Differential Equations

ORDINARY DIFFERENTIAL EQUATIONS based on the course-contents of MMB-352 (A course of Mathematics (Main/Subsidiary)

prescribed for B. Sc. (Hons.) III Semester)

UNIT-2

Higher Order Differential Equations

Differential Equations

A linear differential equation of order n is an equation of the form a_n(x)dⁿy

dxⁿ +an−1(x)dⁿ⁻¹y

dxⁿ⁻¹ +· · ·+a₁(x)dy

dx +a₀(x)y =Q(x) (1) wherea_n, an−1,· · · , a₀ and Q(x) are continuous functions anda_n(x)6= 0.

If Q(x) 6≡ 0, then equation (1) is called the Non-homogeneous linear differential equations of order n.

IfQ(x)≡0, then equation (1) is called the homogeneous linear differential equations.

Solution of Homogeneous linear Differential equations with constant coefficients:- Consider the homogeneous linear differential equation of order n

an

dⁿy

dxⁿ +an−1

dⁿ⁻¹y

dxⁿ⁻¹ +· · ·+a1

dy

dx+a0y = 0 (2) wherean, an−1,· · · , a0 are all constants providedan6= 0.

Rewrite equation (2) in symbolic form as

P(D)y≡(a_nDⁿ+an−1Dⁿ⁻¹+· · ·+a₁D+a₀)y= 0, where D= d dx (3) Definition 1. An equation of order n in m is given by

a_nmⁿ+an−1mⁿ⁻¹+· · ·+a₁m+a₀ = 0; a_n6= 0 (4) is called auxiliary equation (A.E.) of differential equation (2).

1

Clearly, Auxiliary equation is obtained by P(m) = 0.

Notice that A.E. (4) is an algebraic equation in m of degreen. So, it will haven roots. There are the following three cases:

1. Case-IAll the roots are real and distinct.

2. Case-IIAll the roots are real but some of them are repeated.

3. Case-IIIRoots are imaginary.

Case-I:- If the roots are real and distinct; that isn rootsm₁, m₂,· · · , m_n of A.E. (4) are real and distinct. The general solution of (2) or (3) is

y=c₁e^m1x+c₂e^m2x+· · ·+c_ne^mnx

Example 5.1/p.161 Solve d³y

dx³ + 6d²y

dx² + 11dy

dx+ 6y= 0 Solution. The A.E. is

m³+ 6m² + 11m+ 6 = 0

m²(m+ 1) + 5m(m+ 1) + 6(m+ 1) = 0, As m=−1 satisfy the above relation.

(m+ 1)(m² + 5m+ 6) = 0 (m+ 1)(m²+ 3m+ 2m+ 6) = 0 (m+ 1)(m+ 2)(m+ 3) = 0

⇒ m = −1,−2,−3 Therefore, the general solution is

y=c₁e^−x+c₂e^−2x+c₃e^−3x

Case-II:- If the k roots of A.E. (4) such that each equal to m₁ and the remaining (n−k) roots are all different. Then, the general solution of (2) or (3) is

y= (c1+c2x+c3x²+· · ·+ckx^k−1)e^m1x+ck+1e^mk+1x+· · ·+cne^mnx

In particular, if all n roots are equal, i.e., m₁ =m₂ = · · ·= m_n =m (say), then the general solution is

y = (c₁+c₂x+c₃x²+· · ·+c_nxⁿ⁻¹)e^mx

Example 5.4/p.162 Solve d³y

dx³ −3dy

dx+ 2y= 0 Solution. The A.E. is

m³−3m+ 2 = 0 m²(m−1) +m(m−1)−2(m−1) = 0,

(m−1)(m²+m−2) = 0 (m−1)(m²+ 2m−m−2) = 0 (m−1)(m+ 2)(m−1) = 0

⇒ m = 1,1,−2 Therefore, the general solution is

y= (c₁x+c₂)e^x+c₃e^−2x

Example 5.5/p.162 Solve 16d²y

dx² + 24dy

dx + 9y= 0 Solution. The A.E. is

16m²+ 24m+ 9 = 0 (4m)²+ 2·3·4m+ (3)² = 0 (4m+ 3)² = 0,

⇒ m = −3

4,−3 4 Therefore, the general solution is

y= (c1x+c2)e^−34x

Case-III:- If (α±iβ) are two imaginary roots of an Auxiliary equation of a second order linear differential equation, then the solution is

y=e^αx[c₁cosβx+c₂sinβx]

Ifα+iβ and α−iβ each occurs twice as a root, then the general solution is y=e^αx[(c₁+c₂x) cosβx+ (c₃+c₄x) sinβx]

Example 5.8/p.163 Solve d⁴y

dx⁴ + 8d²y

dx² + 16y= 0 Solution. The A.E. is

m⁴+ 8m²+ 16 = 0 which gives

m = 2i,2i,−2i,−2i or m = 0±2i,0±2i Therefore, the general solution is

y=e^0x[(c₁+c₂x) cos 2x+ (c₃+c₄x) sin 2x]

or,

y= (c₁+c₂x) cos 2x+ (c₃+c₄x) sin 2x

Example 5.9/p.163 Solve d³y

dx³ +y= 0 Solution. The A.E. is

m³+ 1 = 0 (m+ 1)(m²−m+ 1) = 0

⇒ m = −1,1±√ 3i 2

Therefore, the general solution is

y=c₁e^−x+e^12x

"

c₂cos

√3

2 x+c₃sin

√3 2 x

#

Example 5.10/p.163 Solve dy

dx −y 2

d²y dx² +y

2

= 0 Solution. The A.E. is

(m−1)²(m²+ 1)² = 0

⇒ m = 1,1,±i,±i Therefore, the general solution is

y= (c₁x+c₂)e^x+e^0x[(c₃x+c₄) cos 1x+ (c₅x+c₆) sin 1x]

i.e.,

y = (c₁x+c₂)e^x+ (c₃x+c₄) cosx+ (c₅x+c₆) sinx

Home Assignment Solve the following differential equations:

1. (Ex.5.2/ p-161) d²y

dx² −3dy

dx + 2y= 0 with y= 0, x= 0 and ^dy_dx = 0 [Ans : y=0]

2. (Ex.5.7/ p-163) d²y

dx² + 4y= 0 [Ans : y=c1cos2x+c2sin2x]

3. (Q.6/ p-239) (D⁴−2D³ + 2D²−2D+ 1)y= 0 [Ans : y= (c₁ +c₂x)e^x+c₃cosx+c₄sinx]

4. (Q.8/ p-239) (D⁴−4D³ + 8D²−8D+ 4)y= 0 [Ans : y=e^x[(c₁+c₂x) cosx+ (c₃+c₄x) sinx]]

Solution of Non-Homogeneous linear differential equations with con- stant coefficients:- The complete solution of non-homogeneous linear differ- ential equation

a_ndⁿy

dxⁿ +a_n−1dⁿ⁻¹y

dxⁿ⁻¹ +· · ·+a₁dy

dx +a₀y=Q(x) (1) wherea_n6= 0;Q(x)6= 0; and a₁, a₂,· · · , a_n are constants, is

y =y_c+y_p

where y_c is called complementary function and y_p is called particular integral.

1. Complementary Function (C.F.): The general solution of homoge- neous equation corresponding to (1) is called complimentary function of non-homogeneous equation (1). In other words, y_c can be found by taking Q(x) = 0 as discussed in Lecture-01.

2. Particular Integral (P.I.): Particular integral of P(D)y = Q(x) is defined as

y_p = 1

P(D)Q(x).

As P(D) is polynomial in D, its inverse _P(D)¹ is rational expression in D.

Remark. 1. P.I. does not contain any arbitrary constant.

1

2. If P(D) =D, then

y_p = 1

DQ(x) = Z

Q(x)dx 3. If P(D) =D^k, then

y_p = 1

D^kQ(x) = Z Z

· · · Z

| {z }

k times

Q(x)dx·dx· · ·dx

| {z }

k times

4. _P(D)¹ is linear operator, i.e., 1

P(D)[Q₁(x) +Q₂(x)] = 1

P(D)Q₁(x) + 1

P(D)Q₂(x) 1

P(D)[cQ(x)] = c 1

P(D)Q(x)

Now, we consider general case, i.e., P(D) is a polynomial operator of order n of the form:

P(D) = a_nDⁿ+an−1Dⁿ⁻¹+· · ·+a₁D+a₀.

In this case, we give formulae for finding the particular integrals of some elementary functions as follows:

(i) WhenQ(x) =x^k with a₀ 6= 0:- Then, P.I. is yp = 1

P(D)x^k

= 1

(a₀+a₁D+· · ·+a_nDⁿ)x^k

= 1

a₀ 1 + ^a_a¹

0D+· · ·+ ^a_aⁿ

0Dⁿx^k

= 1

a₀

1 + a1

a₀D+· · ·+an

a₀Dⁿ −1

x^k Ifk = 0, then P.I. is

y_p = 1

P(D)1 = 1

a₀, a₀ 6= 0

Example 5.15/p.170 Solve (D³−2D+ 4)y =x⁴+ 3x²−5x+ 2.

Solution. Here, the Auxiliary Equation is m³−2m+ 4 = 0 (m+ 2)(m²−2m+ 4) = 0

⇒ m = −2,1±i Therefore, C.F. is

y_c=c₁e^−2x+e^x(c₂cosx+c₃sinx) Also, P.I. is

y_p = 1

P(D)Q(x)

= 1

D³−2D+ 4(x⁴+ 3x²−5x+ 2)

= 1 4

1

1− ¹₂D+¹₄D³(x⁴+ 3x²−5x+ 2)

= 1 4

1− 1

2

D− 1 2D³

−1

(x⁴+ 3x²−5x+ 2)

= 1 4

1 + 1 2

D−1

2D³

+ 1 4

D− 1

2D³ 2

+1 8

D− 1

2D³ 3

+· · ·

(x⁴+ 3x²−5x+ 2) After simplification, we have

y_p = 1 4

x⁴+ 2x³+ 6x²−5x− 7 2

Therefore, the complete solution is

y=y_c+y_p =c₁e^−2x+e^x(c₂cosx+c₃sinx) + 1 4

x⁴+ 2x³+ 6x²−5x− 7 2

Home Assignment Solve the following differential equations:

1. (Ex 5.12/p.169) d²y

dx² −2dy

dx −3y= 5.

[Ans : y=c₁e^−x+c₂e^3x− ⁵₃] 2. (Ex 5.13/p.169) (D²−4)y=x².

[Ans : y=c₁e^2x+c₂e^−2x− ¹₄ x²+ ¹₂ ] 3. (Ex 5.14/p.170) (D²+ 2D+ 1)y = 2x+x².

[Ans : y= (c₁x+c₂)e^−x+x²−2x+2]

(ii) When Q(x) = x^k and a₀ = 0:- Then P.I. is y_p = Q(x)

P(D) = x^k

(a_nDⁿ+a_n−1Dⁿ⁻¹+· · ·+a₁D)

= x^k

D(a₁+a₂D+· · ·+a_nDⁿ⁻¹); when a₁ 6= 0

= x^k

a₁D(1 + ^a_a²

1D+· · ·+^a_aⁿ

1Dⁿ⁻¹) If,a₁ = 0, then

y_p = x^k

a2D²(1 + ^a_a³

2D+· · ·+^a_aⁿ

2Dⁿ⁻¹) and so on.

Example 5.17/p.172 Solve d³y

dx³ + 3d²y

dx² + 2dy dx =x². Solution. Here, the A.E. is

m³+ 3m²+ 2m = 0

⇒ m(m²+ 3m+ 2) = 0

⇒ m(m+ 2)(m+ 1) = 0

⇒ m = 0,−1,−2

Hence, C.F. is

y_c =c₁+c₂e^−x+c₃e^−2x Now, P.I. is

y_p = 1

P(D)Q(x) = 1

D³+ 3D²+ 2D(x²)

= 1

2D

1 + ³₂D+^D₂²(x²)

= 1

2D

1 + 3

2D+D² 2

−1

(x²)

= 1

2D

"

1− 3

2D+ D² 2

+

3

2D+ D² 2

2

− · · ·

# (x²)

= 1

2D

x²−(3x+ 1) +9 2

= 1

2D

x²−3x+7 2

= 1 2

x³

3 −3x² 2 +7

2x

= 1

12

2x³ −9x²+ 21x Thus, the complete solution is

y=y_c+y_p =c₁+c₂e^−x+c₃e^−2x+ 1 12

2x³−9x²+ 21x

Home Assignment

1. (Ex 5.16/p.171) Solve (D³−D²−6D)y=x²+ 1.

[Ans : y=c1+c2e^3x+c3e^−2x− ¹₆

25

18x+ ^x₃³ − ^x₆² ] (iii) When Q(x) =e^ax with P(a)6= 0:- Then P.I. is

y_p = 1

P(D)e^ax = e^ax

P(a); provided P(a)6= 0.

Example 5.19/p.173 Solve (D³−D²−4D+ 4)y =e^3x.

Solution. The A.E. is

m³−m²−4m+ 4 = 0 m²(m−1)−4(m−1) = 0 (m−1)(m²−4) = 0

⇒ m = 1,2,−2 Thus, C.F. is

yc =c1e^x+c2e^2x+c3e^−2x Also, P.I. is

y_p = 1

P(D)Q(x) = 1

D³−D²−4D+ 4e^3x

= e^3x

3³ −3²−4·3 + 4

= e^3x 10

Therefore, the complete solution is

y=yc+yp =c1e^x+c2e^2x+c3e^−2x+ 1 10e^3x

Home Assignment 1. (Ex 5.18/p.173) Solve (D² −2D+ 5)y=e^−x.

[Ans : y=e^−x(c1cos2x+c2sin2x) + ¹₈e^−x]

2. (Q. 17/p.240) Find the particular Integral of y⁰⁰+ 3y⁰+ 2y= 12e^x. [Ans : yp =2e^x]

(iv) When Q(x) = e^ax with P(a) = 0:- In this case, there exists some nat- ural number n such that (D−a)ⁿ is a factor ofP(D) so that

P(D) = (D−a)ⁿf(D), where f(D)6= 0.

Then, P.I. is

y_p = 1

(D−a)ⁿf(D)e^ax = xⁿe^ax

n!f(a); provided f(a)6= 0.

Example 5.30/p.180 Solve (D²+ 4D+ 4)y=e^2x−e^−2x. Solution. The A.E. is

m² + 4m+ 4 = 0 (m+ 2)² = 0

⇒ m = −2,−2 Thus, C.F. is

y_c= (c₁+c₂x)e^−2x Now, P.I. is

y_p = 1

P(D)Q(x) = 1

D²+ 4D+ 4(e^2x−e^−2x)

= 1

D²+ 4D+ 4e^2x− 1

D² + 4D+ 4(e^−2x)

= e^2x

(2)²+ 4(2) + 4 − 1

(D+ 2)²·1(e^−2x) Here f or 2nd expression a =−2, n= 2, f(D) = 1

= e^2x

16 −x²e^−2x 2!·1

= 1

16e^2x−e^−2xx² 2 Thus the required solution is

y= (c₁+c₂x)e^−2x+ 1

16e^2x− 1 2x²e^−2x

Home Assignment Solve the following differential equations:

1. (Ex 5.28/p.179) (D²+ 6D+ 9)y = 2e^−3x. [Ans : y= (c₁+c₂x)e^−3x+x²e^−3x]

2. (Ex 5.29/p.180) (D²−4D+ 4)y= 8(x²+e^2x+ sin 2x).

[Ans : y= (c₁+c₂x)e^2x+2x²+4x+4x²e^2x+ cos2x+3]

(v) WhenQ(x) = sinax or cosax:- If P(−a²)6= 0, then P.I. are y_p = 1

P(D²)sinax= sinax P(−a²) and

y_p = 1

P(D²)cosax= cosax P(−a²)

IfP(−a²) = 0, then consider special case P(D) =D²+a². In this case, P.I.

are

y_p = 1

D² +a² sinax=−x

2acosax and

y_p = 1

D² +a² cosax= x

2asinax Example 5.21/p.175 Solve (D³+D²−D−1)y= cos 2x.

Solution. The A.E. is

m³+m²−m−1 = 0 m²(m+ 1)−(m+ 1) = 0 (m+ 1)(m²−1) = 0

⇒ m = −1,−1,1 Thus, C.F. is

y_c=c₁e^x+ (c₂+c₃x)e^−x

Now, P.I. is

y_p = 1

(D³+D²−D−1)cos 2x

= 1

(−(2)²D−(2)²−D−1)cos 2x

= 1

(−4D−4−D−1)cos 2x

= 1

(−5D−5)cos 2x

= −1 5

1

(D+ 1) × D−1 D−1cos 2x

= −1 5

D−1

D²−1cos 2x

= −1 5

(D−1) cos 2x

−(2)²−1

= −1 5

(−2 sin 2x−cos 2x)

−5

= − 1

25(2 sin 2x+ cos 2x) Thus, the complete solution is

y=c1e^x+ (c2+c3x)e^−x− 1

25(2 sin 2x+ cos 2x) Example 5.60(iii)/p.224 Solve (D³−D²+ 4D−4)y= sin 3x.

Solution. The A.E. is

m³−m²+ 4m−4 = 0 m²(m−1) + 4(m−1) = 0 (m−1)(m²+ 4) = 0

⇒ m = 1,±2i Thus, C.F. is

y_c=c₁e^x+c₂cos 2x+c₃sin 2x

Now, P.I. is

y_p = 1

(D³−D²+ 4D−4)sin 3x

= 1

(−(3)²D+ (3)²+ 4D−4)sin 3x

= 1

(−9D+ 9 + 4D−4)sin 3x

= 1

(−5D+ 5)sin 3x

= −1 5

1

(D−1)× D+ 1 D+ 1sin 3x

= −1 5

D+ 1

D²−1sin 3x

= −1 5

(D+ 1) sin 3x

−(3)²−1

= −1 5

(3 cos 3x+ sin 3x)

−10

= 1

50(3 cos 3x+ sin 3x) Thus, the complete solution is

y=c₁e^x+c₂cos 2x+c₃sin 2x+ 1

50(3 cos 3x+ sin 3x) Example 5.20/p.175 Solve (D²−3D+ 2)y= 3 sin 2x.

Solution. The A.E. is

m²−3m+ 2 = 0 (m−2)(m−1) = 0

⇒ m = 1,2 Thus, C.F. is

y_c=c₁e^x+c₂e^2x

Also, P.I. is

yp = 1

(D²−3D+ 2)3 sin 2x

= 3 1

(−(2)²−3D+ 2)sin 2x

= 3 1

(−3D−2)sin 2x

= −3 1

(3D+ 2) × 3D−2 3D−2sin 2x

= −33D−2

9D²−4sin 2x

= −3(3D−2) sin 2x 9× −(2)²−4

= −3(3×2 cos 2x−2 sin 2x)

−40

= 3

20(3 cos 2x−sin 2x) Thus, the complete solution is

y =c₁e^x+c₂e^2x+ 3

20(3 cos 2x−sin 2x)

Home Assignment 1. (Ex 5.22/p.176) Solve (D³+ 1)y= cos 2x.

[Ans : y=c₁e^−x+e^x2 c₂cos

√ 3

2 x+c₃sin

√ 3 2 x

+ ₆₅¹ (cos2x−8sin2x)]

2. (Ex 5.24/p.177) Solve (D⁴−1)y= sinx.

[Ans : y=c₁e^x+c₂e^−x+c₃cosx+c₄sinx+ ¹₄xcosx]

3. (Ex 5.60(ii)/p.224) Solve (D²−4D−5)y= 3 cos(4x+ 3).

[Ans : y=c₁e^−x+c₂e^5x− ₆₉₇³ [16sin(4x+3)−21cos(4x+3)]]

Find the particular integral of the following differential equation:

4. (Q.19/p.240) d²y

dx² −y= sinx.

[Ans : y_p =−¹₂sinx]

5. (Q.20/p.240) d²y

dx² −y= cosx.

[Ans : y_p =−¹₂cosx]

6. (Q.21/p.240) d²y

dx² −3dy

dx + 2y= 3 sinx.

[Ans : y_p = ₁₀³ (sinx+3cosx)]

(vi) When Q(x) = e^axV(x):- Then P.I. is

y_p = 1

P(D)e^axV(x) =e^ax 1

P(D+a)V(x)

Example 5.27/p.179 Solve (D²−2D+ 5)y=e^2xsinx.

Solution. The A.E. is

m²−2m+ 5 = 0

⇒ m = 2±√

4−4·5 2

⇒ m = 2±4i 2

⇒ m = 1±2i Thus, C.F. is

y_c=e^x(c₁cos 2x+c₂sin 2x)

P.I. is

y_p = 1

P(D)Q(x) = 1

D²−2D+ 5e^2xsinx

= e^2x 1

(D+ 2)²−2(D+ 2) + 5sinx

= e^2x 1

D²+ 4D+ 4−2D−4 + 5sinx

= e^2x 1

D²+ 2D+ 5sinx

= e^2x sinx

−(1)²+ 2D+ 5

1

P(D²)sinax= sinax P(−a²)

= e^2x sinx 2D+ 4

= e^2x 2

1

D+ 2 × D−2 D−2sinx

= e^2x 2

D−2 D²−4sinx

= e^2x 2

(cosx−2 sinx) {−(1)²−4}

= e^2x 2

(cosx−2 sinx)

−5

= − 1

10e^2x(cosx−2 sinx) Therefore, the general solution is

y=e^x(c₁cos 2x+c₂sin 2x)− 1

10e^2x(cosx−2 sinx)

Home Assignment 1. (Ex 5.25/p.178) Solve (D²−2D+ 1)y=e^xx².

[Ans : y= (c₁ +c₂x)e^x+ ₁₂¹ e^xx⁴]

2. (Ex 5.26/p.178) Solve (D²+ 4D−12)y= (x−1)e^2x. [Ans : y=c₁e^2x+c₂e^−6x+ ₆₄¹ e^2x(4x²−9x)]

3. (Q.24/p.240) Find the Particular integral of d³y

dx³ + 3d²y

dx² + 3dy dx+y= e^−x(2−x²).

[Ans : y_p = ^x3₆₀^e−x(20−x²)]

(vii) WhenQ(x) = xV(x); where V is any function of x:- Here, P.I. is y_p = 1

P(D)xV(x) =x 1

P(D)V(x)− P⁰(D) [P(D)]²V(x) Example 5.31/p.181 Solve d²y

dx² −2dy

dx +y =xsinx.

Solution. The A.E. is

m² −2m+ 1 = 0 (m−1)² = 0

m = 1,1 Thus, C.F. is

y_c= (c₁+c₂x)e^x Now, P.I. is

yp = 1

P(D)Q(x) = 1

D²−2D+ 1(xsinx)

= x 1

D²−2D+ 1sinx− 2D−2

(D²−2D+ 1)² sinx 1

P(D)xV(x) =x 1

P(D)V(x)− P⁰(D) [P(D)]²V(x)

= x sinx

[−(1)²−2D+ 1] − 2(D−1) sinx [−(1)²−2D+ 1]²

= −x 2

1

Dsinx− 2

4D²(cosx−sinx)

= x

2cosx− 1

2D(sinx+ cosx)

= x

2cosx− 1

2(−cosx+ sinx) Therefore, the complete solution is

y= (c₁+c₂x)e^x+x

2cosx−1

2(sinx−cosx)

Home Assignment

Solve the following differential equations:

1. (Ex 5.33/p.182) (D²−2D+ 1)y =xe^xsinx.

[Ans : y= (c₁ +c₂x)e^x−e^x(xsinx+2cosx)]

2. (Q.36/p.240) (D²+ 1)y=xe^2x.

[Ans : y=c₁cosx+c₂sinx+₂₅¹ e^x(5x−4)]

Mixed Problems:

Example 5.61(ii)/p.225 Solve (D²+ 2)y=x²e^3x+e^xcos 2x.

Solution. The A.E. is

m²+ 2 = 0

⇒ m = ±√

2i

Thus, C.F. is

y_c =c₁cos√

2x+c₂sin√ 2x

Now, P.I. is

y_p = 1

P(D)Q(x) = 1

(D²+ 2)(x²e^3x+e^xcos 2x)

= 1

(D²+ 2)x²e^3x+ 1

(D² + 2)e^xcos 2x

= e^3x 1

(D+ 3)²+ 2x² +e^x 1

(D+ 1)²+ 2cos 2x 1

P(D)e^axV(x) = e^ax 1

P(D+a)V(x)

= e^3x 1

D²+ 6D+ 9 + 2x²+e^x 1

D²+ 2D+ 1 + 2cos 2x

= e^3x 1

D²+ 6D+ 11x²+e^x 1

D²+ 2D+ 3cos 2x

= e^3x 1

11 1 + ₁₁⁶ D+₁₁¹ D²x²+e^x 1

−(2)²+ 2D+ 3cos 2x 1

P(D²)cosax= cosax P(−a²)

= 1

11e^3x

1 + 6

11D+ 1 11D²

−1

x²+e^x 1

2D−1× 2D+ 1 2D+ 1cos 2x

= 1

11e^3x

"

1− 6

11D+ 1 11D²

+

6

11D+ 1 11D²

2

+· · ·

#

x²+e^x 2D+ 1

4D²−1cos 2x

= 1

11e^3x

x²− 12

11x

+ 25

121

+e^x(2D+ 1) cos 2x 4(−2²)−1

= 1

11e^3x

x²− 12

11x+ 25 121

+e^x(−4 sin 2x+ cos 2x)

−17

= 1

11e^3x

x²− 12

11x+ 25 121

− 1

17e^x(cos 2x−4 sin 2x) Therefore the complete solution is

y =c₁cos√

2x+c₂sin√

2x+ 1 11e^3x

x²− 12

11x+ 25 121

− 1

17e^x(cos 2x−4 sin 2x)

Example 5.62(ii)/p.227 Solve (D² −4D+ 4)y= 8x²e^2xsin 2x.

Solution. The A.E. is

m²−4m+ 4 = 0 (m−2)² = 0

⇒ m = 2,2 Hence, C.F. is

y_c= (c₁+c₂x)e^2x P.I. is

yp = 1

P(D)Q(x) = 1

D²−4D+ 4(8x²e^2xsin 2x)

= 1

D²−4D+ 4(8e^2x·x²sin 2x)

= 8e^2x 1

(D+ 2)²−4(D+ 2) + 4(x²sin 2x) 1

P(D)e^axV(x) = e^ax 1

P(D+a)V(x)

= 8e^2x 1

D²+ 4 + 4D−4D−8 + 4(x²sin 2x)

= 8e^2x 1

D²(x²sin 2x)

= 8e^2x(−2x²sin 2x+ 3 sin 2x−4 cos 2x) Therefore, the required solution is

y= (c₁+c₂x)e^2x+ 8e^2x(−2x²sin 2x+ 3 sin 2x−4 cos 2x)

Home Assignment

1. (Q.29/p.240) Find the particular integral ofy⁰⁰+ 3y⁰+ 2y= 8 + 6e^x+ 2 sinx.

[Ans : y_p =4+e^x+¹₂(sinx−3cosx)]

Solve the following differential equations:

2. (Q.37/p.240) (D²+ 1)y=e^−x+ cosx+x³+e^xcosx.

[Ans : y=c₁cosx+c₂sinx+¹₂e^−x+¹₂xsinx+x³−6x+ ¹₅e^x(2sinx+ cosx)]

3. (Q.42/p.240) (D²−4D+ 4)y=x²+e^x+ sin 2x.

[Ans : y= (c₁+c₂x)e^2x+¹₄ x²+2x+³₂

+e^x+¹₈cos2x]

4. (Q.43/p.240) (D²+ 1)y = cosx+xe^2x+e^xsinx.

[Ans : y=c₁cosx+c₂sinx+¹₂xsinx+ ₂₅¹ e^2x(5x−4)−¹₅e^x(2cosx−sinx)]

Variation of Parameters

In this lecture, we discuss yet another technique called ‘Variation of Pa- rameters’ to find the particular solution of a second order linear differential equation with constant of the form:

a2

d²y dx² +a1

dy

dx +a0y=Q(x); a2 6= 0, Q(x)6= 0. (2) The complementary function (yc) contains linearly independent solutions y1(x) and y2(x) so that

y_c=c₁y₁(x) +c₂y₂(x).

By Variation of Parameters, the particular solution will be of the form y_p =u(x)y₁(x) +v(x)y₂(x);

where u(x) andv(x) are to be determined by u(x) = −1

a₂

Z y₂(x)Q(x) W(y₁, y₂)dx;

v(x) = 1 a₂

Z y₁(x)Q(x) W(y₁, y₂)dx;

so that

W(y₁, y₂) =

y₁(x) y₂(x) y⁰₁(x) y⁰₂(x)

Example 5.40/p.189 Solve (D²−3D+ 2)y= sine^−x.

Solution. The Auxiliary equation is

m² −3m+ 2 = 0 (m−2)(m−1) = 0

m = 1,2 Therefore, Complimentary function is

y_c =c₁e^x+c₂e^2x

Hence, here y₁(x) =e^x and y₂(x) =e^2x. Therefore, the particular integral is y_p =u(x)y₁(x) +v(x)y₂(x) (3) Now, we have

W(y1, y2) =

y₁(x) y₂(x) y⁰₁(x) y₂⁰(x)

=

e^x e^2x e^x 2e^2x

= 2e^3x−e^3x

= e^3x Thus,

u(x) = −1 a₂

Z y₂(x)Q(x) W(y₁, y₂)dx;

= −1 1

Z e^2xsine^−x

e^3x dx; here a₂ = 1

= −

Z

e^−xsine^−xdx P ut e^−x=t⇒ −e^−xdx=dt⇒ Z

sintdt⇒ −cost

= −cose^−x

and

v(x) = 1 a₂

Z y₁(x)Q(x) W(y₁, y₂)dx;

= 1 1

Z e^xsine^−x

e^3x dx; here a₂ = 1

= Z

e^−2xsine^−xdx P ut e^−x =t⇒ −e^−xdx=dt

= −

Z

tsintdt

= −

t(−cost)− Z

1·(−cost)dt

= −[−tcost+ sint]

= e^−xcose^−x−sine^−x Putting these values in (3), we get

y_p = u(x)y₁(x) +v(x)y₂(x)

= (−cose^−x)e^x+ (e^−xcose^−x−sine^−x)e^2x

= −e^xcose^−x+e^xcose^−x−e^2xsine^−x

= −e^2xsine^−x Thus, the complete solution is

y=y_c+y_p =c₁e^x+c₂e^2x−e^2xsine^−x Example 5.41/p.190 Solve (D²+ 4D+ 4)y = 3xe^−2x. Solution. The Auxiliary equation is

m²+ 4m+ 4 = 0 (m+ 2)² = 0

m = −2,−2 Therefore, Complimentary function is

y_c = (c₁+c₂x)e^−2x ⇒c₁e^−2x+c₂xe^−2x

Hence, herey₁(x) =e^−2x andy₂(x) =xe^−2x. Therefore, the particular integral is

y_p =u(x)y₁(x) +v(x)y₂(x) (4) Now, we have

W(y1, y2) =

y₁(x) y₂(x) y⁰₁(x) y⁰₂(x)

=

e^−2x xe^−2x

−2e^−2x e^−2x−2xe^−2x

= e^−2x(e^−2x−2xe^−2x)−xe^−2x(−2e^−2x)

= e^−4x−2xe^−4x+ 2xe^−4x

= e^−4x Thus,

u(x) = −1 a₂

Z y2(x)Q(x) W(y₁, y₂)dx;

= −1 1

Z xe^−2x·3xe^−2x

e^−4x dx; here a₂ = 1

= −

Z

3x²dx

= −3 x³

3

= −x³ and

v(x) = 1 a₂

Z y1(x)Q(x) W(y₁, y₂)dx;

= 1 1

Z e^−2x·3xe^−2x

e^−4x dx; here a₂ = 1

= Z

3xdx

= 3x² 2

Putting these values in (4), we get

y_p = u(x)y₁(x) +v(x)y₂(x)

= (−x³)e^−2x+ 3x²

2

xe^−2x

= −x³e^−2x+ 3x³ 2 e^−2x

= 1 2x³e^−2x Thus, the complete solution is

y =y_c+y_p =c₁e^−2x+c₂xe^−2x+ 1 2x³e^−2x

Home Assignment

Using the method of variation of parameter, solve the following equations.

1. (Q.53/ p-241) y⁰⁰+y = 4xsinx.

[Ans : y=c₁cosx+c₂sinx−x²cosx+xsinx]

2. (Q.55/ p-241) y⁰⁰−2y⁰+y=e^xlogx.

[Ans : y= (c₁+c₂x)e^x+x²e^{x 1}₂logx− ³₄ ]

Reduction of Order Method for Homogeneous Equation

Consider the general linear differential equation an(x)dⁿy

dxⁿ +an−1(x)dⁿ⁻¹y

dxⁿ⁻¹ +· · ·+a1(x)dy

dx +a0(x)y=Q(x) (5) And its related homogeneous equation

a_n(x)dⁿy

dxⁿ +a_n−1(x)dⁿ⁻¹y

dxⁿ⁻¹ +· · ·+a₁(x)dy

dx +a₀(x)y= 0 (6)

wherea_n(x), an−1(x),· · ·, a₁(x), a₀(x) andQ(x) are each continuous functions of x and a_n(x)6= 0.

When the coefficients are constant, the methods for finding the solutions of equation (5) and (6) are known. Here we shall give a method for solving equation (5) and (6) when the coefficients a_n(x) are not constant.

Equation (5) don’t always have solution and finding solution can be much more difficult than constant coefficients. However, if we knew one linearly independent solution (non-zero) of homogeneous part, then we can find the general solution (other independent solution of (6) & particular solution).

This method is known as reduction of order method.

Reduction of Order method for Homogeneous Equation:Consider the homogeneous equation of second order with variable coefficients

a2(x)d²y

dx² +a1(x)dy

dx +a0(x)y = 0. (7) Suppose that a linearly independent solution (y₁ 6= 0) of equation (7) is given.

Then second linearly independent solution (y₂) can be found by reduction of order method as:

y₂(x) =y₁(x) Z

u(x)dx (8)

where

u= exp

−R a₁(x) a₂(x)dx

y₁² (9)

Then, the general solution of (7) is

y=c₁y₁+c₂y₂

Example (5.42/P-192)Solvex²y⁰⁰−xy⁰+y= 0. Giveny₁ = xas a solution.

Solution. Given homogeneous differential equation is

x²y⁰⁰−xy⁰+y= 0, with given solution y₁ =x.

Then, another linear independent solution is given by

y₂(x) = y₁(x) Z

u(x)dx where

u= exp

−R a₁(x) a₂(x)dx

y₁²

Here, a₂(x) = x², a₁(x) =−x and y₁ =x, then we have

u = exp

−R a₁(x) a₂(x)dx

y₁²

= exp

−R −x x² dx

(x)²

= e^{R 1xdx}

x² = e^logx x² = x

x² u = 1

x Thus

y₂ = y₁ Z

u(x)dx

= x Z 1

xdx

= xlogx Therefore the required solution is

y =c1x+c2xlogx.

Home Assignment

Use the reduction of order method to find the solution of the following equation; one solution of the homogeneous equation is given.

1. (Q.58/ p-241) y⁰⁰−_x²y⁰ +_x²2y= 0; y₁ =x.

[Ans : y=c₁x+c₂x²]

2. (Q.59/ p-241) (2x²+ 1)y⁰⁰−4xy⁰ + 4y= 0; y1 =x.

[Ans : y=c₁x+c₂(2x²−1)]

Reduction of Order Method for Non-Homogeneous Equations

Reduction of Order method for Non-Homogeneous Equation: We consider a second order non-homogeneous differential equation

a₂(x)d²y

dx² +a₁(x)dy

dx +a₀(x)y =Q(x) (10) and its related homogeneous equation

a₂(x)d²y

dx² +a₁(x)dy

dx +a₀(x)y= 0 (11) Suppose that a linearly independent solution (y1 6= 0) of equation (11) is known. Then, the second independent solution of equation (10), as well as a particular integral of equation (10) can be obtain by reduction of order method.

Working Rule:

1. Step-1: Let ¯y(x) be another solution of equation (10) which includes second linearly independent solution as well as the particular solution, then

¯

y(x) = y1(x) Z

u(x)dx, (12)

whereu(x) is a function of x to be determined.

2. Step-2: Differentiate (12) twice, we get

¯

y⁰(x) = y₁u+y₁⁰ Z

u(x)dx (13)

¯

y⁰⁰(x) = y₁u⁰+ 2y₁⁰u+y₁⁰⁰ Z

u(x)dx (14)

3. Step-3: Substitute the values of ¯y,y¯⁰,y¯⁰⁰ in equation (10) and simplify- ing, we obtain the value of u.

4. Step-4: Substituting this value of u in equation (12), we get ¯y, the required remaining part of the solution.

5. Step-5: Then the general solution of (10) is given by y=c₁y₁+ ¯y

Example (5.43/P-192)Given that y=x as a solution of x²d²y

dx² +xdy

dx −y = 0, x6= 0.

Find the general solution of x²d²y

dx² +xdy

dx −y=x.

Solution. Here, given solution of homogeneous differential equation isy₁(x) = x. Thus

¯

y(x) = y₁(x) Z

u(x)dx, where u(x) is to be determined.

= x Z

u(x)dx (15)

Differentiating it twice, we get

¯

y⁰ = xu+ Z

u(x)dx

¯

y⁰⁰ = [xu⁰ +u] +u=xu⁰ + 2u

Since, y¯ is another solution; Substituting the values of y,¯ y¯⁰,y¯⁰⁰ in the given

non-homogeneous differential equation and simplifying, we obtain

x²y⁰⁰+xy⁰−y = x x²[xu⁰+ 2u] +x

xu+

Z udx

−x Z

udx = x x³u⁰+ 2x²u+x²u+x

Z

udx−x Z

udx = x x³u⁰+ 3x²u = x

u⁰+ 3

xu = 1 x²

which is linear equation in u, so

I.F. = e^{RP dx}=e^{R x3dx}

= e^{3 logx}

= x³,

thus, the solution is

u·(I.F.) = Z

(Q·I.F.)dx+c u·x³ =

Z 1 x² ·x³

dx+c u·x³ =

Z

xdx+c u·x³ = x²

2 +c u = 1

2x + c x³

Put this value of u in equation (15) to obtain

¯

y(x) = x Z

u(x)dx

= x Z

1 2x + c

x³

dx

= x 1

2logx− c 2x⁻²

= 1

2xlogx− c 2

1 x

= 1

2xlogx+c2x⁻¹, where c2 = −c 2 , therefore, the required solution is

y =c1x+c2x⁻¹+ 1 2xlogx

Home Assignment

Use the reduction of order method to find the solution of the following equation; one solution of the homogeneous equation is given.

1. (Q.60/ p-241) y⁰⁰− ²_xy⁰+_x²2y =xlogx; y₁ =x.

[Ans : y=c1x+c2x²+¹₂x³logx−³₄x³]

2. (Q.61/ p-241) x²y⁰⁰+xy⁰−y=x²e^−x; y1 =x.

[Ans : y=c1x+c2x⁻¹+e^−x(1+x⁻¹)]

Cauchy-Euler Equation

An equation of the form a_nxⁿdⁿy

dxⁿ +a_n−1xⁿ⁻¹dⁿ⁻¹y

dxⁿ⁻¹ +· · ·+a₁xdy

dx +a₀y=Q(x)

wherea₀, a₁,· · · , an−1, a_n are constant, is called aCauchy-Euler Equation of order n.

For the sake of convenience, we consider the Cauchy-Euler Equation of order 2 as

a₂x²d²y

dx² +a₁xdy

dx +a₀y=Q(x)

a₂x²y⁰⁰+a₁xy⁰ +a₀y=Q(x) (16) To solve (16), we have the substitution x=e^t⇒logx=t⇒ _x^1dx_dt = 1.

⇒ dy

dx = dy dt · dt

dx

= 1 x

dy dt i.e.,

xdy dx = dy

dt Also,

d²y

dx² = d dx

dy dx

= d dx

1 x

dy dt

= 1 x

d dx

dy dt

+ d

dx 1

x dy

dt

= 1 x

d dt

dy dt

dt dx − 1

x² dy dt

= 1

x² d²y

dt² − 1 x²

dy dt

= 1

x² d²y

dt² − dy dt

i.e.,

x²d²y

dx² = d²y dt² −dy

dt

therefore (16) reduces to

a₂x²y⁰⁰+a₁xy⁰ +a₀y=Q(x) a₂

d²y dt² − dy

dt

+a₁ dy

dt

+a₀y=Q(e^t) a₂d²y

dt² + (a₁−a₂)dy

dt +a₀y=Q(e^t) A₂d²y

dt² +A₁dy

dt +A₀y=R(t) Which is linear differential equation with constant coefficients.

Remark. Observe that

xy⁰ −→ D1y

x²y⁰⁰ −→ D₁(D₁−1)y

x³y⁰⁰⁰ −→ D1(D1−1)(D1−2)y and so on. Where D₁ = _dt^d.

Example (5.44/p. 194)Solve x^2d_dx^2y2 +x^dy_dx −4y =x². Solution. Given Differential equation is

x²y⁰⁰+xy⁰−4y=x² (17) Put x=e^t⇒logx=t with

xy⁰ = D₁y

x²y⁰⁰ = D₁(D₁−1)y⇒(D₁²−D₁)y where D₁ = _dt^d, we have from (17),

x²y⁰⁰+xy⁰ −4y = x² (D²₁−D₁)y+D₁y−4y = e^2t

D₁²y−D₁y+D₁y−4y = e^2t (D₁²−4)y = e^2t,

which is linear differential equation with constant coefficients, so the Auxiliary Equation is

m² −4 = 0 m² = 4

m = ±2 Therefore

C.F. =y_c=c₁e^2t+c₂e^−2t And

P.I.=yp = R(t)

f(D₁) = e^2t D₁²−4

= e^2t 1

(D1+ 2)²−4(1) ∵ e^ax

f(D) =e^ax 1

f(D+a)(1)

= e^2t 1

D₁²+ 4D₁+ 4−4(1)

= e^2t 1

D₁²+ 4D₁(1)

= e^2t 1 4D₁

1 + D₁ 4

−1

(1)

= e^2t 1 4D₁

1− D₁ 4 +· · ·

(1) ∵(1 +x)ⁿ= 1 +nx+n(n+ 1)

2! x²+· · ·

= e^2t 1 4D₁(1)

= e^2t 4 t Thus solution

y = y_c+y_p

= c₁e^2t+c₂e^−2t+te^2t 4

= c₁x²+c₂x⁻²+ logx·x² 4

The required solution is

y=c1x²+c2x⁻²+ x²·logx 4

Q.62/ p-241x²d²y

dx² −xdy

dx +y= 2 logx

Solution. Put x=e^t⇒logx=t in the given differential equation with xy⁰ = D1y

xy⁰⁰ = D₁(D₁−1)y We have

x²d²y

dx² −xdy

dx +y = 2 logx [D₁(D₁−1)−D₁+ 1]y = 2t

D²₁−D1−D1+ 1

y = 2t (D²₁ −2D₁+ 1)y = 2t

which is a linear differential equation with constant coefficients, so the auxiliary equation is

m² −2m+ 1 = 0 (m−1)² = 0

m = 1,1 Therefore, the complementary solution is

C.F. =y_c = (c₁+tc₂)e^t

= (c₁+c₂logx)x

And, the particular solution is

P.I.=y_p = = R(t)

f(D1) = 2t (D1−1)²

= 2(1−D₁)⁻²(t)

= 2 1 + 2D1+ 3D²₁+· · · (t)

= 2(t+ 2(1) + 3(0) +· · ·)

= 2(t+ 2)

= 2(logx+ 2)

= 2 logx+ 4 The required general solution is

y = y_c+y_p

y = (c1+c2logx)x+ 2 logx+ 4 i.e.,

y= (c₁+c₂logx)x+ 2 logx+ 4

Home Assignment Solve the following differential equation:

1. (Ex. 5.45/p. 194) x²D²y−xDy−3y=x²logx.

[Ans : y=c1x⁻¹+c2x³−¹₃x² logx+²₃ ]

2. (Ex 5.46/ p-195) x³D³y+ 3x²D²y+xDy+y=x+ logx.

[Ans : y=c₁x⁻¹+√ xh

c₂cos^√

3

2 logx

+c₃sin^√

3

2 logxi

+¹₂x+ logx]

3. (Q.63/ p-241) x⁴d³y

dx³ + 2x³d²y

dx² −x²dy

dx +xy= 1 [Ans : y=c₁x⁻¹+c₂x+c₃xlogx+ ¹₄x⁻¹logx]

4. (Q.64/ p-241) x²d²y

dx² −xdy

dx + 2y=xlogx

[Ans : y=x[c₁cos(logx) +c₂sin(logx)] +xlogx]