equations 101 and

(a₀ + a₁D + + a D )f (x) is a P. S. of F (D)y = f (x):

pf.

* F ¹(D) = a₀ + a₁D + + a_rD^r + D ⁺¹ (D) F (D) ) 1 = F(x) F ¹(x)

= F (x)(a₀ + a₁x + + a x ) + x ⁺¹ (x) and

I = F (D) F ¹(D)

= F (D)(a₀ + a₁D + + a D ) + D ⁺¹ (D);

so

f (x) = (F (D)F ¹(D))f (x)

= F (D)(a₀ + a₁D + + a D )f (x) + (D)D ⁺¹f (x):

Again,

* D ⁺¹f (x) = 0

) f(x) = F(D)(a⁰ + a₁D + + a D )f (x);

i. e.

(a₀ + a₁D + + a D )f (x)

Differential Equations 102 is a P. S. of F (D)y = f (x) ,

1

F (D)f (x) = (a₀ + a₁D + + a_rD^r)f (x):

Def. f (x) = O(g(x)) if

xlim!c

f (x)

g(x) = L; jLj < 1 and c 2 R [ f 1g: ^(Big-O.)

Def. f (x) = o(g(x)) if

xlim!c

f (x)

g(x) = 0;

and c 2 R [ f 1g: (Little-o.) Ex.

1

(D 1)²x²e^2x =?

Solution:

1

(D 1)²x²e^2x = e^2x 1

(D + 2 1)²x²

= e^2x 1

(D + 1)²x²:

Differential Equations 103

* 1

F (x) = 1

(x + 1)² = 2

0 x⁰ + 2

1 x¹ +

= 1 2x + 3x² + O(x³)

= (1 2x + 3x²) + x³ (x) (x + 1)²;

) 1

F (D)x² = 1

(D + 1)²x²

= (1 2D + 3D²)x² + (D)

(D + 1)²(D³x²)

= (1 2D + 3D²)x²

= x² 4x + 6:

) 1

(D 1)²(x²e^2x) = e^2x(x² 4x + 6):

Ex. Find the G. S. of

D²(D + 1)²y = x² (1):

Solution: Let y_c be the G. S. of D²(D + 1)²y = 0 ) y_c = C₁ + C₂x + (C₃ + C₄x)e ^x:

Let y_p be a P. S. of D²(D + 1)²y = x², i. e.

y_p = x²

D²(D + 1)²;

Differential Equations 104

Differential Equations 105

Differential Equations 106 and

y_G = y_c + y_p

= C₁ + C₂cos x + C₃ sin x + 1

2(e^x x sin x):

Note: The formulas for the inverse D operator are only applicable for D. E.s of

F (D)y = f (x)

with f (x) being e^ax; sin ax; cos ax; polynomial functions or their product.

Homework.:

1. Find the G. S. of

D(D 1)²y = e^x cos x + x²: 2. Solve the G. S. of

(D² 1)y = x²e^x + x sin x:

Differential Equations 107 5.4 Solutions of Linear D. E. by Variation of Parameters

Let

L[y] = f (x) (1) and

L[y] P₀y^[n] + + P_ny;

where P_i:functions of x; i = 0; 1; ; n: Let y_c be the G. S. of L[y] = 0 (2);

then

y_c =

Xn i=1

C_iy_i (3);

where C_i are arbitrary constants, i = 1; 2; ; n and y_i are L. I.

P. S. of L[y] = 0; i. e. W (y₁; ; y_n) 6= 0^{, or}

y₁ y₂ y_n

y₁⁰ y₂⁰ y_n⁰ ... ... ... ...

y₁^{[n 1]} y₂^{[n 1]} y_n^{[n 1]}

6= 0:

Now assume

y =

Xn i=1

C_i(x)y_i (4)

be the G. S. of (1), where C_i(x) are functions of x; i =

Differential Equations 108

Differential Equations 109

Differential Equations 110 ) ^{We have} n equations:

8>

Differential Equations 111 where k_i:arbitrary constants.

) y =

Differential Equations 112 Ex.

(D² + 1)y = sec x (1):

Please solve y(x).

solution Let y_c be the G. S. of (D² + 1)y = 0;

) y^c = C₁ cos x + C₂sin x Set

y_p = C₁(x) cos x + C₂(x) sin x : P. S. of

(D² + 1)y = sec x:

Then

= W (y₁; y₂)(x)

= y₁; y₂ y₁⁰; y₂⁰

= cos x; sin x sin x; cos x ;

1 = 0; sin x sec x; cos x ;

2 = cos x; 0

sin x; sec x ; and

(y₁; y₂)(x) = cos² x + sin²x = 1:

Differential Equations 113 ) C¹(x) =

Z x

1dx =

Z x

tan xdx = ln j cos xj;

C₂(x) =

Z x

2dx = x:

) y^p = ln j cos xj cos x + x sin x;

and

y = y_c + y_p

= (C₁ + ln j cos xj) cos x + (x + C²) sin x:

Homework.

xy⁰⁰ (x + 1)y⁰ + y = x² (1) Find the G. S. of (1) by variation of parameters.

Differential Equations 114 5.5 Cauchy-Euler Linear D. E.

Def. Linear D. E. of the form Xn

r=0

(ax + b)^{n r} _rD^{n r}y = f (x) (1) is called ”Euler Linear D. E.”

Def. When a=1,b=0,then (1) is called ”Cauchy Linear D. E.”

i: e:

Xn r=0

rx^{n r}D^{n r}y = f (x):

Note.a; b and _r are constants, r = 0; 1; 2; n:

Discussion:

In (1), let

ax + b = e^u; i. e.

x = e^u b a : By implicit differentiation,

a = e^udu dx ) du

dx = ae ^u = a ax + b;

Differential Equations 115

Differentiate (1)⁰ w.r.t. x;

d²y holds, we claim that

(ax + b)^k+1D^k+1y = a^k+1D(D 1) (D k)y

Differential Equations 116 is also true.

D[(ax + b)^kD^ky]

= D[a^kD (D k + 1)y]

= d

du[a^kD(D 1) (D k + 1)y]du dx

= a^k+1[D²(D 1) (D k + 1)y] 1 ax + b

* D[(ax + b)^kD^ky] = k(ax + b)^{k 1} aD^ky +(ax + b)^kD^k+1y ) (ax + b)^k+1D^k+1y

= a^k+1DD(D 1) (D k + 1)y ka(ax + b)^kD^ky

= a^k+1D²(D 1) (D k + 1)y

ka^k+1D (D k + 1)y

= a^k+1(D k)(D(D 1) (D k + 1)y)

= a^k+1D(D 1) (D k)y:

By mathematical induction, 8m 2 N;

(ax + b)^mD^my = a^mD (D k + 1)y )

Xn r=0

r(ax + b)^{n r}D^{n r}y = f (x) can be rewritten as

Differential Equations 117

(A)

Xn r=0

ra^{n r}D(D 1) [D (n r) + 1]y = f(e^u b a ):

Let

G(D) =

Xn 1 r=0

ra^{n r}D(D 1) [D (n r) + 1] + _n; (A) ) G(D)y = f(e^u b

a ) g(u) (B) i. e.

G(D)y = g(u); D d du: Ex.

(3x 1)y⁰⁰ + y⁰ = 1 + x (a) Please solve y(x).

Solution: (a) (3x 1) :

(3x 1)²y⁰⁰ + (3x 1)y⁰ = 3x² + 2x 1;

i. e.

a = 3; b = 1;

r = 1; r = 0; 1;

2 = 0:

Differential Equations 118 Hence (a) (3x 1) is an Euler linear D. E. of order 2: Let

(3x 1) = e^u; x = e^u + 1 3 ; and

D d

du: Then (a) (3x 1) :

X1 r=0

3^{2 r}D(D 1) (D (2 r) + 1)y + 0 y

= e^u(1 + e^u + 1 3 ):

) (9D² 9D + 3D)y = e^2u + 4e^u

3 ;

) D(D 2

3)y = e^2u + 4e^u

27 (b):

To solve (b), let

y = y_c + y_p: i).

y_c = C₁ + C₂e^23u:

Differential Equations 119 Solution: (a) is a Cauchy linear D. E. Let

x = e^u; (i. e. ln x = u; ) D d

Differential Equations 120 ) (a) ^becomes

x²( 1

x²D(D 1)y) + x(1

xDy) y = tan u:

) (D² + 1)y = tan u (b):

Let y = y_c + y_p; then:

i).

y_c = C₁ cos u + C₂sin u:

ii). Let y₁ = cos u; y₂ = sin u and

W (y₁; y₂) = cos u; sin u

sin u; cos u = 1;

1 = 0; sin u

tan u; cos u = sin² u cos u ;

2 = cos u; 0

sin u; tan u = sin u:

So

C₁(u) =

Z u sin² u cos u du

= sin u ln j sec u + tan uj + C¹; C₂(u) = cos u + C₂:

Differential Equations 121 i. e.

y_p = (sin u ln j sec u + tan uj) cos u +( cos u) sin u

= ( ln j sec u + tan uj) cos u:

Hence

y = (C₁ ln j sec u + tan uj) cos u + C²sin u

= (C₁ ln j sec(ln x) + tan(ln x)j) cos(ln x) +C₂ sin(ln x):

Homework. Find G. S. of the following:

(1).

(x + 2)²y⁰⁰ + (x + 2)y⁰ + y = 1 + x:

(2).

(2x 1)²y⁰⁰ + 2(2x 1)y⁰ 4y = sin(ln(2x 1)):

Differential Equations 122

Please nd its particular solution.

Differential Equations 123 Solution:

*

X3 i=0

P_i(x) = x + ( x 1) + 1 = 0;

we have exp x as one of its P. S.

(II).

L[x^k] = 0 , P⁰k(k 1) (k n + 1)x^{k n} + + kP_{n 1}x^{k 1} + P_nx^k = 0:

pf.

L[x^k] =

Xn i=0

P_i(x^k)^{[n i]}

=

Xn i=0

P_i[k(k 1) (k (n i) + 1)x^{k (n i)}]

= P₀k(k 1) (k n + 1)x^{k n} + +kP_{n 1}x^{k 1} + P_nx^k

= 0:

(?). In particular, when k = 1,if

P_{n 1} + xP_n = 0 i. e. x is a P. S. of L[y] = 0

Ex.

x²y⁰⁰ xy⁰ + y = 0 (1):

Differential Equations 124 Find its P. S.

Solution: P₀ = x²; P₁ = x; P₂ = 1:

* P^{n 1} + xP_n = P₁ + xP₂ = x x = 0;

) x is a P. S. of (1).

(III). When

aP_{n 1} + (ax + b)P_n = 0;

then (ax + b) is a P. S. of L[y] = 0:

Ex.

xy⁰⁰ (x + 1)y⁰ + y = 0 (1)

Solution: P₀ = x; P₁ = (x + 1); P₂ = 1: Consider aP₁ + (ax + b)P₂ = 0;

i. e.

a(x + 1) + (ax + b) 1 = 0;

, a + b = 0; ^{i. e.} a = b:

Then

ax + a = a(x + 1)

: a P. S. of (1), and hence (x + 1) is a P. S. of (1). We also know, from previous example, that e^x is also a P. S. of (1). Therefore

y₁ = e^x; y₂ = x + 1;

Differential Equations 125 then

W (y₁; y₂) = e^x; x + 1

e^x; 1 = xe^x 6= 0:

i. e. e^x and (x + 1) are 2 L. I. P. S.s.

Differential Equations 126 5.7 The Method of Undetermined Coef cients

The method of undetermined coef cients is used when we want to compute a P. S. of

F (D)y = f (x) (1);

where f (x) is a linear combination of e^ax, cos ax, sin ax, polynomial functions and their products.

Ex. (D² 2D 3)y = 2e^4x (1)

Solution: To nd y_p of (1), set y_p = Ce^4x: a P. S. of (1).

) yp⁰ = 4Ce^4x; y_p⁰⁰ = 16Ce^4x: Plug them into (1), we nd

(16 8 3)C = 2 ) C = 2 5: i. e.

y_p = 2 5e^4x : a P. S. of (1).

Ex.

(D² 2D 3)y = 2e^3x (2) Solution:

i). Let y_p = Ce^3x be a P. S. of (2): Substitute y_p; y_p⁰ & y_p⁰⁰ into (2), we have

(9C 6C 3C)e^3x = 2e^3x = 0 Ce^3x: (! )

Differential Equations 127 i. e. y_p = Ce^3x is not a P. S. of (2):

ii). e^3x and xe^3x are L. I. ,since W (e^3x; xe^3x) 6= 0: ^Let y_p = Cxe^3x be a P. S. of (2),

) C = 1 2 can be found. i. e. y_p = ^x₂e^3x: P. S. of (2).

Ex.

(D² 3D + 2)y = 2x² + e^x + 2xe^x + 4e^3x: Find its general solution.

Solution:

y_c = C₁e^x + C₂e^2x:

Consider 8

>>

>>

<

>>

>>

:

x² ! fx²; x; 1g;

e^x ! fe^xg(^or x);

xe^x ! fxe^x; e^xg(^or x);

e^3x ! fe^3xg:

Firstly, let

y_p = A₁x² + A₂x + A₃ + A₄e^x +A₅xe^x + A₆e^3x;

Differential Equations 128 and plug it back in. We nd that it doesn't hold. Secondly, let

y_p = A₁x² + A₂x + A₃ + A₄xe^x + A₅x²e^x +A₆e^3x

and we nd it ts.Finally,we nd y_p = x² + 3x + 7

2 + 2e^3x x²e^x 3xe^x: Ex. Find the proper P. S. from the D. E.

D(D + 1)y = x²e ^x + sin x (3):

Solution:

i).

y_c = C₁ + C₂e ^x:

ii). Consider any derivative of x²e ^x + sin x : ( sin x ! fsin x; cos xg;

x²e ^x ! fx²e ^x; xe ^x; e ^xg(^or x):

We then set

y_p = (A₁x³ + A₂x² + A₃x)e ^x +A₄sin x + A₅ cos x:

After substituting y_p; y_p⁰ & y_p⁰⁰into (3), we nd y_p = e ^x(x³

3 + x² + 2x) 1

2(sin x + cos x):

Differential Equations 129 Homework. Solve the D. E.s by using undetermined

coef cients:

(1).

D²(D + 1)y = x² (2).

y⁰⁰ + 3y⁰ + 2y = 4x (3).

y⁰⁰ + y = x cos x cos x (4).

y⁰⁰ 4y⁰ + 4y = x³e^2x + xe^2x (5).

y⁰⁰ + 4y = sin²x:

Differential Equations 130 5.8 Simultaneous Linear D. E.s with Constant Coef cients

Ex. (

d

dtx + _dt^d y + y x = e^2t (1)

d²

dt²x + _dt^d y = 3e^2t (2) De ne D _dt^d, (1) and (2) become

( (D 1)x + (D + 1)y = e^2t (A) D²x + Dy = 3e^2t (B)

)

( D((D 1)x + (D + 1)y) = De^2t = 2e^2t (D + 1)(D²x + Dy) = (D + 1)(3e^2t) = 9e^2t

) (D³ + D)x = 7e^2t (3)

* x(t) = x^c + x_p, where

( x_c = c₁ + c₂cos t + c₃ sin t x_p = _D3¹_+D(7e^2t) = ₁₀⁷ e^2t ) x(t) = c¹ + c₂ cos t + c₃sin t + 7

10e^2t (4):

Substitute (4) into (B):

Dy = 1

5e^2t + c₂cos t + c₃ sin t;

) y = 1

10e^2t + c₂sin t c₃ cos t + c⁰₁ (5):

Differential Equations 131 Substitute (4) and (5) into (A), we nd

c₁ + c⁰₁ = 0 , c¹ = c⁰₁ (6) Finally,

x(t) = c₁ + c₂ cos t + c₃sin t + 7 10e^2t and

y(t) = c₁ + c₂sin t c₃cos t + e^2t

10: #

Ex. Solve the simultaneous D. E.s given in (1) and (2).

ax + by = c a⁰x + b⁰y = c⁰

(D + 1)x (D 1)y = cos t (1);

(D² + D)x + Dy = e^t (2):

Solution:

a b

a⁰ b⁰ x = c b c⁰ b⁰ ;

) D + 1 (D 1)

(D² + D) D x = cos t (D 1)

e^t D :

) (D³ + D²)x = D cos t + (D 1)e^t = sin t ) x = x^c + x_p;

where

x_c = c₁ + c₂t + c₃e ^t;

Differential Equations 132 and

x_p = 1

D²(D + 1)( sin t) = 1 D + 1

1

D² sin t

= 1

D + 1 sin t = (D 1) 1

D² 1(sin t)

= (D 1)( 1

2 sin t) = 1

2(sin t cos t):

) x(t) = c¹ + c₂t + c₃e ^t + 1

2(sin t cos t) (3):

Plug (3) back into (2), we have

Dy = e^t (D² + D)x = e^t c₂ cos t;

) y(t) = e^t c₂t sin t + ce₃ (4) Again, plug (3) & (4) into (1), we obtain that

(5) c₂ + c₁ + ce₃ + c₂ = 0

, ec₃ = (c₁ + 2c₂): # Finally, we solve x & y as

x(t) = c₁ + c₂t + c₃e ^t + 1

2(sin t cos t);

y(t) = e^t c₂(t + 2) c₁ sin t:

Homework.

Differential Equations 133

(1) (D + 1)x (D 1)y = 2e^t t;

(D² + D)x + Dy = sin t:

(2) Dx + (D² + 1)y = e^t;

Dx (D² 3)y = 1 + 2t:

Differential Equations 134 5.9 Exact Linear D. E.s

Def. A linear D. E.

p₀y^[n] + + p_{n 1}y⁰ + p_ny = f (x) (1) , where p_i: functions of x; is called "exact" if

9 Q⁰y^{[n 1]} + Q₁y^{[n 2]} + + Q_{n 1}y

Differential Equations 135 Ex.

x²y⁰⁰ + (3x + 1)y⁰ + y = 2x (1) Solution: Consider

x²y⁰⁰ + (3x + 1)y⁰ + y = 0;

p₀ = x²; p₁ = (3x + 1); p₂ = 1; n = 2:

and

p₂ p⁰₁ + p⁰⁰₀ = 1 3 + 2 = 0;

i. e.

x²y⁰⁰ + (3x + 1)y⁰ + y = 2x is exact. Set

Q₀ = p₀ = x²;

Q₁ = p₁ Q⁰₀ = 3x + 1 2x = x + 1;

and then

(x²y⁰ + (x + 1)y)⁰ = x²y⁰⁰ + (3x + 1)y⁰ + y:

i. e.

(x²y⁰ + (x + 1)y)⁰ = 2x (2):

) x²y⁰ + (x + 1)y = x² + c₁ ) y⁰ + (1

x + 1

x²)y = 1 + c₁ x² ) y = e^{Rx(x1+x21 )dx}

Z x

(1 + c₁

t²)e^{Rt(1u+u21 )du}dt + c₂ :

Differential Equations 136 Ex. Solve the following D. E.

x²y⁰⁰⁰ + (3x² 1)y⁰⁰ + (7x + 3)y⁰ + y = 3x² + 1 (1):

Solution:

p₀ = x²; p₁ = 3x² 1; p₂ = 7x + 3; p₃ = 1:

) 1 7 + 6 0 = 0;

i. e. (1) is exact.

) Q⁰ = p₀ = x²; Q₁ = p₁ Q⁰₀ = 3x² 2x 1;

Q₂ = p₂ Q⁰₁ = 7x + 3 6x + 2 = x + 5;

) (x²y⁰⁰ + (3x² 2x 1)y⁰ + (x + 5)y)⁰ = 3x² + 1 (2);

) x²y⁰⁰ + (3x² 2x 1)y⁰ + (x + 5)y = x³ + x + c₁ (3):

Note that (3) is not exact, since

(x + 5) (6x 2) + 2 6= 0: \ (You may try to solve (3) by yourselves !)

Homework.

(1).

Differential Equations 137 (2).

(x 1)y⁰⁰ + (x + 1)y⁰ + y = 2x:

(3).

D²(D² + 1)y = cos x + x²: (4).

x³y⁰⁰⁰ + 2x²y⁰⁰ + xy⁰ y = 0:

(5).

y^[4] + 2y^[3] + 2y⁰⁰ = 3e^x + 2xe ^x + e ^x sin x:

(6).

y⁰⁰⁰ 3y⁰⁰ + 2y⁰ = x + e^x; and

y⁰(0) = 1; y⁰⁰(0) = 1

4; y⁰⁰⁰(0) = 3 2: (7).

y^[4] y⁰⁰⁰ y⁰⁰ + y⁰ = x² + 4 + x sin x:

Differential Equations 138

6 Particular Method for Solving 2nd

在文檔中 LECTURE NOTES OF DIFFERENTIAL EQUATIONS Nai-Sher Yeh (頁 61-138)